Re: [Boost-users] [lambda] Variable assignment from lambda expression

27 Sep 2005

      On 9/27/05, Alexander Borghgraef  wrote:
...
On 9/26/05, jarvi  wrote:
...
the var function makes a lambda functor out of a variable, so you can
write the expression as:
var(n) = _1
Thanks, I should have seen that one in the manual. Posting at the end of
a work day is
 obviously not a good idea :-) I've got another lambda related question
though. My example was
 a useless simplification, what I'm really trying to do is to read data from
a container v into a legacy
 matrix class which doesn't have an iterator. I can do this with a for loop:
std::vector<char> v;
 MyMatrix<char> m;   // Has h and w members for height and width, and
element access using operator()( int i, int j )
 std::vector<char>::iterator v_it = v.begin();
 for ( long c = 0; c < m.h * m.w; ++c )
 {
       m( c / m.w , c % m.w ) =  *v_it;
       ++v_it;
  }
This works perfectly. I figured it should be possible to do the same thing
using for_each and boost::lambda, so I tried
 this approach:
long c = 0;
 for_each( v.begin(),
                v.end(),
                ( var( m( c / m.w, c % m.w ) ) = _1,
                  ++var( c )
                 )
               );
This also fills MyMatrix, but with different values than the for loop.
Which seems surprising to me.
 As far as I can tell, these two approaches should be equivalent, but
apparently they're not. Anyone
 care to explain to me why?
P.S. Sorry for the unfinished post which got through. Please ignore that
one.
--
Alex Borghgraef
That's not going to work because the operator() call in the loop (i.e.
m(...)) is evaluated when the lambda expression is created, not each
time the lambda expression is evaluated.

Try replacing

var( m( c / m.w, c % m.w ) )

with

bind(&MyMatrix::operator(), var(m), var(c) / m->*&MyMatrix::w, var(c)
% m->*&MyMatrix::w)

This makes all accesses to m & c lazy (I think), so should update m as you want.

Stuart Dootson