Re: [Boost-users] [mpl] C++ Template Metaprogramming book - Question

From: David Abrahams (dave_at_[hidden])
Date: 2006-05-24 10:58:43

"Kobi Cohen-Arazi" <kobi.cohenarazi_at_[hidden]> writes:

> Hi,
>
> I've started reading Abrahams' and Gurtovoy's book, since I'm very
> interested in boost.mpl. Seems like a great book. I have a question
> regarding the 3rd chapter (3.1.4 section).
> Eventually, it says that operator*  multiplies the runtime values
> (resulting in 6.0f) and the code uses transform to sum the meta-seq.
>
> template <class T, class D1, class D2>
> quantity<T, typename mpl::transform<D1,D2, plus_f>::type>
> operator*(quantity<T,D1> x, quantity<T,D2> y)
> {
>    typedef  mpl::transform<D1,D2, plus_f>::type dim;
>    return quantity<T,dim>(x.value() * y.value());
> }
>
> 1. Why passing x and y by value? why not const&

a. It adds extra tokens that distract from the point of the example.

b. T is probably float or double, which is usually fine to pass by
   value, and thus quantity<T,...> probably has the same size and is
   also fine to pass by value

> 2. How did we got 6f ?

It's an erratum:
http://boost-consulting.com/mplbook/errata.html#metafunctions-title

> 3. How and when the mpl::transform is getting into action here? I
> cannot spot it.

It gets into action at compile-time, when the operator is
instantiated.  The compiler has to figure out what type the function
returns, and to do so, it instantiates the transform template to
discover what its nested ::type is.

HTH,

-- 
Dave Abrahams
Boost Consulting
www.boost-consulting.com