Re: [Boost-users] about enable_if for member functions

From: Jaakko Järvi (jajarvi_at_[hidden])
Date: 2005-04-01 12:25:55

Hello Yutaka,

The problem is that T is not a template parameter of you function f1.
It is a template parameter of the class A. So when the class A
is instantiated with some type that is not arithmetic, say void*,
it is as if you would directly write a function:

enable_if<boost::is_arithmetic<void*>, void>::type

which is

enable_if<false, void>::type

which is an error, since enable_if<false,void> is an empty class.

You can write your code as follows to make enable_if work, but it may 
not be a solution that works:

class A
{
...
       template <class T>
       enable_if< boost::is_arithmetic<T>, void > :: type  f1()
       {
             ...
       }
};

Also, here you need to explicitly instantiate f1 as

A a;

a.f1<int>();

Best, Jaakko Järvi

On Apr 1, 2005, at 3:27 AM, Yutaka Leon Suematsu wrote:

> Dear booster,
>
> I have a question related to enable_if. I would like to enable or 
> disable
> some member functions from a template class, but enable_if is -not 
> enable-
> to do it :-)
>
> Here is the code.
>
> template <class T>
> class A
> {
> ...
>       enable_if< boost::is_arithmetic<T>, void > :: type  f1()
>       {
>             ...
>       }
> };
>
> In class A, I want to define function f1 only if the template 
> parameter is
> arithmetic. It works fine except when the parameter is not arithmetic. 
> I
> have a compile error "error C2039: 'type' : is not a member of
> 'boost::enable_if<Cond,T>'" (Working with VC7.1, and boost 3.12)
>
> Thank you in advance for any suggestions and help.
>
> Sincerely yours,
>
> Yutaka Leon Suematsu
>
>
>
> _______________________________________________
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> Boost-users_at_[hidden]
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>
-- Jaakko Järvi, Assistant Professor
-- Texas A&M University, Computer Science
-- jarvi_at_[hidden]