为什么在c ++中的函数可以返回堆栈值 [英] Why function in c++ can return stack values

问题描述

请考虑以下程序：

class C {
  ...
};

const C f() {
    C ret;
    cout << &ret << endl;
    return ret;
}

int main() {
    C value = f();
    cout << &value << endl;
}


result: // note the address are the same
0x7ffdd24b26e0
0x7ffdd24b26e0

函数f（）和变量'value'中的变量'ret'具有相同的内存地址，因此'value'不是'ret' 。变量'ret'是一个堆栈变量，所以它应该在f（）返回后无效。那么为什么c ++允许在函数中返回一个堆栈值？

The variable 'ret' in function f() and variable 'value' has the same memory address so it seems 'value' is not a copy of 'ret'. The variable 'ret' is a stack variable so it should be invalidated after f() returns. So why c++ allow returning a stack value inside a function?

g ++版本：

g++ (Ubuntu 4.8.4-2ubuntu1~14.04.3) 4.8.4

推荐答案

ret和value的地址相同的原因是所谓的返回值优化（Return Value Optimization，RVO）。这意味着在这种情况下不会执行副本。注意，你不能依赖这个，因为它不会发生（虽然这将随着C + + 17 [至少当前的草案]）。

The reason for the address of ret and value being the same is the so called Return Value Optimization (RVO). It means that in this case a copy is not going to be performed. Note however, that you can not rely on this, as it is not bound to happen (although that will change with C++17 [at least the current draft]).

https://en.wikipedia.org/wiki/Return_value_optimization

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