Thread: operator overloading problem~

hi all~

i built a class which supports operator overloading, here is my code:

Code:

class Test
{
	public:
		Test() { a=2; b=3; };
		Test operator * (Test t);
		Test operator [] (Test t);
		void trace() { std::cout << a << "\t" << b << "\n"; };
		int a;
	private:
		int b;
};
Test Test::operator * (const Test t)
{
	a = a * t.a;
	b = b * t.b;
	return *this;
}
Test Test::operator [] (const Test t)
{
	std::cout << t.a << "\t" << t.b;
	return *this;
}

This works fine like this: testInstance1 = testInstance2 * testInstance3
but, it didnt work under this occasion: testInstance1 = 5 * testInstance3
i have no idea how to overcome this, any suggestions ?

You need to define an operator that works with integer values. Since the integer in your example is on the left hand side (lhs) you should make a friend function so it will work in that ordering. Remember your methods of that class have the implied this pointer, so if you defined a method as

Code:

Test Test::operator *(int rhs)

Then your example wouldn't work, but

Code:

testInstance1 = testInstance3 * 5;

would work fine.

no no no guys, i know how to make that, my question is how to calculate when that integer appear left side, like: 5 * testInstance1 and not testInstance1 * 5, how to inplements a method which cope with this situation ?

Why don't you re-read my reply, I clearly told you how if you were paying attention.

Originally Posted by MrWizard

You need to define an operator that works with integer values. Since the integer in your example is on the left hand side (lhs) you should make a friend function so it will work in that ordering. Remember your methods of that class have the implied this pointer, so if you defined a method as

Code:

Test Test::operator *(int rhs)

Then your example wouldn't work, but

Code:

testInstance1 = testInstance3 * 5;

would work fine.

hey thanx man ! and why we should consult friend method here ? can any example be showed plz ?