c gcc编译器选项用于指针算术警告 [英] c gcc compiler options for pointer arithmetic warning

问题描述

我使用以下标志，但仍然我不能得到这个警告：
类型void *的指针使用在算术

I m using the following flags , but still I m not able to get this warning: "pointer of type 'void *' used in arithmetic"

使用的标志：
-O2 -Werror -Wall -Wno-main -Wno-format-zero-length -Wpointer-arith -Wmissing-prototypes -Wstrict-prototypes -Wswitch -Wshadow -Wcast-qual- strings -Wno-sign-compare -Wno-pointer-sign -Wno-attributes -fno-strict-aliasing

Flags used: -O2 -Werror -Wall -Wno-main -Wno-format-zero-length -Wpointer-arith -Wmissing-prototypes -Wstrict-prototypes -Wswitch -Wshadow -Wcast-qual -Wwrite-strings -Wno-sign-compare -Wno-pointer-sign -Wno-attributes -fno-strict-aliasing

-Wpointer-arith应该捕获这种类型的警告，

-Wpointer-arith should catch this type of warning, but I m not able to get this warning."pointer of type 'void *' used in arithmetic"

应该使用哪个特定的cflag来获取这个警告？

Which specific cflag should be used to get this warning?

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推荐答案

> -Wpointer-arith 应根据

You're right. -Wpointer-arith should give you a warning as per the documentation.

我刚刚试过下面的程序（有意错误）：

I have just tried the following program (with intentional error):

~/code/samples$ cat foo.c
#include <stdio.h>
int main (int argc, char **argv)
{
  void * bar;
  void * foo;
  foo = bar + 1;
  return 0;
}

我只用 -Wpointer-arith 选项，以及上面列出的所有选项两个尝试都提出了所需的警告。我使用gcc版本4.3.4（Debian 4.3.4-6）：

I have compiled the program with just the -Wpointer-arith option, and all your options as listed above. Both attempts threw up the desired warning. I am using gcc version 4.3.4 (Debian 4.3.4-6).:

~/code/samples$ gcc -Wpointer-arith foo.c
foo.c: In function ‘main’:
foo.c:6: warning: pointer of type ‘void *’ used in arithmetic

和

~/code/samples$ gcc -O2 -Werror -Wall -Wno-main -Wno-format-zero-length -Wpointer-arith -Wmissing-prototypes -Wstrict-prototypes -Wswitch -Wshadow -Wcast-qual -Wwrite-strings -Wno-sign-compare -Wno-pointer-sign -Wno-attributes -fno-strict-aliasing foo.c
cc1: warnings being treated as errors
foo.c: In function ‘main’:
foo.c:6: error: pointer of type ‘void *’ used in arithmetic

如果你给它'正确的'代码，编译器会抛出警告。因此，我建议您检查 为什么 ，您会收到此警告。也许编译的代码已经改变了？

The compiler does throw up the warning if you give it the 'right' code. So, I would recommend you examine why it is you expect this warning. Maybe the code you're compiling has changed?

一个可能的线索我可以给你： foo = bar + 1; 在上面的代码中触发警告。但 foo = bar ++; 不会（你得到一个不同的警告）。因此，如果你的代码对指针使用了增量（或递减）运算符，它可能不会触发警告。

One possible clue I can give you: foo = bar + 1; in the code above triggers the warning. But foo = bar ++; will not (You get a different warning). So if your code uses increment (or decrement) operators on pointers, it will probably not trigger the warning.

我知道这不是一个直接的答案，可帮助您集中调查。

I know this is not a direct answer, but I hope this helps you focus your investigation.

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