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      1. 什么是悬空引用?

        时间:2023-12-03
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                  本文介绍了什么是悬空引用?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跟版网的小编来一起学习吧!

                  问题描述

                  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 nbar() 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.

                  这篇关于什么是悬空引用?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持跟版网!

                  上一篇:当 push_back 新元素到 std::vector 时,C++ 引用发生变化 下一篇:为什么可以将非常量引用参数绑定到临时对象?

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