CODEWITHSUNDEEP
c++templatesc++11
excalibur
excalibur

Reputation: 944

Function template, disallowing certain types

I have a function template that must be allow only certain types. I've seen other questions but they used boost and primitve types. In this case, no boost, and it's a user defined class.

Ex:

template<typename T>
myfunc(T&)
{ ... }

template<>
myfunc(Foo&)
{
   static_assert(false, "You cannot use myfunc with Foo");
}

Problem is static_assert gets called regardless of whether I call myfunc with a Foo object or not.

I just want some way for compile to stop when myfunc is called with Foo.
How can I achieve this functionality?

Upvotes: 4

Views: 745

Answers (2)

Kerrek SB
Kerrek SB

Reputation: 476990

With a return type R, say:

#include <type_traits>

template <typename T>
typename std::enable_if<!std::is_same<T, Foo>::value, R>::type my_func(T &)
{
    // ...
}

If you really don't want to use the standard library, you can write the traits yourself:

template <bool, typename> struct enable_if { };
template <typename T> struct enable_if<true, T> { typedef T type; };

template <typename, typename> struct is_same { static const bool value = false; };
template <typename T> struct is_same<T, T> { static const bool value = true; };

Upvotes: 1

R. Martinho Fernandes
R. Martinho Fernandes

Reputation: 234424

You can use std::is_same for this:

#include <type_traits>

template<typename T>
return_type myfunc(T&)
{
   static_assert(std::is_same<T, Foo>::value, "You cannot use myfunc with Foo");

   // ...
}

Upvotes: 4

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