"Kobi Cohen-Arazi" 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 quantity::type> operator*(quantity x, quantity y) { typedef mpl::transform::type dim; return quantity(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 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