Re: [boost] [static_if] Is there interest in a `static if` emulation library?

Subject: Re: [boost] [static_if] Is there interest in a `static if` emulation library?
From: pfultz2 (pfultz2_at_[hidden])
Date: 2014-08-31 23:35:28

> Another option is to always pass a dummy arg, and required the lambda to
> accept a param. e.g. `[](auto){....}`, so the verbosity of std::bind can
> be
> avoided.

Its not enough for it to be lazy, since accessing non-dependent types still
have to compile. That is why Lorenzo passes them in as template parameters.
Another way accomplishing this is to pass in an identity function:

    namespace aux {
    struct identity
    {
        template<class T>
        T operator()(T&& x) const
        {
            return std::forward<T>(x);
        }
    };

    template< bool Condition >
    struct static_if_statement 
    {
        template< typename F > void then ( F const& f ) { f(identity()); }
        template< typename F > void else_ ( F const& ) { }
    };

    template< >
    struct static_if_statement<false> 
    {
        template< typename F > void then ( F const& ) { }
        template< typename F > void else_ ( F const& f ) { f(identity()); }
    };
    }

    template< bool Condition, typename F >
    aux::static_if_statement<Condition> static_if ( F const& f ) 
    {
        aux::static_if_statement<Condition> if_;
        if_.then(f);
        return if_;
    }

Then you can call it like this:

    template< typename T >
    void assign ( T& x, T const& y ) 
    {
        x = y;

        static_if<boost::has_equal_to&lt;T>::value>([](auto f) 
        {
            assert(f(x) == f(y));
            std::cout << "asserted for: " << typeid(T).name() << std::endl;
        })
        .else_([] (auto) 
        {
            std::cout << "cannot assert for: " << typeid(T).name() <<
std::endl;
        });
    } 

The identity function will make non-dependent types dependent, so this will
compile.

Paul

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