Re: [boost] [type_traits?] enable_if an overload for a function exists

From: Marco (mrcekets_at_[hidden])
Date: 2007-04-19 19:00:57

On Thu, 19 Apr 2007 23:10:31 +0200, Michael Marcin  
<mmarcin_at_[hidden]> wrote:

> I want to enable a function templated on a type T if a function it  
> relies on overloaded on that type.
>
.....

>
>
>     template< typename T > inline
>     typename detail::enable_if_overloaded< T (T), std::asin, radians<T>  
> >::type
>     my_asin( T x )
>     {
>         return radians<T>( std::asin( x ) );
>     }
>

I tried your code on gcc 4.1.1 and compiler crashes, too.
I found that if the function symbol passed as argument for Signature*  
template parameter isn't overloaded, that is there is only one signature  
bounded to the function name, your code works correctly.
IMO the problem is : if a function is overloaded what is its type ?
example:

float std::asin(float )
long double std::asin(long double )

tipeof( std::asin ) is "float (float )" or "long double (long double )" ?

I don't know what the standard says about.

Anyway, I don't understand why vc and gcc crash.

Marco

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