C++ ambiguous overload for ‘operator ’

Question

I'v read several posts here about this kind of errors, but I wasn't able to solve this one... Has soon I define the operator int and the function f, fails to compile. I tested several things by I wasn't able to solve the issue.... Thanks

ex1.cpp: In function ‘int main(int, char**)’:
ex1.cpp:35:13: error: ambiguous overload for ‘operator+’ in ‘a + 4’
ex1.cpp:35:13: note: candidates are:
ex1.cpp:35:13: note: operator+(int, int) <built-in>
In file included from ex1.cpp:3:0:
Fraccao.h:41:9: note: Fraccao operator+(const Fraccao&, const Fraccao&)
ex1.cpp:38:13: error: ambiguous overload for ‘operator+’ in ‘4 + a’
ex1.cpp:38:13: note: candidates are:
ex1.cpp:38:13: note: operator+(int, int) <built-in>
In file included from ex1.cpp:3:0:
Fraccao.h:41:9: note: Fraccao operator+(const Fraccao&, const Fraccao&)

The class:

class Fraccao {
    int numerador;
    int denominador;

public:
    Fraccao(int num = 0, int deno = 1) : numerador(num), denominador(deno) {}

    Fraccao & operator+=(const Fraccao &fra);

    Fraccao & operator*=(const Fraccao &fra);

    operator int() const;

    const Fraccao & operator++();
    const Fraccao operator++(int);

    string getAsString() const;
};

Fraccao operator +(const Fraccao &a, const Fraccao &b);
ostream & operator<<(ostream & saida, const Fraccao & fra);

And on my main:

void func(int n) {
    cout << n; // 
}

int main(int argc, char** argv) {
    //...
    d = a + b;
    const Fraccao h(7, 3);
    func(h);

    return 0;
}

Answer 1

You don't seem to have posted the code that actually causes the error. I guess it looks something like

Fraccao a;
Fraccao b = a + 4;
Fraccao c = 4 + a;

The problem is that your class allows implicit conversions both to and from int ; so a + 4 could be either

int(a) + 4

or

a + Fraccao(4)

with no reason to choose one over the other. To resolve the ambiguity, you could either:

• make the conversion explicit, as above; or
• declare either the constructor or the conversion operator (or even both) explicit so that only one (or even neither) conversion can be done implicitly.

Answer 2

The compiler sees two ways to interpret a + 4 . It can convert the 4 to an object of type Fraccao and use operator+(const Fraccao&, const Fraccao&) or it can convert a to type int with the member conversion operator, and add 4 to the result. The rules for overloading make this ambiguous, and that's what the compiler is complaining about. In general, as @gx_ said in a comment, this problem comes up because there are conversions in both directions. If you mark the operator int() with explicit (C++11) the code will be okay.

Answer 3

You have several options to fix your problem:

• Forbid implicit conversion from int to Fraccao : Make the constructor explicit .

• Forbid implicit conversion from Fraccao to int : Make the conversion operator explicit .

• Convert manually on the callers side, given Fraccao f; int i; Fraccao f; int i; either int(f)+i or f+Fraccao(i) .

• Provide additional overloads to resolve the ambiguity:

Fraccao operator +(const Fraccao &a, const Fraccao &b);
Fraccao operator +(const int a, const Fraccao &b);
Fraccao operator +(const Fraccao &a, const int b);

The latter probably means you also want:

Fraccao & operator+=(const Fraccao &fra);
Fraccao & operator+=(const int i);

And finally, if you want the latter, you can use libraries like Boost.Operators or my df.operators to support you and avoid writing the same forwarders over and over again.