Hicham,

it does not create a new new type it only declares a type.

Consider the following example:

template<int ()>
struct X
{};


This means your type X expects a function pointer type with return value int and no parameters. This can be used in a meta programming to make some type assumptions or inspection. This is how the result_of works. Think of result_of as of type X but a bit more complex and which can give an assumption about the function type passed. We can write:

template<int ()>
struct X
{
   typedef int        type;
};

A more generic form would be if int would be any valid result type T, so that we could write smth like this:

template<T ()>
struct X
{
   typedef T        type;
};

Actually this is not a legal C++ struct, since T was not declared as template parameter, but think of it as if it would.

template<T MyFunc()>
struct X
{
   typedef T             result_type;
   typedef MyFunc   function_type;
};

Here MyFunc is a name of the function pointer type, which allows us to reference this type from inside the template.


Hope that helps.


With Kinds regards,
Ovanes

On Thu, May 1, 2008 at 2:32 PM, Hicham Mouline <hicham@mouline.org> wrote:
But "floatfct" was defined earlier as the type of a pointer to a function with only 1 arg float.
just passing the template argument to result_of creates a new type?
is the new type also called floatfct ?

----- Original Message -----
From: "Ovanes Markarian"
To: boost-users@lists.boost.org
Subject: Re: [Boost-users] boost::result_of error?
Date: Thu, 1 May 2008 13:58:21 +0200

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,
_______________________________________________
Boost-users mailing list
Boost-users@lists.boost.org
http://lists.boost.org/mailman/listinfo.cgi/boost-users


_______________________________________________
Boost-users mailing list
Boost-users@lists.boost.org
http://lists.boost.org/mailman/listinfo.cgi/boost-users


_______________________________________________
Boost-users mailing list
Boost-users@lists.boost.org
http://lists.boost.org/mailman/listinfo.cgi/boost-users