templates - In C++, how can I get an arbitrary function's type from its declaration? -

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possible duplicate:
extract return type of function without calling (using templates?)

starting (provided else):

int my_function(int, int *, double);

i want this:

typedef boost::function_types::result_type< my_function_type >::type my_result; typedef boost::function_types::parameter_types< my_function_type >::type my_parameters;

how my_function_type?

note: know boost_typeof(), seems bit scary, in "perhaps not totally portable"?

decltype. examples:

char foo(int) {} decltype (foo(3)) const *frob = "hello foo"; typedef decltype (foo(3)) typeof_foo; using typeof_foo = decltype(foo(3));

the expression decltype evaluated @ compile time , must resolvable. pass constexpr integer it.


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