C++ variable scope

问题

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);
}

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

回答1:

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

That is instead of

A(){int x = 5 ;}

you should write either

A(){ x = 5 ;}

or

A(){ ::x = 5 ;}

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

For example

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

//...

int A::x = 0;

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

回答2:

The constructor body of A is a no-op and is therefore a red herring. (You are declaring an int in local scope.)

The global variable x is always 8.

That is shadowed by the parameter name passed to print.

It would have been less uninteresting had the constructor body been x = 5;, then indeed the result would have been 5 and 7.

来源：https://stackoverflow.com/questions/44261802/c-variable-scope