C++ 覆盖赋值运算符

Question

为了理解构造函数和赋值，我写了一个非常简单的测试代码，如下所示:

class A {
public:
    A() { std::cout<<"This is default cstr."; }
    A(int i) { std::cout<<"This is int cstr. value is "<<i; }
    A(const A &a) { std::cout<<"This is copy cstr."; }
    A operator=(const A &a) { std::cout<<"This is assignment operator."; return *this;// this line is tricky }
};
int _tmain(int argc, _TCHAR* argv[]) {
    std::cout<<"line 1 "; A a1; std::cout<<std::endl;
    std::cout<<"line 2 "; A a2 = A(1); std::cout<<std::endl;
    std::cout<<"line 3 "; a1 = a2; std::cout<<std::endl;
    return 0;
}

对于第 3 行，我得到:

line 3 This is assignment operator.This is copy cstr.

但是如果我将 return *this; 更改为 return NULL，我得到:

line 3 This is assignment operator.This is int cstr. value is 0

有人可以为我解释一下里面发生了什么吗？

Answer 1

您的运营商正在返回 A 而不是 A&:

A operator=(const A &a)

因此，当您返回 NULL 时，您正在调用隐式构造函数 A(int) 并将 NULL 传递给它。