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        对静态变量 c++ 的未定义引用

        时间:2023-12-03
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                  本文介绍了对静态变量 c++ 的未定义引用的处理方法,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

                  我在以下代码中遇到未定义的引用错误:

                  Hi i am getting undefined reference error in the following code:

                  class Helloworld{
                    public:
                       static int x;
                       void foo();
                  };
                  void Helloworld::foo(){
                       Helloworld::x = 10;
                  };
                  

                  我不想要 static foo() 函数.如何在类的非 static 方法中访问类的 static 变量?

                  I don't want a static foo() function. How can I access static variable of a class in non-static method of a class?

                  推荐答案

                  我不想要一个 static foo() 函数

                  好吧,foo() 在你的类中不是静态的,你不需要让它staticcode> 以访问您的类的 static 变量.

                  Well, foo() is not static in your class, and you do not need to make it static in order to access static variables of your class.

                  您需要做的只是为您的静态成员变量提供一个定义:

                  What you need to do is simply to provide a definition for your static member variable:

                  class Helloworld {
                    public:
                       static int x;
                       void foo();
                  };
                  
                  int Helloworld::x = 0; // Or whatever is the most appropriate value
                                         // for initializing x. Notice, that the
                                         // initializer is not required: if absent,
                                         // x will be zero-initialized.
                  
                  void Helloworld::foo() {
                       Helloworld::x = 10;
                  };
                  

                  这篇关于对静态变量 c++ 的未定义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!

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