C++ 奇怪的构造函数行为 [英] C++ Strange constructor behaviour

问题描述

谁能给我解释一下Complex a;和Complex b();之间的区别吗?

Can anybody explain to me the difference between Complex a; and Complex b();?

#include<iostream>

class Complex
{
public:

    Complex()
    {
        std::cout << "Complex Constructor 1" << std::endl;
    }

    Complex(float re, float im)
    {
        std::cout << "Complex Constructor 2" << std::endl;
    }

    ~Complex()
    {
        std::cout << "Complex Destructor" << std::endl;
    }    
};

int main()
{
    Complex a;
    std::cout << "--------------------------" << std::endl;
    Complex b();
    std::cout << "--------------------------" << std::endl;
    Complex c(0,0);
    std::cout << "--------------------------" << std::endl;

    return 0;
}

输出:

Complex Constructor 1
--------------------------
--------------------------
Complex Constructor 2
--------------------------
Complex Destructor
Complex Destructor

如您所见，Complex a; 确实调用了它的默认构造函数，Complex b(); 没有，Complex c(0,0); 调用重载的构造函数.

As you can see, Complex a; does call its default constructor, Complex b(); doesn't and Complex c(0,0); calls an overloaded constructor.

这里发生了什么?我想，Complex b(); 会创建一个堆栈变量并调用它的默认构造函数来初始化它?

What is going on here? I thought, that Complex b(); would create a stack-variable and call it's default constructor to initialize it?

推荐答案

Complex b(); 是函数声明.这是不带参数并返回 Complex 对象的函数.

Complex b(); is function declaration. That is function taking no arguments and returning Complex object.

这是一个很常见的错误并且有自己的名字:最令人头疼的解析

This is very common mistake and has its own name: most vexing parse

C++11 通过引入统一初始化语法帮助解决了这个问题

C++11 helped with this issue by introducing uniform initialization syntax

Complex b{};

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