C ++中的转换运算符 [英] Conversion Operators in C++

问题描述

请帮助我了解C ++中的转换运算符的工作原理。
我有一个简单的例子，我试图理解，虽然不是很清楚如何转换实际上是由编译器发生。

Please help me understand how exactly the conversion operators in C++ work. I have a simple example here which I am trying to understand, though it is not very clear how the conversion actually happens by the compiler.

class Example{
public:
    Example();
    Example(int val);
    operator unsigned int();
    ~Example(){}
private:
    int itsVal;
};

Example::Example():itsVal(0){}

Example::Example(int val):itsVal(val){}

Example::operator unsigned int (){
    return (itsVal);
}

int main(){
    int theInt = 5;
    Example exObject = theInt; // here 
    Example ctr(5);
    int theInt1 = ctr; // here
    return 0;
}

推荐答案

使用调试器（和/或在每个构造函数和运算符上放置断点），以查看哪些构造函数和运算符由哪些行调用。

You can walk through that code with a debugger (and/or put a breakpoint on each of your constructors and operators) to see which of your constructors and operators is being invoked by which lines.

没有明确定义它们，编译器也为你的类创建了一个隐藏/默认拷贝构造函数和赋值运算符。

Because you didn't define them explicitly, the compiler also created a hidden/default copy constructor and assignment operator for your class. You can define these explicitly (as follows) if you want to use a debugger to see where/when they are being called.

Example::Example(const Example& rhs)
: itsVal(rhs.itsVal)
{}

Example& operator=(const Example& rhs)
{
    if (this != &rhs)
    {
        this->itsVal = rhs.itsVal;
    }
    return *this;
}

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