Re: Should/can I use GitHub to fix issue #7902 Division by scalar should use enable_if<>

Joaquim Duran <[email protected]>
Newsgroups gmane.comp.lib.boost.ublas
Message-ID <CANdawSHQ+vjTo0CE0We9=0E++YPZNk=0S_gcZb5j1UziOgAQ-A@mail.gmail.com>
I wrote the function. I executed the tests with g++ 4.8 (Fedora 20)
and I didn't detect any issue.

As work around, you should rename the function equal_to.

To fix the issue, is it really needed have to sets of functions to
compare vectors and matrices? Maybe, we could have just a jest of
functions named 'compare'.

Thanks and Best Regards,
Joaquim Duran


2014-12-09 20:40 GMT+01:00 Sean Reilly <[email protected]>:
> Still having a problem getting the tests to running on CentOS 6.6 with gcc
> 4.4.7 compiler.
>
> /usr/lib/gcc/x86_64-redhat-linux/4.4.7/../../../../include/c++/4.4.7/bits/stl_function.h:199:
> error: ‘template<class _Tp> struct std::equal_to’ is not a function,
> common/testhelper.hpp:126: error:   conflict with ‘template<class M1, class
> M2> bool equal_to(const boost::numeric::ublas::vector_expression<E>&, const
> boost::numeric::ublas::vector_expression<E2>&, double)’
> test_assignment.cpp:30: error:   in call to ‘equal_to’
> test_assignment.cpp:601:   instantiated from here
>
>
> Basically, there seems to be a name conflict between the equal_to() template
> function defined in testhelper.h
>
> template < class M1, class M2 >
> bool equal_to( const boost::numeric::ublas::vector_expression<M1> & m1,
>                const boost::numeric::ublas::vector_expression<M2> & m2,
>                double tolerance = 0.0 ) {
>     if (m1().size() != m2().size()) {
>         return false;
>     }
>
>     return mean_square(m2() - m1()) <= tolerance;
> }
>
>
> And the equal_to() template function defined in stl_function.h
>
>   template<typename _Tp>
>     struct equal_to : public binary_function<_Tp, _Tp, bool>
>     {
>       bool
>       operator()(const _Tp& __x, const _Tp& __y) const
>       { return __x == __y; }
>     };
>
>
> Anybody else having this issue?
>
> Sean Reilly
>
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