什么是悬空引用?

2022-01-05 00:00:00 reference c++ undefined-behavior

以下程序给我运行时错误(Segmentation Fault (SIGSEGV)).

Following program give me runtime error(Segmentation Fault (SIGSEGV)).

#include <iostream>
using namespace std;

int& bar()
{
    int n = 10;
    return n;
}

int main() {
    int& i = bar();
    cout<<i<<endl;
    return 0;
}

我的老师告诉我这是一个未定义的行为，因为悬空引用.他说得对吗?如果是，那么如何避免它?

My teacher told me it is a undefined behavior because dangling reference. Is he right? If yes then how to avoid it?

推荐答案

是的，这确实是一个未定义的行为，因为您正在返回对自动变量的引用，该变量将在执行 bar() 时被销毁完成

Yes it is indeed an undefined behavior because you are returning a reference to automatic variable which will be destroyed when execution of bar() completes

你可以通过写来避免它:

You can avoid it by writing:

#include <iostream>
using namespace std;

int& bar()
{
    static int n = 10;
    return n;
}

int main() {
    int& i = bar();
    cout<<i<<endl;
    return 0;
}

在这种情况下，静态变量 n 在 bar() 执行完成时不会被销毁，它会在您的程序终止时被销毁.

In this case static variable n will not be destroyed when execution of bar() completes, it will be destroyed when your program terminates.

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