Re: [Boost-users] boost::result_of error?
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
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
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AMDG Ovanes Markarian wrote:
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; };
<snip>
Here MyFunc is a name of the function pointer type, which allows us to reference this type from inside the template.
Hope that helps.
This has very little to do with result_of. The expression result_of<floatfct(float, float)>::type is the return type of calling an object of type floatfct with two float arguments. floatfct(float, float) is a function type that takes two floats and returns a floatfct. result_of is an ordinary class template, that uses metaprogramming to decompose the function type and treats the return type as the function that will be called. template<class T> struct result_of; template<class F> struct result_of<F()> {}; template<class F, class Arg1> struct result_of<F(Arg1)> {}; ... In Christ, Steven Watanabe
Ovanes Markarian <om_boost <at> keywallet.com> writes: exactly. thanks for the hint. a.
participants (4)
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Steven Watanabe