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      如何在 C++ 中通过引用返回类对象?

      时间:2023-12-03
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              • 本文介绍了如何在 C++ 中通过引用返回类对象?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

                我有一个名为 Object 的类,用于存储一些数据.

                I have a class called Object which stores some data.

                我想使用这样的函数通过引用返回它:

                I would like to return it by reference using a function like this:

                    Object& return_Object();
                

                然后,在我的代码中,我会这样称呼它:

                Then, in my code, I would call it like this:

                    Object myObject = return_Object();
                

                我已经编写了这样的代码并且它可以编译.但是,当我运行代码时,我始终遇到段错误.通过引用返回类对象的正确方法是什么?

                I have written code like this and it compiles. However, when I run the code, I consistently get a seg fault. What is the proper way to return a class object by reference?

                推荐答案

                您可能正在返回堆栈上的对象.也就是说,return_Object() 可能看起来像这样:

                You're probably returning an object that's on the stack. That is, return_Object() probably looks like this:

                Object& return_Object()
                {
                    Object object_to_return;
                    // ... do stuff ...
                
                    return object_to_return;
                }
                

                如果这是你正在做的事情,那你就不走运了 - object_to_return 已经超出范围并在 return_Object 的末尾被破坏,所以 myObject 指的是一个不存在的对象.您要么需要按值返回,要么返回在更广泛范围内声明的 Objectnew 到堆上.

                If this is what you're doing, you're out of luck - object_to_return has gone out of scope and been destructed at the end of return_Object, so myObject refers to a non-existent object. You either need to return by value, or return an Object declared in a wider scope or newed onto the heap.

                这篇关于如何在 C++ 中通过引用返回类对象?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!

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