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

1 May 2008

      Thanks for your clarifications, In your example, if "MyFunc" was
typedef'd before its use in the template argument as say typedef  int
(*MyFunc)(int);
it would not conflict with its use in the template argument ? best
regards,

  ----- Original Message -----
  From: "Ovanes Markarian"
  To: boost-users@lists.boost.org
  Subject: Re: [Boost-users] boost::result_of error?
  Date: Thu, 1 May 2008 17:13:40 +0200

  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,
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