need explanation on boost::mpl

Question

typedef vector<long,float,short,double,float,long,long double> types;
typedef fold<
      types
    , int_<0>
    , if_< is_float<_2>,next<_1>,_1 >
    >::type number_of_floats;

BOOST_MPL_ASSERT_RELATION( number_of_floats::value, ==, 4 );

I don't understand the way fold works, why int_<0>? why is_float<_2> ?

Could someone give me some clues to understand this "function" ? thanks

pmr · Accepted Answer · 2011-04-23T12:42:33.573

3

int_<0> is the starting value of the accumulator used for the fold. Try to use int_<1> and see what happens.

The third argument is the operator used to fold the sequence. This needs to be a binary metafunction. if_< is_float<_2>,next<_1>,_1 > is turned into a lambda expression with two arguments where _1 and _2 refer to the first and second argument this lambda expression takes.

The predicate is_float<_2> returns true if the second argument to if_ is a float. _2 is a placeholder. Placeholders refer to the n-th argument of the template specialization.

next<_1> simply returns the next value of the current state (e.g. next<int_<0>> == int_<1>).

If the predicate returns false we simply return _1 which is the unaltered state.

Try to understand what a fold is first, then try to understand the boost::mpl way of doing it.

A simple exercise is to write a fold that return the length of the vector.

edited Apr 23 '11 at 12:42

answered Apr 23 '11 at 11:05

pmr

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just a precision about placeholer _2 refer to the n-th arguement of if_ or of fold ?and I think it is of fold.... – Guillaume Paris Apr 23 '11 at 11:20
@guillaume07: No, it refers to the 2nd argument of `if_`. – pmr Apr 23 '11 at 11:32
and _1 refers to the first argument of for ?? – Guillaume Paris Apr 23 '11 at 11:53
1

No, placeholders turn something in a lambda expression. Thus `if_` takes further arguments. I try to clarify that. – pmr Apr 23 '11 at 12:38

need explanation on boost::mpl

1 Answers1