如果忽略返回值，如何发出警告?

问题描述

我想查看我的代码 (C++) 中所有忽略函数返回值的地方.我该怎么做 - 使用 gcc 或静态代码分析工具?

I'd like to see all the places in my code (C++) which disregard return value of a function. How can I do it - with gcc or static code analysis tool?

错误代码示例:

int f(int z) {
    return z + (z*2) + z/3 + z*z + 23;
}


int main()
{
  int i = 7;
  f(i); ///// <<----- here I disregard the return value

  return 1;
}

请注意:

• 即使函数及其用法在不同的文件中也应该可以工作
• 免费静态检查工具

• it should work even if the function and its use are in different files
• free static check tool

推荐答案

你想要 GCC 的 warn_unused_result 属性:

You want GCC's warn_unused_result attribute:

#define WARN_UNUSED __attribute__((warn_unused_result))

int WARN_UNUSED f(int z) {
    return z + (z*2) + z/3 + z*z + 23;
}

int main()
{
  int i = 7;
  f(i); ///// <<----- here i disregard the return value
  return 1;
}

尝试编译此代码会产生:

Trying to compile this code produces:

$ gcc test.c
test.c: In function `main':
test.c:16: warning: ignoring return value of `f', declared with
attribute warn_unused_result

您可以在 Linux内核;他们有一个 __must_check 宏可以做同样的事情；看起来你需要 GCC 3.4 或更高版本才能工作.然后你会发现内核头文件中使用的那个宏:

You can see this in use in the Linux kernel; they have a __must_check macro that does the same thing; looks like you need GCC 3.4 or greater for this to work. Then you will find that macro used in kernel header files:

unsigned long __must_check copy_to_user(void __user *to,
                                        const void *from, unsigned long n);

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