Re: [Boost-users] boost::result_of error?

From: Hicham Mouline (hicham_at_[hidden])
Date: 2008-05-01 08:32:21

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_at_[hidden]
  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_at_[hidden]>
  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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