A question regarding convertion constructor

[LarryChen]

Here is the code snippet. In the definition of class Rational, I specified convertion constructor explicit so that I can prevent a int converted implicitly into a Rational object. But to my surprise, when I call Rational r = r1*2, 2 is still converted to a Rational object. Why?

class Rational
{
public:
	Rational(int x = 0, int y = 1)
	{
		num = x;
		den = y;
	}

	explicit Rational(const int& x)
	{
		
	}

	const Rational& operator*(const Rational& rhs)
	{
		num*=rhs.num;
		den*=rhs.den;
		return *this;
	}

private:
	int num;
	int den;
};



int  main()
{
	Rational r1(2, 3);
	Rational r = r1*2;
	
	return 0;
}

[ZuK]

Rational(int x = 0, int y = 1)

Because of the default value for y that constructor can convert an int to a Rational.
Kurt

[LarryChen]

So constructor Rational(int x = 0, int y = 1) could also be a convertion constructor. It is good to learn. Thanks.

Rational(int x = 0, int y = 1)

Because of the default value for y that constructor can convert an int to a Rational.
Kurt