Overload a C++ function according to the return value

Asked 2008-10-22T15:02:00.823

We all know that you can overload a function according to the parameters:

int mul(int i, int j) { return i*j; }
std::string mul(char c, int n) { return std::string(n, c); }

Can you overload a function according to the return value? Define a function that returns different things according to how the return value is used:

int n = mul(6, 3); // n = 18
std::string s = mul(6, 3); // s = "666"
// Note that both invocations take the exact same parameters (same types)

You can assume the first parameter is between 0-9, no need to verify the input or have any error handling.

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3 Answers

If you wanted to make mul be a real function instead of a class, you could just use an intermediate class:

class StringOrInt
{
public:
    StringOrInt(int p1, int p2)
    {
        param1 = p1;
        param2 = p2;
    }
    operator int ()
    {
        return param1 * param2;
    }

    operator std::string ()
    {
        return std::string(param2, param1 + '0');
    }

private:
    int param1;
    int param2;
};

StringOrInt mul(int p1, int p2)
{
    return StringOrInt(p1, p2);
}

This lets you do things like passing mul as a function into std algorithms:

int main(int argc, char* argv[])
{
    vector<int> x;
    x.push_back(3);
    x.push_back(4);
    x.push_back(5);
    x.push_back(6);

    vector<int> intDest(x.size());
    transform(x.begin(), x.end(), intDest.begin(), bind1st(ptr_fun(&mul), 5));
    // print 15 20 25 30
    for (vector<int>::const_iterator i = intDest.begin(); i != intDest.end(); ++i)
        cout << *i << " ";
    cout << endl;

    vector<string> stringDest(x.size());
    transform(x.begin(), x.end(), stringDest.begin(), bind1st(ptr_fun(&mul), 5));
    // print 555 5555 55555 555555
    for (vector<string>::const_iterator i = stringDest.begin(); i != stringDest.end(); ++i)
        cout << *i << " ";
    cout << endl;

    return 0;
}

answered 2009-01-09T21:02:03.143

As far as I know, you can't (big pity, though...). As a workaround, you can define an 'out' parameter instead, and overload that one.

answered 2008-10-22T15:04:32.397

You can use the functor solution above. C++ does not support this for functions except for const. You can overload based on const.

answered 2008-10-22T15:48:56.707

Overload a C++ function according to the return value

3 Answers

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