为什么赋值操作调用的是单参构造函数？

[Damn_boy]

class WhoWillBeCall
{
public:
	WhoWillBeCall()
	{
		printf("default param Constructor\n");
		m_i=1;
	};
	~WhoWillBeCall(){};

	WhoWillBeCall(int i)
	{
		printf("single param Constructor\n");
	};

	WhoWillBeCall &operator=(int &ref)
	{
		printf("operator = ");
		return *this;
	}
	int m_i;
};

WhoWillBeCall test;
test = 1;

结果是

default param Constructor
single param Constructor

把

WhoWillBeCall &operator=(int &ref)

改为

WhoWillBeCall &operator=(const int &ref)

则是

default param Constructor
operator = 

这个...什么原理？

const int &ref 能指代任何的值？
而int ref只能指代int类型的变量？

记得primer里提过
const类型的引用可以指向任何的值
因为编译器会为其生成中间代码
const int& ref = 100.12;
{
int temp = 100.12;
const int&ref = temp;
}
不过忘了这么做的含义了。

[Fn_1088]

这个还真奇怪了~~~~

[开发者说]

const int &ref能够接受右值，而int &ref只能接受左值；
WhoWillBeCall test;
test = 1;
1是右值，在const int &ref时，以上语句刚好对应了重载的=；而int &ref时找不到对应的=运算，以上语句就被编译器解释成为了拷贝构造函数的调用进行初始化。

[Fn_1088]

class WhoWillBeCall
{
public:
WhoWillBeCall()
{
printf("default param Constructor\n");
m_i=1;
}
~WhoWillBeCall(){}

WhoWillBeCall(int i)
{
printf("single param Constructor\n");
}

WhoWillBeCall(const WhoWillBeCall &sc)
{
this->m_i = sc.m_i;
printf("copy constructor \n");
}

WhoWillBeCall& operator=(const WhoWillBeCall &sc)
{
this->m_i = sc.m_i;
printf("=========\n");
return *this;
}

WhoWillBeCall &operator=(/*const */int &ref)
{
printf("operator = \n");
return *this;
}

int m_i;
};

[Damn_boy]

operator= 针对int的重载版本

WhoWillBeCall &operator=(int &ref)

WhoWillBeCall &operator=(const int &ref)
WhoWillBeCall &operator=(const int val)
WhoWillBeCall &operator=(int val)

比较特殊的是

WhoWillBeCall &operator=(int &ref)版本

int &ref只能处理左值
而其他三个

WhoWillBeCall &operator=(const int &ref)
WhoWillBeCall &operator=(const int val)
WhoWillBeCall &operator=(int val)

都能接收右值

测试代码中

test = 1;

1明显为右值
所以不会触发该重载的operator=

而会触发另一个隐式的类型转换
既将1转化为WhoWillBeCall类型
然后赋值给test

其中就涉及到了单参构造函数的调用
可以把该int的单参构造函数声明为explicit
在编译时就可以看出问题

此外还涉及到了operator=(const WhoWillBeCall &ref)的调用