warning assignment discards 'const' qualifier from pointer target type

Understanding 'Assignment Discards Const Qualifier from Pointer Target Type': Solutions and Best Practices

This documentation aims to provide a comprehensive guide on understanding and solving the warning message "assignment discards 'const' qualifier from pointer target type" in the C programming language. We will cover the cause of this warning and provide step-by-step solutions and best practices to avoid this issue in the future.

Table of Contents

• What is 'const' qualifier?
• Why does the warning occur?
• Solutions and Best Practices
• Using const-correctness
• Casting away const-ness

What is 'const' qualifier? {#what-is-const-qualifier}

In C, the const keyword is used to declare a variable that cannot be modified after initialization. It is a form of compile-time enforcement to guarantee that the value of a variable remains unchanged throughout the program. The const keyword is also used to indicate that a function does not modify the object pointed to by a pointer argument.

Consider the following example:

Why does the warning occur? {#why-does-the-warning-occur}

The "assignment discards 'const' qualifier from pointer target type" warning occurs when you try to assign a pointer to a non-const object to a pointer to a const object. This is because doing so would allow the modification of the const object through the non-const pointer, which is not allowed.

For example:

In the above example, the compiler will generate a warning because assigning the address of a const object (x) to a non-const pointer (p) could lead to the modification of the const object, which is not allowed.

Solutions and Best Practices {#solutions-and-best-practices}

Using const-correctness {#using-const-correctness}.

The best practice to avoid this warning is to ensure const-correctness in your code. When working with const objects, always use const pointers. This will prevent you from accidentally modifying const objects and will help you catch potential issues during compilation.

For example, the following code is const-correct and will not generate any warning:

Casting away const-ness {#casting-away-const-ness}

In some cases, you might need to cast away the const-ness of an object. This should be done with caution and only when you are sure that the object will not be modified. You can use a typecast to remove the const qualifier from a pointer:

Keep in mind that casting away const-ness can lead to undefined behavior if you attempt to modify the object through the non-const pointer. Therefore, it is recommended to only use this approach when absolutely necessary and when you are sure that the object will not be modified.

What is const-correctness? {#what-is-const-correctness}

Const-correctness is a programming practice that ensures the proper use of const qualifiers in your code. It involves using const pointers when working with const objects and preventing the modification of const objects through non-const pointers.

Why should I use const pointers? {#why-should-i-use-const-pointers}

Using const pointers helps to maintain the integrity of const objects and prevents accidental modifications. It also helps to catch potential issues during compilation by generating warnings when you try to assign a const object to a non-const pointer.

Can I modify a const object using a non-const pointer? {#can-i-modify-a-const-object-using-a-non-const-pointer}

Modifying a const object through a non-const pointer leads to undefined behavior. While it is possible to cast away the const-ness of an object, doing so should be done with caution and only when you are sure that the object will not be modified.

When should I cast away const-ness? {#when-should-i-cast-away-const-ness}

Casting away const-ness should be done cautiously and only when absolutely necessary. It is recommended to use this approach when you are sure that the object will not be modified and when there is no other way to accomplish your task without casting away const-ness.

How can I ensure const-correctness in my code? {#how-can-i-ensure-const-correctness-in-my-code}

To ensure const-correctness in your code, always use const pointers when working with const objects, and never attempt to modify const objects through non-const pointers. Make sure to follow the proper use of const qualifiers in your code and pay attention to compiler warnings related to const-ness.

Additional Resources

• C Programming: const Keyword
• Understanding const in C: A Complete Guide
• A Guide to Undefined Behavior in C and C++

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AOverflow.com

warning: assignment discards 'const' qualifier from pointer target type

I am implementing the strchr function with the following prototype:

and whose code is:

with a small main to prove it.

When executing the program it gives me the following warning:

I understand that it is because of working with a constant variable and making an assignment, but I can't understand what is the root of the problem and how to solve it.

Isn't doing the assignment like making a "copy" of the pointer *s and storing it in *aux, later working with the "copy"?

First of all, you have to understand what this statement means:

The code above is interpreted like this: " s it's a pointer to a constant character", in other words, you're telling the compiler: "Hey man, s it points to a character that is a constant, its value shouldn't change".

Since pa it points to a constant integer, it is illegal to dereference the pointer , for that reason this expression leads to an error: *pa = 1 , because pa it cannot modify the variable a (it doesn't matter if a it wasn't declared as a constant variable, the expression is still illegal).

Once you know this, it's easy to see why the compiler throws a warning on this line:

I go back and repeat: s it is a pointer that points to a constant character, instead it aux points to a character (it can be modified, because there is no such restriction), for that reason the compiler complains, because aux now it will be able to modify the character to which the pointer points s .

Now you may be wondering, is this really a problem? If we execute your code it will surely work but the code will be confusing, because first you indicate that the value it points to s should not be modified and then aux if it can be modified, then it is not clear what the intention of the code is.

In fact, the purpose of this statement:

It is being able to indicate to other developers that they s should NEVER modify the value to which it refers. This helps to avoid possible errors, since in this way the developer will not intentionally modify (as you know, programmers are clueless) the value that the pointer points to.

Answering questions...

Of course yes.

This line is simple to interpret:

You're copying the memory address pointed s to by aux , at the end aux and s they point to the same thing. The problem with that assignment is that you aux can modify the value you point to s . This warning is simple to solve, put the const in the declaration:

In fact, the pointer aux is unnecessary, you can remove it and work directly with s .

A LOT OF EYE: Do not be confused with this statement:

In that case s it is a constant pointer that points to a character, this means that you cannot modify the content of the pointer.

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The qualifier const can be applied to the declaration of any variable to specify that its value will not be changed (which depends upon where const variables are stored, we may change the value of the const variable by using a pointer). The result is implementation-defined if an attempt is made to change a const.

Using the const qualifier in C is a good practice when we want to ensure that some values should remain constant and should not be accidentally modified.

In C programming, the const qualifier can be used in different contexts to provide various behaviors. Here are some different use cases of the const qualifier in C:

1. Constant Variables

In this case, const is used to declare a variable var as a constant with an initial value of 100. The value of this variable cannot be modified once it is initialized. See the following example:

2. Pointer to Constant

We can change the pointer to point to any other integer variable, but cannot change the value of the object (entity) pointed using pointer ptr. The pointer is stored in the read-write area (stack in the present case). The object pointed may be in the read-only or read-write area. Let us see the following examples.

Example 2: Program where variable i itself is constant.

Down qualification is not allowed in C++ and may cause warnings in C. Down qualification refers to the situation where a qualified type is assigned to a non-qualified type.

Example 3: Program to show down qualification.

3. Constant Pointer to Variable

The above declaration is a constant pointer to an integer variable, which means we can change the value of the object pointed by the pointer, but cannot change the pointer to point to another variable.

4. Constant Pointer to Constant

The above declaration is a constant pointer to a constant variable which means we cannot change the value pointed by the pointer as well as we cannot point the pointer to other variables. Let us see with an example. 

Advantages of const Qualifiers in C

The const qualifier in C has the following advantages:

• Improved Code Readability: By marking a variable as const, you indicate to other programmers that its value should not be changed, making your code easier to understand and maintain.
• Enhanced Type Safety : By using const, you can ensure that values are not accidentally modified, reducing the chance of bugs and errors in your code.
• Improved Optimization: Compilers can optimize const variables more effectively, as they know that their values will not change during program execution. This can result in faster and more efficient code.
• Better Memory Usage: By declaring variables as const, you can often avoid having to make a copy of their values, which can reduce memory usage and improve performance.
• Improved Compatibility : By declaring variables as const, you can make your code more compatible with other libraries and APIs that use const variables.
• Improved Reliability : By using const, you can make your code more reliable, as you can ensure that values are not modified unexpectedly, reducing the risk of bugs and errors in your code.

This article is compiled by “ Narendra Kangralkar “.

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Warning: initialization discards 'const' qualifier from pointer target type #11142.

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Compiler warnings: assignment discards 'const' qualifier from pointer target type

Description.

Compiling LVGL (v8.0.2) code (RISC-V GCC, 8.3.0) I see these warnings:

lvgl/src/widgets/lv_checkbox.c:123:13: warning: assignment discards ‘const’ qualifier from pointer target type [-Wdiscarded-qualifiers]

lvgl/src/core/lv_obj.c:132:18: warning: initialization discards ‘const’ qualifier from pointer target type [-Wdiscarded-qualifiers]

lvgl/src/extra/widgets/tabview/lv_tabview.c:84:25: warning: assignment discards ‘const’ qualifier from pointer target type [-Wdiscarded-qualifiers]

lvgl/src/extra/widgets/tabview/lv_tabview.c:93:24: warning: assignment discards ‘const’ qualifier from pointer target type [-Wdiscarded-qualifiers]

lvgl/src/extra/widgets/tabview/lv_tabview.c:95:37: warning: assignment discards ‘const’ qualifier from pointer target type [-Wdiscarded-qualifiers]

lvgl/src/extra/widgets/tabview/lv_tabview.c:97:37: warning: assignment discards ‘const’ qualifier from pointer target type [-Wdiscarded-qualifiers]

lvgl/src/extra/widgets/tabview/lv_tabview.c:196:21: warning: assignment discards ‘const’ qualifier from pointer target type [-Wdiscarded-qualifiers]

What MCU/Processor/Board and compiler are you using?

RISC-V (GCC 8.3.0), custom platform

What do you want to achieve?

Clean LVGL build with no discarded-qualifiers warnings. I’d like to treat them as errors (-Werror), but currently LVGL prevents me from doing this. It should be easy to fix these.

What have you tried so far?

Adding -Wno-discarded-qualifiers option: the build works, but I’d like to avoid having to add this option.

Code to reproduce

Screenshot and/or video.

(cc @kisvegabor )

Currently, our CI uses -Wno-discarded-qualifiers ; this would need to be changed if we are going to consider this a disallowed warning.

Unfortunately, I can’t see these warnings with GCC 7.5.0, but I agree that it’d be better to enable them.

If @embeddedt also agrees the enable discarded-qualifiers warnings, @marcinh0001 , can you send a PR with the fixes?

I could, but as I explained in another thread, I’m required to add copyright notices to all files I modify, even slightly. I’d like to avoid it, it’s too many files. Would it be possible for someone else to get rid of discarded-qualifiers warnings?

I can do it, however, I need to find a way of getting them to show up first, as when I enabled -Wdiscarded-qualifiers locally last week, they weren’t appearing.

Thanks. I think you’ll need a sufficiently fussy compiler (probably meaning modern?) I don’t have a full view of which GCC versions are / aren’t fussy about it, but these two I use definitely are:

• 8.3.0 for RISC-V
• 10.2.0 for x86 used under mingw

And, needless to say, you need to get rid of the -Wno-discarded-qualifiers option.

I am having the same issues. Environment is

Win 11 pro PlatformIO using Arduino framework lvgl 8.3.4 ESP32DEV module when compiling lvgl, i get several messages like

when using squarline studio 1.2.x, even more of such messages show up. Unfortunately, when I try to disable warning by specifying -Wno-discarded-qualifiers I get informed that this is not allowed in a C++ compile environment. Now, there shouldn’t be any warnings when compiling anyhow, so - what do I have to do to have this situation remedied?

With best regards

Volker Bandke

Same here, I’m working with the latest version of LVGL and PlatformIO using the Arduino Framework. I would like to suppress this warning using the build flag -Wno-discarded-qualifiers However, as previously mentioned, this is not allowed in C++.

Is there another way to suppress this warning globally?

corrects the warning:

src\core\lv_obj.c 143

const uint8_t * txt_u8; txt_u8 = (const uint8_t *)txt;

src\extra\widgets\tabview 81 const char ** new_map;

src\extra\widgets\tabview 107 tabview->map = (char **)new_map;

src\extra\widgets\tabview 228 tabview->map[0] = (char *)"";

\src\widgets\lv_checkbox.c 125 cb->txt = (char *)"Check box";

Great! Are you going to issue a pull request?

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C Warning - assignment discards qualifiers from pointer target type

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#22 warning: assignment discards ‘const’ qualifier from pointer target type Closed: fixed 5 years ago Opened 5 years ago by emaldonado.

This is what we get:

Triggered by the '-Wdiscarded-qualifiers' uses in building.

This patch fixes the discarded 'const' qualifier warnings.

(Edited description for formatting)

This is fixed by jss pr#25 . The upstream commit is cf19f3 .

Metadata Update from @cipherboy : - Custom field component adjusted to None - Custom field feature adjusted to None - Custom field origin adjusted to None - Custom field proposedmilestone adjusted to None - Custom field proposedpriority adjusted to None - Custom field reviewer adjusted to None - Custom field type adjusted to None - Custom field version adjusted to None

Metadata Update from @cipherboy : - Issue close_status updated to: fixed - Issue status updated to: Closed (was: Open)

Login to comment on this ticket.

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質問をすることでしか得られない、回答やアドバイスがある。

15分調べてもわからないことは、質問しよう！.

C言語は、1972年にAT&Tベル研究所の、デニス・リッチーが主体となって作成したプログラミング言語です。 B言語の後継言語として開発されたことからC言語と命名。そのため、表記法などはB言語やALGOLに近いとされています。 Cの拡張版であるC++言語とともに、現在世界中でもっとも普及されているプログラミング言語です。

const修飾子での警告発生理由

投稿 2016/11/12 13:22

###前提・実現したいこと c言語でハッシュテーブルのプログラムを組んでいます、見たことのない警告が出たので、出た理由をしりたく、質問します ###発生している問題・エラーメッセージ ChainHash.c:14: warning: assignment discards qualifiers from pointer target type

###試したこと インターネットで調べてみたところ、なにやらconst修飾子が問題のようですが、しっくりくる、情報が見つけられませんでした。

###補足情報(言語/FW/ツール等のバージョンなど) 実行環境 : Centos6.5(final) コンパイラ : gcc-4.4.7-4.el6.x86_64

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Node構造体のnextという変数が、constがついていないNode*型であることが理由です。 const Nodeへのポインタを、Nodeへのポインタに代入しようとしているのです。

上記のコードを見てください。pの中身は、constがついているので変わらない「はず」ですよね? ですが、qを使うことで、中身を操作できてしまいます。 これが意図しないバグを生むことが多いため、constがついたポインタをそうでないポインタに代入しようとすると、警告が出るのです。

投稿 2016/11/12 13:38

2016/11/12 14:43

ここの238ページにわかりやすく書いているとおりです。 https://www.math.nagoya-u.ac.jp/~naito/lecture/2003_SS/PDF/02/C-1.pdf

投稿 2016/11/12 13:48

質問の解決につながる回答をしましょう。 サンプルコードなど、より具体的な説明があると質問者の理解の助けになります。 また、読む側のことを考えた、分かりやすい文章を心がけましょう。

15分調べてもわからないことは teratailで質問しよう！

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