是否可以在同一个函数中引用范围外的局部变量？ [英] Is it OK to reference an out-of-scope local variable within the same function?

问题描述

在这段代码中，我引用了局部变量 b ，即使它超出范围。但是我从同一个函数里面做，所以它可能还在堆栈上，对吧？我运行的程序，它的工作，但我想知道是否保证在所有实现上工作。

In this code, I reference the local variable b even though it is out of scope. But I do it from within the same function so it's probably still on the stack, right? I ran the program and it worked but I'd like to know if it's guaranteed to work on all implementations.

#include <iostream>

void main()
{
    int* a;
    {
        int b = 5;
        a = &b;
    }
    std::cout << *a;
}

推荐答案

不保证工作。 a 一旦退出内部作用域就会悬挂，因此任何取消引用都会导致未定义的行为，并且不会保证任何东西。

No, that's not guaranteed to work. a is dangling once the inner scope is exited, so any dereference of it results in Undefined Behaviour and nothing whatsoever is guaranteed.

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