Overloaded function evaluated with ternary

Question

I have an overloaded function which can take two argument types : int and double. When I evaluate it with a ternary which can return either an int or a double, it always uses the double version. Why is that?

#include<iostream>
using namespace std;

void f(int a)
{
    cout << "int" << endl;
}

void f(double a)
{
    cout << "double" << endl;
}

int main()
{
    string a;
    cin >> a;
    f(a=="int" ? 3 : 3.14159);
    return 0;
}

[Please see this answer](http://stackoverflow.com/a/8535301/1729885). — Niels Keurentjes, Dec 25 '14 at 22:24
possible duplicate of [Return type of '?:' (ternary conditional operator)](http://stackoverflow.com/questions/8535226/return-type-of-ternary-conditional-operator) — Anonymous, Dec 25 '14 at 22:45

score 8 · Accepted Answer · answered Dec 25 '14 at 22:23

8

Ternary operator always do type promotion (into single type). So if one result is int and another is double, the result of ? operartor will always be double.

answered Dec 25 '14 at 22:23

Anonymous

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Overloaded function evaluated with ternary

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