On 4/22/2011 8:47 PM, Karsten Ahnert wrote:

Hi,

I have a small functor:

struct operation { template<class> struct result;

template< class F> struct result<F(int)> { typedef long type; };

template< class F> struct result<F(float)> { typedef double type; };

template< class T> typename result< operation(T)>::type operator()( T t ) { typedef typename result< operation(T)>::type result_type; return result_type( t * T( 2 ) ); } };

and I wonder why a transform_view works as expected:

typedef fusion::vector< int , float> vector_type; vector_type v( 1 , 3.4 ); fusion::transform_view< vector_type , operation> tv( v , operation() );

while

result_of::transform does not:

fusion::result_of::transform< vector_type , operation>::type tv2 = fusion::transform( v , operation() );

result in compile errors? Any hints?

Try: fusion::result_of::transform< vector_type const, operation >::type tv2 = fusion::transform( v , operation() ); Notice the const. As of now, Fusion many algos accept only const sequences. There's a rationale why: Fusion algos are basically non-mutating and returns views. Anyway, I intend to relax this restriction and accept non-const sequences soon. Regards, -- Joel de Guzman http://www.boostpro.com http://boost-spirit.com