Hicham,

as far as I understand your code you pass to result_of a new type:
typedef typename boost::result_of<floatfct(float, float)>::type resultype;

This type is a pointer to a function type, which has as return type a pointer to floatfct and as params float, float.
That's why it compiles.


Regards,
Ovanes


On Thu, May 1, 2008 at 1:25 PM, Hicham Mouline <hicham@mouline.org> wrote:
Hello,
trying out Pete Becker's "c++ std lib ext" exercises,
ex1 p155

#include <iostream>
#include <typeinfo>
#include <boost/utility/result_of.hpp>

typedef float (*floatfct)(float);
int main(int argc, char* argv[])
{
 typedef typename boost::result_of<floatfct(float, float)>::type resultype;
 std::cout<< typeid(resultype).name() << std::endl;
}

should fail, because result_of is instantiated with a callable type with 2 float args,
while it's been defined as taking 1 float arg only?

with intel10.1-MSVC8-boost1.35, it links.

rds,
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