g ++如何获取忽略函数返回值的警告 [英] g++ How to get warning on ignoring function return value

问题描述

lint产生一些警告，如：

lint produces some warning like:

foo.c XXX Warning 534: Ignoring return value of function bar()

从lint 手动

534忽略函数的返回值

534 Ignoring return value of function

'Symbol'（与位置比较）A
返回一个值的函数是
只是为副作用调用，对于
示例，在一个语句中自己或
逗号
运算符的左侧。 Try：（void）function（）;到
调用一个函数并忽略它的返回
的值。另请参阅§5.5标记选项中的fvr，fvo和fdr
标记。

'Symbol' (compare with Location) A function that returns a value is called just for side effects as, for example, in a statement by itself or the left-hand side of a comma operator. Try: (void) function(); to call a function and ignore its return value. See also the fvr, fvo and fdr flags in §5.5 "Flag Options".

我想获得此警告，如果存在，在编译期间。在gcc / g ++中有没有任何选项来实现这一点？我已经打开了 -Wall ，但是显然没有检测到这一点。

I want to get this warning, if there exists any, during compilation. Is there any option in gcc/g++ to achieve this? I had turned on -Wall but that apparently did not detect this.

推荐答案

p>感谢 WhirlWind 和 paxdiablo 的答案和评论。

Thanks to WhirlWind and paxdiablo for the answer and comment. Here is my attempt to put the pieces together into a complete (?) answer.

-Wunused-result

-Wno-unused-result

标有属性 warn_unused_result 的函数（请参阅
变量属性）不使用其返回值。默认值为 -Wunused-result

Do not warn if a caller of a function marked with attribute warn_unused_result (see Variable Attributes) does not use its return value. The default is -Wunused-result

函数上的 warn_unused_result 属性。

这是一个完整的例子。文件的内容unused_result.c

Here is a full example. The contents of the file unused_result.c

int foo() { return 3; }

int bar() __attribute__((warn_unused_result));
int bar() { return 5; }

int main()
{
    foo();
    bar();    /* line 9 */
    return 0;
}

和相应的编译结果：

$gcc unused_result.c 
unused_result.c: In function ‘main’:
unused_result.c:9: warning: ignoring return value of ‘bar’, declared with attribute warn_unused_result

再次注意，没有必要使用-Wunused-结果，因为它是默认值。人们可能试图明确地提到它来传达意图。虽然这是一个高尚的意图，但在分析情况后，我的选择，但是，将是反对。因为，在编译选项中有 -Wunused-result 可能会产生一个假的安全/满意感，这是不正确的，除非所有该代码库使用 warn_unused_result 限定。

Note again that it is not necessary to have -Wunused-result since it is default. One may be tempted to explicitly mention it to communicate the intent. Though that is a noble intent, but after analyzing the situation, my choice, however, would be against that. Because, having -Wunused-result in the compile options may generate a false sense of security/satisfaction which is not true unless the all the functions in the code base are qualified with warn_unused_result.

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