C ++变量范围 [英] C++ variable scope

问题描述

与预期相比，我的C ++代码输出有所不同，我希望了解其执行方式

I have C++ code different output compared to what I expected, I hope to understand how it's executed

#include <iostream>
#include <string>

int x = 8;

class A {
public:
    A() { 
        int x = 5 ;
    }

    void print (int x = 4) { 
         std::cout << "the scope variable"<< ::x << "passed variable" << x;
    }
};

int main() {
    A a;
    a.print(7);
}

我期望分别是5和7，但结果分别是8和7

I expected to be 5 and 7 but the result is 8 and 7

推荐答案

如果期望输出5和7，则构造函数必须处理全局变量而不是局部变量.

If you expected the output 5 and 7 then the constructor has to deal with the global variable instead of the local variable.

那不是

A(){int x = 5 ;}

您应该写

A(){ x = 5 ;}

或

A(){ ::x = 5 ;}

请注意，最好将变量 x 声明为类 A 的静态数据成员.

Take into account that it would be better to declare the variable x as a static data member of the class A.

例如

class A {
    public :
    //...
    private:
        static int x;  
    };

//...

int A::x = 0;

在这种情况下，只有类的对象才能访问该变量.

In this case only objects of the class could access the variable.

这篇关于C ++变量范围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IT屋！