这两个陈述之间有什么区别

Question

这两个陈述有什么区别？

ob.A::ar[0] = 200;
ob.ar[0] = 200;

哪个ob是类的对象A

class A
{
    public:
        int *ar;
        A()
        {
            ar = new int[100];
        }
};

Answer 1

没有区别.ar在这种情况下,显式名称空间限定是多余的.

在(多个非虚拟)继承重新定义名称的情况下,它可能不是多余的ar.样本(人为):

#include <string>

class A 
{
    public:
        int *ar;
        A() { ar = new int[100]; }
        // unrelated, but prevent leaks: (Rule Of Three)
       ~A() { delete[] ar; }
    private:
        A(A const&);
        A& operator=(A const&);
};

class B : public A
{
    public:
        std::string ar[12];
};


int main()
{
    B ob;
    ob.A::ar[0] = 200;
    ob.ar[0] = "hello world";
}

请参阅http://liveworkspace.org/code/d25889333ec378e1382cb5af5ad7c203

Answer 2

在这种情况下,没有区别.

这种表示法:

obj.Class::member

只是为了解决继承的模糊性:

class A {
public:
  int a;
}

class B {
public:
  int a;
}

class C : public A, B {
   void func() {
      // this objects holds 2 instance variables of name "a" inherited from A and B
      this->A::a = 1;
      this->B::a = 2;
   }

}