Re: [Boost-users] enable_if question

From: Howard Hinnant (hinnant_at_[hidden])
Date: 2004-06-12 17:24:25

On Jun 12, 2004, at 5:20 PM, Agoston Bejo wrote:

> Hi,
> I understand that by using enable_if the following way:
>
> template <class T>
> T foo(T t, typename enable_if<boost::is_arithmetic<T> >::type* dummy = 
> 0);
>
> I can include/exclude the appropriate function overloads.
> When I write:
>
> template<class T>
> T foo(T t, typename enable_if<[condition1]>::type* dummy = 0,
> typename enable_if<[condition2]>::type* dummy = 0);
>
> Does this mean an AND-relationship between condition1 and condition2? 
> (So
> that a particular overload is enabled if and only if both conditions
> evaluate to true?)

Yes, it does mean "and".  But it is a cryptic way to do it.  Instead 
consider:

template<class T>
T foo(T t, typename enable_if<condition1 && condition2>::type* dummy = 
0);

Also, no need for the dummy name:

template<class T>
T foo(T t, typename enable_if<condition1 && condition2>::type* = 0);

And you might prefer to place the condition on the return type for 
readability purposes:

template <class T>
typename enable_if
<
     condition1 && condition2,
     T
 >::type
foo(T t);

-Howard