invalid array assignment in c

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Why does C not support direct array assignment?

In C you cannot assign arrays directly.

At first I thought this might because the C facilities were supposed to be implementable with a single or a few instructions and more complicated functionality was offloaded to standard library functions. After all using memcpy() is not that hard. However, one can directly assign structures/unions which can be arbitrarily sized, and can even contain arrays.

What is/was the reasoning for allowing structures/unions to be directly assigned but not arrays, if the former can contain arrays anyway? We can only assign arrays directly if they are wrapped in a structure/union, so my aforementioned hypothesis as to why this is the case is out.

• Because language features are unimplemented by default . –  Doc Brown Commented May 4, 2023 at 5:59

2 Answers 2

It is arguably a shortcoming, due to missing features around arrays, but one the original designers choose not to resolve.  Add in the (over) emphasis on pointers and pointer arithmetic and keeping track of these things yourself as would be done in assembly.

Arrays in C don't really have sizes.  Yes, you can int A[100]; ... sizeof(A) ... for a declared array, but once A is used in an expression, its type is immediately converted to pointer to element (here int* ) and the original array's size is lost.  This is particularly evident with parameters where an array is passed.

There's not even a convenient built-in for length (e.g. say, lengthof ) that would work for arrays of any element size — instead we have to write something like (sizeof(A)/sizeof(*A)) .

A lot of code in C relies on use of pointers where length is kept by the program using other variables or knowledge rather than having the language keep track.

Often, declaring an array in a struct/union as a last member is secret code for variable length.  See for example:

I can't say for sure but apparently resolving the short comings only for declared arrays but not for pointers seemed not all that useful.  So, some larger built in length mechanism would have been required for general use with pointers.

Still, something could have been done, say by having some syntax that allows the user to specify a number of elements — like A = B[0:n] or A[0:n] = B or some other variant — but they simply stopped short of that, leaving it for memcpy and memmove .  Let's also note that when you choose between memcpy and memmove , with the former you're saying that the source and destination do not overlap, whereas with memmove you're saying they might overlap so backwards copying of the elements may be better than forwards (and it will check at runtime).

So, the various problems:

• pointer oriented rather than array oriented
• no implicit size information on pointers
• aliasing issues
• fixed sized arrays are only useful in narrow contexts

I think punting to the standard library was pretty reasonable here, plus as you note for fixed sized arrays, we can wrap them in a struct/union..

C originally came from a predecessor language called B, made by Ken Thompson . In B, there were no types. Everything was a "word" (basically, an int ).

In B, arrays were just pointers to the first element. If you declared an array:

This would allocate 10 words on the stack (to be freed automatically when the function returned, thus auto ), and arr would be a pointer to the first one.

When the type system was added, Dennis Ritchie didn't want any existing code to break. This is why, for example, on early versions of C if you omitted the type it would default to int . It's also where pointer decay (array arguments to functions being just pointers) came from.

For this reason, arrays ended up being a second-class citizen in C for a long time. Because of the stability of the language, even the more modern C standards (like C23 which is supposed to come out this year) have to try to fix it without breaking anything, and this particular issue (copying arrays) is not really a priority, because you rarely actually want to copy an array (especially a large one).

Source: The Development of the C language

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Arrays in C – Declare, initialize and access

Array is a data structure that hold finite sequential collection of homogeneous data .

To make it simple let’s break the words.

• Array is a collection – Array is a container that can hold a collection of data.
• Array is finite – The collection of data in array is always finite, which is determined prior to its use.
• Array is sequential – Array stores collection of data sequentially in memory.
• Array contains homogeneous data – The collection of data in array must share a same data type .

We can divide arrays in two categories.

• One-dimensional array (Or single-dimensional array)
• Multi-dimensional array

Why we need arrays?

Let us understand the significance of arrays through an example.

Suppose, I asked you to write a program to input 1000 students marks from user. Finally print average of their marks .

To solve the problem you will declare 1000 integer variable to input marks. Call I/O functions to input marks in 1000 variables and finally find the average.

Think for a while how tedious will be to code if solved using above approach. Declare 1000 variables, take input in all variables, then find average and finally print its average. The above increases length of code, complexity and degrades performance. If you don’t believe, try to code solution using above approach.

To solve above problem efficiently we use arrays. Arrays are good at handling collection of data (collection of 1000 student marks). Mind that in programming, we will always use some data structure (array in our case) to handle a collection of data efficiently.

How to use arrays?

Array in memory is stored as a continuous sequence of bytes. Like variables we give name to an array. However unlike variables, arrays are multi-valued they contain multiple values. Hence you cannot access specific array element directly.

For example, you can write sum = 432; to access sum . But, you cannot access specific array element directly by using array variable name. You cannot write marks to access 4 th student marks.

In array, we use an integer value called index to refer at any element of array. Array index starts from 0 and goes till N - 1 (where N is size of the array). In above case array index ranges from 0 to 4 .

To access individual array element we use array variable name with index enclosed within square brackets [ and ] . To access first element of marks array, we use marks[0] . Similarly to access third element we use marks[2] .

How to declare an array?

Syntax to declare an array.

• data_type is a valid C data type that must be common to all array elements.
• array_name is name given to array and must be a valid C identifier .
• SIZE is a constant value that defines array maximum capacity.

Example to declare an array

How to initialize an array.

There are two ways to initialize an array.

• Static array initialization – Initializes all elements of array during its declaration.
• Dynamic array initialization – The declared array is initialized some time later during execution of program.

Static initialization of array

We define value of all array elements within a pair of curly braces { and } during its declaration. Values are separated using comma , and must be of same type.

Example of static array initialization

Note: Size of array is optional when declaring and initializing array at once. The C compiler automatically determines array size using number of array elements. Hence, you can write above array initialization as.

Dynamic initialization of array

You can assign values to an array element dynamically during execution of program. First declare array with a fixed size. Then use the following syntax to assign values to an element dynamically.

Example to initialize an array dynamically

Instead of hard-coding marks values, you can ask user to input values to array using scanf() function.

The array index is an integer value, so instead of hard-coding you can wrap array input code inside a loop .

The above code will run 5 times from 0 to 4 . In each iteration it ask user to input an integer and stores it in successive elements of marks array.

Example program to implement one-dimensional array

Let us write a C program to declare an array capable of storing 10 student marks. Input marks of all 10 students and find their average.

Output –

Array best practices

• Arrays are fixed size, hence always be cautious while accessing arrays. Accessing an element that does not exists is undefined. You may get a valid value or the program may crash.For example, consider the below program. int array[5]; // Declare an integer array of size 5 /* Accessing sixth element of array which is undefined. */ array[5] = 50; /* Accessing -1th element of array is undefined */ array[-1] = 10;
• Always keep in mind that all array element store a value of similar type.

Recommended array example programs

• Program to read and print array element.
• Program to find maximum and minimum element in array.
• Program to insert a new element in array.
• Program to search an element in array.
• Program to sort array elements.

Practice more array programming exercises to learn more.

• An array is a collection of data items, all of the same type, accessed using a common name.
• A one-dimensional array is like a list;  A two dimensional array is like a table;  The C language places no limits on the number of dimensions in an array, though specific implementations may.
• Some texts refer to one-dimensional arrays as vectors , two-dimensional arrays as matrices , and use the general term arrays when the number of dimensions is unspecified or unimportant.

Declaring Arrays

• Array variables are declared identically to variables of their data type, except that the variable name is followed by one pair of square [ ] brackets for each dimension of the array.
• Uninitialized arrays must have the dimensions of their rows, columns, etc. listed within the square brackets.
• In C99, dimensions must still be positive integers, but variables can be used, so long as the variable has a positive value at the time the array is declared. ( Space is allocated only once, at the time the array is declared. The array does NOT change sizes later if the variable used to declare it changes. )
• C99 Only Example:

Initializing Arrays

• Arrays may be initialized when they are declared, just as any other variables.
• Place the initialization data in curly {} braces following the equals sign.  Note the use of commas in the examples below.
• An array may be partially initialized, by providing fewer data items than the size of the array.  The remaining array elements will be automatically initialized to zero.
• If an array is to be completely initialized, the dimension of the array is not required.  The compiler will automatically size the array to fit the initialized data.  ( Variation:  Multidimensional arrays - see below. )

Designated Initializers: In C99 there is an alternate mechanism, that allows you to initialize specific elements, not necessarily at the beginning. This method can be mixed in with traditional iniitalization For example: int numbers[ 100 ] = { 1, 2, 3, [10] = 10, 11, 12, [60] = 50, [42] = 420 };

• In this example,the first three elements are initialized to 1, 2, and 3 respectively.
• Then element 10 ( the 11th element ) is initialized to 10
• The next two elements ( 12th and 13th ) are initialized to 11 and 12 respectively.
• ( Note that the designated initializers do not need to appear in order. )
• As with traditional methods, all uninitialized values are set to zero.
• If the size of the array is not given, then the largest initialized position determines the size of the array.

Using Arrays

• Elements of an array are accessed by specifying the index ( offset ) of the desired element within square [ ] brackets after the array name.
• Array subscripts must be of integer type.  ( int, long int, char, etc. )
• VERY IMPORTANT: Array indices start at zero in C, and go to one less than the size of the array.  For example, a five element array will have indices zero through four.  This is because the index in C is actually an offset from the beginning of the array.  ( The first element is at the beginning of the array, and hence has zero offset. )
• Landmine:   The most common mistake when working with arrays in C is forgetting that indices start at zero and stop one less than the array size.
• Arrays are commonly used in conjunction with loops, in order to perform the same calculations on all ( or some part ) of  the data items in the array.

Sample Programs Using 1-D Arrays

• The first sample program uses loops and arrays to calculate the first twenty Fibonacci numbers.  Fibonacci numbers are used to determine the sample points used in certain optimization methods.
• Exercise: What is the output of the following program:

Multidimensional Arrays

• Multi-dimensional arrays are declared by providing more than one set of square [ ] brackets after the variable name in the declaration statement.
• One dimensional arrays do not require the dimension to be given if the array is to be completely initialized.  By analogy, multi-dimensional arrays do not require the first dimension to be given if the array is to be completely initialized.  All dimensions after the first must be given in any case.
• For two dimensional arrays, the first dimension is commonly considered to be the number of rows, and the second dimension the number of  columns.  We will use this convention when discussing two dimensional arrays.
• Two dimensional arrays are considered by C/C++ to be an array of ( single dimensional arrays ).  For example, "int numbers[ 5 ][ 6 ]"  would refer to a single dimensional array of 5 elements, wherein each element is a single dimensional array of 6 integers.  By extension, "int numbers[ 12 ][ 5 ][ 6 ]" would refer to an array of twelve elements, each of which is a two dimensional array, and so on.
• Another way of looking at this is that C stores two dimensional arrays by rows, with all elements of a row being stored together as a single unit.  Knowing this can sometimes lead to more efficient programs. 
• Multidimensional arrays may be completely initialized by listing all data elements within a single pair of curly {} braces, as with single dimensional arrays.
• It is better programming practice to enclose each row within a separate subset of curly {} braces, to make the program more readable.  This is required if any row other than the last is to be partially initialized.  When subsets of braces are used, the last item within braces is not followed by a comma, but the subsets are themselves separated by commas.
• Multidimensional arrays may be partially initialized by not providing complete initialization data.  Individual rows of a multidimensional array may be partially initialized, provided that subset braces are used.
• Individual data items in a multidimensional array are accessed by fully qualifying an array element.  Alternatively, a smaller dimensional array may be accessed by partially qualifying the array name.  For example, if  "data" has been declared as a three dimensional array of floats, then data[ 1 ][ 2 ][ 5 ] would refer to a float, data[ 1 ][ 2 ] would refer to a one-dimensional array of floats, and data[ 1 ] would refer to a two-dimensional array of floats.  The reasons for this and the incentive to do this relate to memory-management issues that are beyond the scope of these notes.

Sample Program Using 2-D Arrays

• See Attached

Material Below This Line Not Covered in CS 107 Until Later

Passing arrays to functions ( in c++ ).

• Pass-by-Value, in which the function receives a copy and all changes are local, and
• Pass-by-Reference, in which the names of the variables in the called and calling functions become aliases, and changes made in the called function DO affect the variables in the calling function.
• So if nums was declared as a one-dimensional array of ints, then passing nums[ i ] to a function would behave the exact way as passing any other int - Either pass-by-value or pass-by-reference, depending on how the function is written.
• ( It is actually passed by pointer/address, which is not covered here, but functionally it is pass-by-reference. )
• The net result, is that when an entire array is passed to a function, and the function changes variables stored in that array, it does affect the data in the calling function's array.
• Because arrays are passed by reference, there is generally no need for a function to "return" an array. - It merely needs to fill in an array provided by the calling function. ( It is possible for functions to return arrays but it requires the use of pointers and addresses, and frequently dynamic memory allocation, all of which is beyond the scope of this course. )
• In order to prevent the function from changing the array values, the array parameter can be modified with the keyword const.
• When an entire array is passed to a function, the size of the array is usually passed as an additional argument.
• For a one dimensional array, the function's formal parameter list does not need to specify the dimension of the array.  If such a dimension is provided, the compiler will ignore it.
• For a multi-dimensional array, all dimensions except the first must be provided in a functions formal parameter list.  The first dimension is optional, and will be ignored by the compiler.
• A partially qualified multi-dimensional array may be passed to a function, and will be treated as an array of lower dimension.  For example, a single row of a two dimensional array may be passed to a function which is expecting a one dimensional array.  ( Question:  Why is it possible to pass a row this way, but not a column? )
• Examples:  Note the use of arrays and functions in the following sample program.  Note that in the calculation of max4, we have passed a two dimensional array containing two rows of three elements as if it were a single dimensional array of six elements.  This is cheating, but it happens to work because of the way that the rows of a multidimensional array are stored.

Write the following functions:

• Calculates the average of the values in x.
• nValues is how many elements to calculate
• errorCode should be set to 0 if no errors or -1 if errors are encountered.
• Calculates the dot product of x and y.
• Calculates the sin of each element of x, and stores the results in sinX.
• The return value is the actual number of values calculated. It should be equal to nValues if all goes well, or some other value ( e.g. zero ) if there is a problem.

Character Strings as Arrays of Characters

• The "string" class type is a new addition to C++, which did not exist in traditional C.
• The traditional method for handling character strings is to use an array of characters.
• A null byte ( character constant zero, '\0' ) is used as a terminator signal to mark the end of the string.
• Ordinary quoted strings ( "Please enter a number > " ) are stored as null-terminated arrays of characters.
• The cstring library contains a number of functions for dealing with traditional arrays of characters.
• string prompt = "Please enter your choice > ";

cppreference.com

Array declaration.

Array is a type consisting of a contiguously allocated nonempty sequence of objects with a particular element type . The number of those objects (the array size) never changes during the array lifetime.

[ edit ] Syntax

In the declaration grammar of an array declaration, the type-specifier sequence designates the element type (which must be a complete object type), and the declarator has the form:

[ edit ] Explanation

There are several variations of array types: arrays of known constant size, variable-length arrays, and arrays of unknown size.

[ edit ] Arrays of constant known size

If expression in an array declarator is an integer constant expression with a value greater than zero and the element type is a type with a known constant size (that is, elements are not VLA) (since C99) , then the declarator declares an array of constant known size:

Arrays of constant known size can use array initializers to provide their initial values:

[ edit ] Arrays of unknown size

If expression in an array declarator is omitted, it declares an array of unknown size. Except in function parameter lists (where such arrays are transformed to pointers) and when an initializer is available, such type is an incomplete type (note that VLA of unspecified size, declared with * as the size, is a complete type) (since C99) :

[ edit ] Qualifiers

[ edit ] Assignment

Objects of array type are not modifiable lvalues , and although their address may be taken, they cannot appear on the left hand side of an assignment operator. However, structs with array members are modifiable lvalues and can be assigned:

[ edit ] Array to pointer conversion

Any lvalue expression of array type, when used in any context other than

• as the operand of the address-of operator
• as the operand of sizeof
• as the operand of typeof and typeof_unqual (since C23)
• as the string literal used for array initialization

undergoes an implicit conversion to the pointer to its first element. The result is not an lvalue.

If the array was declared register , the behavior of the program that attempts such conversion is undefined.

When an array type is used in a function parameter list, it is transformed to the corresponding pointer type: int f ( int a [ 2 ] ) and int f ( int * a ) declare the same function. Since the function's actual parameter type is pointer type, a function call with an array argument performs array-to-pointer conversion; the size of the argument array is not available to the called function and must be passed explicitly:

[ edit ] Multidimensional arrays

When the element type of an array is another array, it is said that the array is multidimensional:

Note that when array-to-pointer conversion is applied, a multidimensional array is converted to a pointer to its first element, e.g., pointer to the first row:

[ edit ] Notes

Zero-length array declarations are not allowed, even though some compilers offer them as extensions (typically as a pre-C99 implementation of flexible array members ).

If the size expression of a VLA has side effects, they are guaranteed to be produced except when it is a part of a sizeof expression whose result doesn't depend on it:

[ edit ] References

• C23 standard (ISO/IEC 9899:2024):
• 6.7.6.2 Array declarators (p: TBD)
• C17 standard (ISO/IEC 9899:2018):
• 6.7.6.2 Array declarators (p: 94-96)
• C11 standard (ISO/IEC 9899:2011):
• 6.7.6.2 Array declarators (p: 130-132)
• C99 standard (ISO/IEC 9899:1999):
• 6.7.5.2 Array declarators (p: 116-118)
• C89/C90 standard (ISO/IEC 9899:1990):
• 3.5.4.2 Array declarators

[ edit ] See also

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An array is a variable that can store multiple values. For example, if you want to store 100 integers, you can create an array for it.

How to declare an array?

For example,

Here, we declared an array, mark , of floating-point type. And its size is 5. Meaning, it can hold 5 floating-point values.

It's important to note that the size and type of an array cannot be changed once it is declared.

Access Array Elements

You can access elements of an array by indices.

Suppose you declared an array mark as above. The first element is mark[0] , the second element is mark[1] and so on.

Few keynotes :

• Arrays have 0 as the first index, not 1. In this example, mark[0] is the first element.
• If the size of an array is n , to access the last element, the n-1 index is used. In this example, mark[4]
• Suppose the starting address of mark[0] is 2120d . Then, the address of the mark[1] will be 2124d . Similarly, the address of mark[2] will be 2128d and so on. This is because the size of a float is 4 bytes.

How to initialize an array?

It is possible to initialize an array during declaration. For example,

You can also initialize an array like this.

Here, we haven't specified the size. However, the compiler knows its size is 5 as we are initializing it with 5 elements.

Change Value of Array elements

Input and output array elements.

Here's how you can take input from the user and store it in an array element.

Here's how you can print an individual element of an array.

Example 1: Array Input/Output

Here, we have used a  for loop to take 5 inputs from the user and store them in an array. Then, using another  for loop, these elements are displayed on the screen.

Example 2: Calculate Average

Here, we have computed the average of n numbers entered by the user.

Access elements out of its bound!

Suppose you declared an array of 10 elements. Let's say,

You can access the array elements from testArray[0] to testArray[9] .

Now let's say if you try to access testArray[12] . The element is not available. This may cause unexpected output (undefined behavior). Sometimes you might get an error and some other time your program may run correctly.

Hence, you should never access elements of an array outside of its bound.

Multidimensional arrays

In this tutorial, you learned about arrays. These arrays are called one-dimensional arrays.

In the next tutorial, you will learn about multidimensional arrays (array of an array) .

Table of Contents

• C Arrays (Introduction)
• Declaring an Array
• Access array elements
• Initializing an array
• Change Value of Array Elements
• Array Input/Output
• Example: Calculate Average
• Array Elements Out of its Bound

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Assignment Operators in C

In C language, the assignment operator stores a certain value in an already declared variable. A variable in C can be assigned the value in the form of a literal, another variable, or an expression.

The value to be assigned forms the right-hand operand, whereas the variable to be assigned should be the operand to the left of the " = " symbol, which is defined as a simple assignment operator in C.

In addition, C has several augmented assignment operators.

The following table lists the assignment operators supported by the C language −

Simple Assignment Operator (=)

The = operator is one of the most frequently used operators in C. As per the ANSI C standard, all the variables must be declared in the beginning. Variable declaration after the first processing statement is not allowed.

You can declare a variable to be assigned a value later in the code, or you can initialize it at the time of declaration.

You can use a literal, another variable, or an expression in the assignment statement.

Once a variable of a certain type is declared, it cannot be assigned a value of any other type. In such a case the C compiler reports a type mismatch error.

In C, the expressions that refer to a memory location are called "lvalue" expressions. A lvalue may appear as either the left-hand or right-hand side of an assignment.

On the other hand, the term rvalue refers to a data value that is stored at some address in memory. A rvalue is an expression that cannot have a value assigned to it which means an rvalue may appear on the right-hand side but not on the left-hand side of an assignment.

Variables are lvalues and so they may appear on the left-hand side of an assignment. Numeric literals are rvalues and so they may not be assigned and cannot appear on the left-hand side. Take a look at the following valid and invalid statements −

Augmented Assignment Operators

In addition to the = operator, C allows you to combine arithmetic and bitwise operators with the = symbol to form augmented or compound assignment operator. The augmented operators offer a convenient shortcut for combining arithmetic or bitwise operation with assignment.

For example, the expression "a += b" has the same effect of performing "a + b" first and then assigning the result back to the variable "a".

Run the code and check its output −

Similarly, the expression "a <<= b" has the same effect of performing "a << b" first and then assigning the result back to the variable "a".

Here is a C program that demonstrates the use of assignment operators in C −

When you compile and execute the above program, it will produce the following result −

Next: Unions , Previous: Overlaying Structures , Up: Structures   [ Contents ][ Index ]

15.13 Structure Assignment

Assignment operating on a structure type copies the structure. The left and right operands must have the same type. Here is an example:

Notionally, assignment on a structure type works by copying each of the fields. Thus, if any of the fields has the const qualifier, that structure type does not allow assignment:

See Assignment Expressions .

When a structure type has a field which is an array, as here,

structure assigment such as r1 = r2 copies array fields’ contents just as it copies all the other fields.

This is the only way in C that you can operate on the whole contents of a array with one operation: when the array is contained in a struct . You can’t copy the contents of the data field as an array, because

would convert the array objects (as always) to pointers to the zeroth elements of the arrays (of type struct record * ), and the assignment would be invalid because the left operand is not an lvalue.

How to solve “Error: Assignment to expression with array type” in C?

In C programming, you might have encountered the “ Error: Assignment to expression with array type “. This error occurs when trying to assign a value to an already initialized  array , which can lead to unexpected behavior and program crashes. In this article, we will explore the root causes of this error and provide examples of how to avoid it in the future. We will also learn the basic concepts involving the array to have a deeper understanding of it in C. Let’s dive in!

What is “Error: Assignment to expression with array type”? What is the cause of it?

An array is a collection of elements of the same data type that are stored in contiguous memory locations. Each element in an array is accessed using an index number. However, when trying to assign a value to an entire array or attempting to assign an array to another array, you may encounter the “Error: Assignment to expression with array type”. This error occurs because arrays in C are not assignable types, meaning you cannot assign a value to an entire array using the assignment operator.

First example: String

Here is an example that may trigger the error:

In this example, we have declared a char array “name” of size 10 and initialized it with the string “John”. Then, we are trying to assign a new string “Mary” to the entire array. However, this is not allowed in C because arrays are not assignable types. As a result, the compiler will throw the “Error: Assignment to expression with array type”.

Initialization

When you declare a char array in C, you can initialize it with a string literal or by specifying each character separately. In our example, we initialized the char array “name” with the string literal “John” as follows:

This creates a char array of size 10 and initializes the first four elements with the characters ‘J’, ‘o’, ‘h’, and ‘n’, followed by a null terminator ‘\0’. It’s important to note that initializing the array in this way does not cause the “Error: Assignment to expression with array type”.

On the other hand, if you declare a char array without initializing it, you will need to assign values to each element of the array separately before you can use it. Failure to do so may lead to undefined behavior. Considering the following code snippet:

We declared a char array “name” of size 10 without initializing it. Then, we attempted to assign a new string “Mary” to the entire array, which will result in the error we are currently facing.

When you declare a char array in C, you need to specify its size. The size determines the maximum number of characters the array can hold. In our example, we declared the char array “name” with a fixed size of 10, which can hold up to 9 characters plus a null terminator ‘\0’.

If you declare a char array without specifying its size, the compiler will automatically determine the size based on the number of characters in the string literal you use to initialize it. For instance:

This code creates a char array “name” with a size of 5, which is the number of characters in the string literal “John” plus a null terminator. It’s important to note that if you assign a string that is longer than the size of the array, you may encounter a buffer overflow.

Second example: Struct

We have known, from the first example, what is the cause of the error with string, after that, we dived into the definition of string, its properties, and the method on how to initialize it properly. Now, we can look at a more complex structure:

This struct “struct_type” contains an integer variable “id” and a character array variable “string” with a fixed size of 10. Now let’s create an instance of this struct and try to assign a value to the “string” variable as follows:

As expected, this will result in the same “Error: Assignment to expression with array type” that we encountered in the previous example. If we compare them together:

• The similarity between the two examples is that both involve assigning a value to an initialized array, which is not allowed in C.
• The main difference between the two examples is the scope and type of the variable being assigned. In the first example, we were dealing with a standalone char array, while in the second example, we are dealing with a char array that is a member of a struct. This means that we need to access the “string” variable through the struct “s1”.

So basically, different context, but the same problem. But before dealing with the big problem, we should learn, for this context, how to initialize a struct first. About methods, we can either declare and initialize a new struct variable in one line or initialize the members of an existing struct variable separately.

Take the example from before, to declare and initialize a new struct variable in one line, use the following syntax:

To initialize the members of an existing struct variable separately, you can do like this:

Both of these methods will initialize the “id” member to 1 and the “struct_name” member to “structname”. The first one is using a brace-enclosed initializer list to provide the initial values of the object, following the law of initialization. The second one is specifically using strcpy() function, which will be mentioned in the next section.

How to solve “Error: Assignment to expression with array type”?

Initialize the array type member during the declaration.

As we saw in the first examples, one way to avoid this error is to initialize the array type member during declaration. For example:

This approach works well if we know the value of the array type member at the time of declaration. This is also the basic method.

Use the strcpy() function

We have seen the use of this in the second example. Another way is to use the strcpy() function to copy a string to the array type member. For example:

Remember to add the string.h library to use the strcpy() function. I recommend going for this approach if we don’t know the value of the array type member at the time of declaration or if we need to copy a string to the member dynamically during runtime. Consider using strncpy() instead if you are not sure whether the destination string is large enough to hold the entire source string plus the null character.

Use pointers

We can also use pointers to avoid this error. Instead of assigning a value to the array type member, we can assign a pointer to the member and use malloc() to dynamically allocate memory for the member. Like the example below:

Before using malloc(), the stdlib.h library needs to be added. This approach is also working well for the struct type. In the next approach, we will talk about an ‘upgrade-version’ of this solution.

Use structures with flexible array members (FAMs)

If we are working with variable-length arrays, we can use structures with FAMs to avoid this error. FAMs allow us to declare an array type member without specifying its size, and then allocate memory for the member dynamically during runtime. For example:

The code is really easy to follow. It is a combination of the struct defined in the second example, and the use of pointers as the third solution. The only thing you need to pay attention to is the size of memory allocation to “s”. Because we didn’t specify the size of the “string” array, so at the allocating memory process, the specific size of the array(10) will need to be added.

This a small insight for anyone interested in this example. As many people know, the “sizeof” operator in C returns the size of the operand in bytes. So when calculating the size of the structure that it points to, we usually use sizeof(*), or in this case: sizeof(*s).

But what happens when the structure contains a flexible array member like in our example? Assume that sizeof(int) is 4 and sizeof(char) is 1, the output will be 4. You might think that sizeof(*s) would give the total size of the structure, including the flexible array member, but not exactly. While sizeof is expected to compute the size of its operand at runtime, it is actually a compile-time operator. So, when the compiler sees sizeof(*s), it only considers the fixed-size members of the structure and ignores the flexible array member. That’s why in our example, sizeof(*s) returns 4 and not 14.

How to avoid “Error: Assignment to expression with array type”?

Summarizing all the things we have discussed before, there are a few things you can do to avoid this error:

• Make sure you understand the basics of arrays, strings, and structures in C.
• Always initialize arrays and structures properly.
• Be careful when assigning values to arrays and structures.
• Consider using pointers instead of arrays in certain cases.

The “ Error: Assignment to expression with array type ” in C occurs when trying to assign a value to an already initialized  array . This error can also occur when attempting to assign a value to an array within a  struct . To avoid this error, we need to initialize arrays with a specific size, use the strcpy() function when copying strings, and properly initialize arrays within structures. Make sure to review the article many times to guarantee that you can understand the underlying concepts. That way you will be able to write more efficient and effective code in C. Have fun coding!

“Expected unqualified-id” error in C++ [Solved]

How to Solve does not name a type error in C++

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Programmer guide, tips and tutorial, c++ bug: [error] invalid array assignment.

C++BUG: [Error] invalid array assignment

1. Introduction2. The difference between the return value of memcpy() function prototype function header file and strcpy

1. Introduction

When using array to assign value to array, the above bug will appear. The general chestnut is as follows:

The purpose is to store the data in the structure array object SS [J], but this assignment will report an error.

2. memcpy()

Function prototype

The data of consecutive n bytes with SRC pointing address as the starting address is copied into the space with destination pointing address as the starting address.

Header file

The use of # include & lt; string. H & gt;;

Both # include & lt; CString & gt; and # include & lt; string. H & gt; can be used in C + +

Return value

Function returns a pointer to dest.

The difference from strcpy

1. Compared with strcpy, memcpy does not end when it encounters’ \ 0 ‘, but will copy n bytes. Therefore, we need to pay special attention to memory overflow. 2. Memcpy is used to copy memory. You can use it to copy objects of any data type, and you can specify the length of the data to be copied. Note that source and destination are not necessarily arrays, and any read-write space can be used; however, strcpy can only copy strings, and it ends copying when it encounters’ \ 0 ‘.

For 2D arrays

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Invalid array assignment while shifting struct element

I have to "delete" some elements from struct. For example let it be years. So i should find them in my struct, shift them and display new struct without them.

Our professor said, that we should use pointers to structs. And i don't get it. I mean from my point of view structs works as arrays but with some specifications.

I guess it should shift data to "deleted" space but I get invalid array assignment with all char stuff.

Here are my struct s

• could you please provide the definition of your struct ? what is mbvs ? what is pos ? –  Martin Commented Mar 23, 2019 at 17:27
• My fault. Here: `````` struct date { int day; int month; int year; }; date datebase[100], *dtbs=datebase; struct movies { int udk; char name[10]; char genre[10]; char creator[10]; int releaseYear; struct date movieRental; }; movies mooviebase[100], *mvbs=mooviebase; ```` –  umbrella123 Commented Mar 23, 2019 at 17:28
• You can edit your code for readability –  Martin Commented Mar 23, 2019 at 17:29
• Please edit the code in your question rather than providing it in the comments. Please also make it a fully compilable program. –  brothir Commented Mar 23, 2019 at 17:34
• The variable definitions are wrong, should be struct date datebase[100], *dtbs=datebase; and struct movies mooviebase[100], *mvbs=mooviebase; . The keyword struct is missing. –  Weather Vane Commented Mar 23, 2019 at 17:48

Warning if I indent correctly your code it is :

but this is not what you want because only udk is moved in the loop and for all the other element k value n-1 (I suppose pos <= n - 1 ) so you do for instance (mvbs+n-1) -> name = (mvbs+n) -> name , and n being visible the size you go out of the initialized part of the array or may be out of the array.

Of course you missed to add {} to do :

However first what pos is ? and because you reuse/modify the variable i in the embedded for you cannot manage the case where several entries have the same releaseYear

Probably the first for is supposed to be on pos rather than on i to be more consistent and to remove the entries pos having the search year.

After the embedded for there is one less element so you need to decrease n

Note if there are several consecutive entries having the search year you can optimize to bypass them at the same time rather than to do one by one.

The type of the elements are compatible to a direct copy so the code can be simplified to be :

and finally you can use a memmove rather than to do the copy by yourself element by element :

Our professor said, that we should use pointers to structs

You already use a pointer because you do not do mooviebase[pos] for instance but use mvbs+offset , but that pointer never change and you need the additional offset, so your pointer is not useful, you can do for instance :

• and you too: mvbs[i] = mvbs[i + 1] :D –  Antti Haapala -- Слава Україні Commented Mar 23, 2019 at 18:08
• @AnttiHaapala 1) that changes nothing, 2) I follow the way to write of the OP to help him to see the errors comparing his code and mine, 3) the goal for the OP is to move to the use of pointer , so ... ;-) –  bruno Commented Mar 23, 2019 at 18:14
• Mind you that [] operator is only defined for pointers, about time we teach the people that. –  Antti Haapala -- Слава Україні Commented Mar 23, 2019 at 18:20
• @AnttiHaapala c'mon, using pointer means to do *mvbs and ++mvbs for instance, do you really suppose I am stupid ? I hope not ^^ Did you read the end of my answer ? –  bruno Commented Mar 23, 2019 at 18:22
• I am not sure if you're sarcastic or not, let me reiterate - foo[i] operator requires a pointer as one of the operands, so using *(foo + bar) isn't "more using pointers", it just looks awkward. –  Antti Haapala -- Слава Україні Commented Mar 23, 2019 at 18:52

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Problem with arrays

I'n trying to make a sketch about a led matrix 8x8 with 2 74hc595 shift register, but i have a problem with one array. I have several arrays declared like this:

int M[8][8]= { {0, 1, 1, 1, 0, 1, 1, 1}, {0, 0, 1, 0, 0, 1, 1, 1}, {0, 1, 0, 1, 0, 1, 1, 1}, {0, 1, 0, 1, 0, 1, 1, 1}, {0, 1, 1, 1, 0, 1, 1, 1}, {0, 1, 1, 1, 0, 1, 1, 1}, {0, 1, 1, 1, 0, 1, 1, 1}, {0, 1, 1, 1, 0, 1, 1, 1} };

and i want to make other array like this: const int LETRAS=25; int frase[LETRAS][8][8]={W,W,WW,P,R,A,C,T,I,C,A,N,D,O,A,R,D,U,I,N,OO,C,O,M,SPACE};

but i have this error 200: error: invalid conversion from 'int (*)[8]' to 'int'

If i try this way

int frase[LETRAS][8][8]; frase[0]= M;

i have the follow error: 203: error: invalid array assignment

Does anybody can help me please?

First things first - if all you saving in each element is a 1 or a 0, that's a huge waste of memory. Make the elements "byte" or better still, pack the bits.

And post your code - use code tags.

You've got two dimensions to that array, but you're only specifying one. That may "work" but it will bite you later.

Your biggest problem is that you are trying to shove a letter in a hole meant for a number. You have an int data type and then try to put M in it.

Maybe you want the ascii representation of M, so you could use 'M'. I'm not real sure what you want to do so I'm not sure how you should change it. But what you've got now is just completely wrong on too many levels to list.

Re-reading it, it isn't just a waste of memory, unless you're running it on a Mega, it's almost certainly more memory than you've got. So, "byte" rather than "int", and then eventually, you'll be moving the whole lot into PROGMEM.

i want to store in an array 25 matrix 8x8, each matrix position is an integer, so i will have 25 matrix like M (with different digits), so my question is how can i make this?

i want to store in an array 25 matrix 8x8, each matrix position is an integer,

You probably can't. That would be 3200 bytes, which is more than exists on an Arduino.

If you want to have an array of references to bitmaps, suitable for displaying on dot-matrix displays, that usually comes out somewhat differently. Keeping with the basics you have now, You could do:

More typically, one would pack bits into a sort of font table, stick that in program memory, and index it more-or-less directly from character values:

undesrtood, thanks to all, i'll try this way

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How to fix SyntaxError – ‘invalid assignment left-hand side in JavaScript’?

In JavaScript, a SyntaxError : Invalid Assignment Left-Hand Side occurs when the interpreter encounters an invalid expression or structure on the left side of an assignment operator ( = ).

This error typically arises when trying to assign a value to something that cannot be assigned, such as literals, expressions, or the result of function calls. Let’s explore this error in more detail and see how to resolve it with some examples.

Understanding an error

An invalid assignment left-hand side error occurs when the syntax of the assignment statement violates JavaScript’s rules for valid left-hand side expressions. The left-hand side should be a target that can receive an assignment like a variable, object property or any array element, let’s see some cases where this error can occur and also how to resolve this error.

Case 1: Error Cause: Assigning to Literals

When you attempt to assign a value to a literal like a number, string or boolean it will result in SyntaxError: Invalid Assignment Left-Hand Side .

Resolution of error

In this case values should be assigned to variables or expressions which will be on the left side of an equation and avoid assigning values directly to literals.

Case 2: Error Cause: Assigning to Function Calls

Assigning a value directly to the result of function call will give an invalid assignment left-hand side error.

Explanation : In this example, getX() returns a value but is not a valid target for assignment. Assignments should be made to variables, object properties, or array elements, not to the result of a function call.

Therefore, store it into a variable or at least place it on the left-hand side that is valid for assignment.

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