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::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
wrote:
Hello,
trying out Pete Becker's "c++ std lib ext" exercises,
ex1 p155
#include <iostream>
#include <typeinfo>
#include
typedef float (*floatfct)(float);
int main(int argc, char* argv[])
{
typedef typename boost::result_of::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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