Terje Slettebų <[EMAIL PROTECTED]> writes:
>>From: "David Abrahams" <[EMAIL PROTECTED]>
>
>> "Jaap Suter" <[EMAIL PROTECTED]> writes:
>>
>> > Mm, I still don't quite understand. Consider the following function:
>> >
>> > void foo( int_c< 0 > );
>> >
>> > Shouldn't the following code:
>> >
>> > foo( int_c< minus< int_c< 4 >, int_c< 4 > >::type > );
>> >
>> > Behave the same as this:
>> >
>> > foo( minus< int_c< 4 >, int_c< 4 > >::type );
>>
>> We'd like it to. However, as Aleksey said, until we get typedef
>> templates, there's no way to make int_c<0> into the same type as
>> integral_c<int,0>. The result of
>>
>> minus< int_c< 4 >, int_c< 4 > >::type
>> is
>> integral_c<int,0>
>
> You could get the above code to compile, if they were related by
> inheritance, and you use "void foo( integral_c<0> )", instead. For example:
>
> template<int N>
> struct int_c : integral_c<int, N> {};
or
template <class T, T N>
struct integral_c
: select_int_base<T,N>::type
{};
To let Jaap keep using his convenient int_c arguments
--
David Abrahams
[EMAIL PROTECTED] * http://www.boost-consulting.com
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