A question regarding convertion constructor

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?

Code:

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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;
}

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Code:

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

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

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

Quote:

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Originally Posted by ZuK

Code:

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

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

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