缺少返回值的函数，运行时的行为 [英] Function with missing return value, behavior at runtime

问题描述

如预期，编译器（VisualStudio 2008）会发出警告

As expected, the compiler (VisualStudio 2008) will give a warning

警告C4715：'doSomethingWith'：not
所有控制路径在编译以下代码时都会返回值

warning C4715: 'doSomethingWith' : not all control paths return a value

：

int doSomethingWith(int value)
{
    int returnValue = 3;
    bool condition = false;

    if(condition)
        // returnValue += value; // DOH

    return returnValue;
}

int main(int argc, char* argv[])
{
    int foo = 10;
    int result = doSomethingWith(foo);
    return 0;
}

但程序运行正常。函数doSomethingWith（）的返回值为0。

But the program runs just fine. The return value of function doSomethingWith() is 0.

Is只是未定义的行为，或者是否有一定的规则在运行时如何创建/计算结果值。非POD数据类型作为返回值会发生什么？

Is is just undefined behavior, or is there a certain rule how the result value is created/computed at runtime. What happens with non-POD datatypes as return value?

推荐答案

它是ISO C ++标准第6.6节中规定的未定义行为。 3：

It is Undefined behaviour as specified in the ISO C++ standard section 6.6.3:

离开函数末尾的是
，相当于没有值的返回;
这会导致
中一个值返回函数的未定义行为。

Flowing off the end of a function is equivalent to a return with no value; this results in undefined behavior in a value-returning function.

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