No implicit conversion in overloaded operator

Question

d1 + 4 works but 4 + d1 doesn't even though 4 can be converted implicitly to a GMan. Why aren't they equivalent?

struct GMan
{
    int a, b;

    GMan() : a(), b() {}
    GMan(int _a) : a(_a), b() {}
    GMan(int _a, int _b) : a(_a), b(_b) {}

    GMan operator +(const GMan& _b)
    {
         GMan d;
         d.a = this->a + _b.a;
         d.b = this->b + _b.b;
         return d;
    }
};

int main()
{
    GMan d1(1, 2), d(2);
    GMan d3;
    d3 = d1 + 4; 
    d3 = 4 + d1;
}

Answer 1

A call x + y is translated by the C++ compiler into either of the following two calls (depending on whether x is of class type, and whether such a function exists):

1. Member function

   x.operator +(y);
   

2. Free function

   operator +(x, y);
   

Now C++ has a simple rule: no implicit conversion can happen before a member access operator (.). That way, x in the above code cannot undergo an implicit conversion in the first code, but it can in the second.

This rule makes sense: if x could be converted implicitly in the first code above, the C++ compiler wouldn’t know any more which function to call (i.e. which class it belongs to) so it would have to search all existing classes for a matching member function. That would play havoc with C++’ type system and make the overloading rules even more complex and confusing.

Answer 2

This answer is correct. Those points then entail the canonical way of implementing such operators:

struct GMan
{
    int a, b;

    /* Side-note: these could be combined:
    GMan():a(),b(){}
    GMan(int _a):a(_a),b(){}
    GMan(int _a, int _b):a(_a),b(_b){}
    */
    GMan(int _a = 0, int _b = 0) : a(_a), b(_b){} // into this

    // first implement the mutating operator
    GMan& operator+=(const GMan& _b)
    {
        // the use of 'this' to access members
        // is generally seen as noise
        a += _b.a;
        b += _b.b;

        return *this;
    }
};

// then use it to implement the non-mutating operator, as a free-function
// (always prefer free-functions over member-functions, for various reasons)
GMan operator+(GMan _a, const GMan& _b)
{
    _a += b; // code re-use
    return _a;
}

And so on for other operators.