Re: [Boost-users] C++ guru required!

Subject: Re: [Boost-users] C++ guru required!
From: Daryle Walker (darylew_at_[hidden])
Date: 2012-02-18 11:58:46

From: ramey_at_rrsd.comDate: Fri, 17 Feb 2012 12:40:12 -0800
I've got a really dumb question.  My question is illustrated by the followingcode snippet. template<class T, class U>T inline operator%(const T & lhs, const U & rhs) {    if(0 == rhs)        throw std::domain_error("Divide by zero");    if(boost::is_same<T, boost::uintmax_t>::value     && boost::numeric::is_signed<U>::value     ){        if(1 == rhs || -1 == rhs)  // is this dropped?            overflow("unsigned type can hold this result");    }    return lhs % rhs;}
I would like to think that the second if is alwaysdropped from the compile for a particular pairof types since this can be evaluated at compile time.
If I'm correct, I can replace some tedioustemplate metaprogramming for some straightforwardand transparent code inserted in a convenient codeinserted in the most convenient place.
Andre Alex... gave a talk at "Going Native" proposinga "static if" for this case.  But I don't see the necessity forfor this since I would assume that the compilerjust optimises away the "dead" code.  I've compiledthe above and it seems to do what I want butstill I wonder.
Basically I see lots of applications of variationson this idea to get the benefits of tmp withoutthe attendent pain.
Am I missing anything here?
If you want to (partially) specialize on types, then use a class template that can act as a function object (i.e. has an operator ()).  I was going to leave at that, but I see some other problems.
Unless you use some sort of name-spacing (and even then), you just wrote a universal operator.  Every type out there gets a modulus operation, whether they like it or not.  Even types that already have one get a new one for type combinations they didn't anticipate.  Never do this.The function ends by passing the real work to a modulus operator.  Since the function itself is a universal modulus operator, you get infinite recursion!
So rewrite this as a function object class template, and specialize for certain combinations of built-in integer types.
Daryle W.