Re: [Boost-users] Can not compile a program that uses typeof package by gcc?

Subject: Re: [Boost-users] Can not compile a program that uses typeof package by gcc?
From: Peng Yu (pengyu.ut_at_[hidden])
Date: 2008-10-21 11:17:50

On Mon, Oct 20, 2008 at 10:19 PM, Steven Watanabe <watanabesj_at_[hidden]> wrote:
> AMDG
>
> Peng Yu wrote:
>>>
>>> In line 51-55 of you attached file, it says the following.
>>>
>>>  template <typename T1, typename T2>
>>>  Y<typename A::multiply_traits<T1, T2>::result_type>
>>>   operator*(const Y<T1> &y, const T2 &t) {
>>>     return Y<T1>(y.the_t() * t);
>>>   }
>>>
>>> This way requires that I have to explicitly say 'multiply_traits' is
>>> from the namespace A. I wanted to use the following instead, which
>>> does not need to specify which the namespaces T1 and T2 are. But the
>>> following code still results in the compiler error.
>>>
>>>  template <typename T1, typename T2>
>>>  Y<BOOST_TYPEOF(T1() * T2())>
>>>   operator*(const Y<T1> &y, const T2 &t) {
>>>     return Y<T1>(y.the_t() * t);
>>>   }
>>>
>>> My question is not resolved. I'm attaching the source file with some
>>> comments to avoid any confusion. Would you please take another look at
>>> the problem?
>>>
>
> Alright.  If you don't want to refer to multiply_traits from namespace
> A, then put a generic multiply_traits somewhere else.  When it is defined
> using Boost.Typeof, it will work for any types.  It does not matter what
> namespace multiply_traits is defined in.  Really.

This is a reasonable solution. But no matter which namespace
multiply_traits puts, it pollutes that namespace. Since namespace is a
way to better organized the code, I'd rather multiply_traits be in the
same namespace as that of its associated types.

Since Koenig lookup can be used to deduce the namespace of a function
based on whether its arugments are from, I'm wondering why C++ does
not allow such look up for templates?  If Koenig lookup can be used
for template, then the problem is solved.

Thanks,
Peng