How can you make sure a C++ function can be called as

Question

How can you make sure a C++ function can be called as e.g. void foo(int, int) but not as any other type like void foo(long, long)?

Are you trying to solve a specific problem, or is this academic? — Stephen Newell, Feb 09 '21 at 18:02

HolyBlackCat · Answer 1 · 2021-02-09T18:11:29.483

17

Add a deleted template overload:

template <typename A, typename B> void foo(A, B) = delete;

void foo(int x, int y) {...}

It will be a better match (causing a error) for any argument types except int, int.

edited Feb 09 '21 at 18:11

answered Feb 09 '21 at 18:04

HolyBlackCat

3

`foo(0LL, {});` calls the `int` version with a `long long` and an `int`. – Aykhan Hagverdili Feb 09 '21 at 18:14
You can do `template void foo(A, A) = delete;` to catch that too. – user975989 Feb 09 '21 at 18:17
2

@user975989 do you then have to do this for all permutations of 3 or more arguments? – Aykhan Hagverdili Feb 09 '21 at 18:19
@AyxanHaqverdili Hmm good point, maybe you can make `foo` a template and specialize it for ``? Seems kludgy, probably better to make a wrapper type that can only be constructed from `int` (in a similar way to this). – user975989 Feb 09 '21 at 18:23

1 Answers1