C++ overloading: switching from friend to member function

Question

I have this code which I wish to switch from friend functions to member functions:

inline bool operator< (const MyClass& left, const MyClass& right)
{
    return (((left.value == 1) ? 14 : left.value) < ((right.value == 1) ? 14 : right.value));
}

inline bool operator> (const MyClass& left, const MyClass& right)
{
    // recycle <
    return  operator< (right, left);
}

I have got this far:

inline bool MyClass::operator< (const MyClass& right)
{

    return (((this->value == 1) ? 14 : this->value) < ((right.value == 1) ? 14 : right.value));
}

inline bool MyClass::operator> (const MyClass& right)
{
    // recycle <
    return  right.operator<(*this);
}

However, VC++ gives me this complain:

cannot convert 'this' pointer from 'const MyClass' to 'MyClass &'

How can I fix this? Beside, is my operator> correctly written?

@Tas, MyClass is a class for card comparison. So Ace is 1 and 14 at the same time. That's how I do it. But if you have better ideas, do let me know. — J. Doe, Apr 17 '18 at 03:36

score 4 · Accepted Answer · answered Apr 17 '18 at 03:10

Both of your operators should be const class methods:

inline bool MyClass::operator< (const MyClass& right) const
{

    return (((this->value == 1) ? 14 : this->value) < ((right.value == 1) ? 14 : right.value));
}

inline bool MyClass::operator> (const MyClass& right) const
{
    // recycle <
    return  right.operator<(*this);
}

Note that in the > overload, right is a const MyClass &.

Therefore, right.operator< requires the < operator to be a const class method, because right is const. When you're playing games with a const object, you can only invoke its const methods. You cannot invoke its non-const methods.

C++ overloading: switching from friend to member function

1 Answers1