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Subject: Re: [boost] review request: addition to type_traits library ofhas_operator_xxx
From: John Maddock (boost.regex_at_[hidden])
Date: 2010-12-10 04:04:30

>1. is there an easy way to get the type of an expression and to pass
>it to is_convertible

Not in C++03, only in C++0x

>2. if not, why not just using the working code I wrote already?

That sounds sensible - I may have misunderstood here - I thought from the
comments I saw that you had written your own is_convertible version, but
rather it looks like you have some implementation details (functions) that
just happen to have a similar name? If so I suggest you leave it as it is -
sorry for the confusion on that one!


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