C 警告“返回"没有值，在函数中返回非空 [英] C warning 'return' with no value, in function returning non-void

问题描述

我收到了这个警告.

警告:'return' 没有值，在函数中返回非空值.

warning : 'return' with no value, in function returning non-void.

推荐答案

你有类似的东西:

int function(void)
{
    return;
}

添加返回值，或将返回类型更改为 void.

Add a return value, or change the return type to void.

错误信息很清楚:

警告:'return' 没有值，在函数中返回非空值.

warning : 'return' with no value, in function returning non-void.

没有价值的回报与我展示的相似.该消息还告诉您，如果该函数返回void"，则不会发出警告.但是因为该函数应该返回一个值，而您的返回"语句却没有，所以您遇到了问题.

A return with no value is similar to what I showed. The message also tells you that if the function returns 'void', it would not give the warning. But because the function is supposed to return a value but your 'return' statement didn't, you have a problem.

这通常表示古代代码.在 C89 标准之前的日子里，编译器不一定支持void".当时接受的风格是:

This is often indicative of ancient code. In the days before the C89 standard, compilers did not necessarily support 'void'. The accepted style was then:

function_not_returning_an_explicit_value(i)
char *i;
{
    ...
    if (...something...)
        return;
}

从技术上讲，该函数返回一个 int，但没有预期的值.这种风格的代码会引发您收到的警告 - C99 正式将其取缔，但出于向后兼容性的原因，编译器继续接受它.

Technically, the function returns an int, but no value was expected. This style of code elicits the warning you got - and C99 officially outlaws it, but compilers continue to accept it for reasons of backwards compatibility.

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