返回对本地或临时变量的引用

本文介绍了返回对本地或临时变量的引用的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

查看下面的代码。我知道它不返回局部变量的地址，但为什么它仍然工作，并分配变量 i 在main（）到'6'？

Look at the code below. I know it doesn't return the address of local variable, but why does it still work and assign the variable i in main() to '6'? How does it only return the value if the variable was removed from stack memory?

#include <iostream> int& foo() { int i = 6; std::cout << &i << std::endl; //Prints the address of i before return return i; } int main() { int i = foo(); std::cout << i << std::endl; //Prints the value std::cout << &i << std::endl; //Prints the address of i after return }

推荐答案

你很幸运。从函数返回不会立即擦除刚刚退出的堆栈框架。

You got lucky. Returning from the function doesn't immediately wipe the stack frame you just exited.

BTW，你怎么确认你有一个6回来？表达式 std :: cout<< & i ... 打印 i 的地址，而不是其值。

BTW, how did you confirm that you got a 6 back? The expression std::cout << &i ... prints the address of i, not its value.