c++ - 无效的用户定义转换

请查看该程序及其产生的错误:

#include <iostream>
using namespace std;
    class A
    {
    public:

        virtual void f(){}
        int i;
    };

    class B : public A
    {
    public:
        B(int i_){i = i_;} //needed
        B(){}              //needed
        void f(){}
    };

    int main()
    {

        //these two lines are fixed(needed)
        B b;
        A & a = b;

        //Assignment 1 works
        B b1(2);
        b = b1;

        //But Assignment 2  doesn't works
        B b2();
        b = b2; // <-- error
    }

编译后，出现以下错误:

$ g++ inher2.cpp
inher2.cpp: In function ‘int main()’:
inher2.cpp:32:10: error: invalid user-defined conversion from ‘B()’ to ‘const B&’ [-fpermissive]
inher2.cpp:14:6: note: candidate is: B::B(int) <near match>
inher2.cpp:14:6: note:   no known conversion for argument 1 from ‘B()’ to ‘int’
inher2.cpp:32:10: error: invalid conversion from ‘B (*)()’ to ‘int’ [-fpermissive]
inher2.cpp:14:6: error:   initializing argument 1 of ‘B::B(int)’ [-fpermissive]

你能帮我找到问题吗？谢谢

最佳答案

您的“B b2();”是C++的'令人讨厌的解析'问题(see here-'最讨厌的令人讨厌的解析'进一步模糊了语法)。

在C++编译器中，您似乎在声明一个函数(预声明)。

一探究竟:

int foo(); //A function named 'foo' that takes zero parameters and returns an int.

B b2(); //A function named 'b2' that takes zero parameters and returns a 'B'.

以后再做时:

b = b2;

似乎您正在尝试将函数( b2 )分配给变量( b )。
要使用零参数调用构造函数，请在不带括号的情况下调用它，这样就可以了:

B b2;

Most vexing parse: why doesn't A a(()); work?