No implicit conversion in overloaded operator
-
09-10-2019 - |
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;
}
Solution
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):
Member function
x.operator +(y);
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.
OTHER TIPS
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.