Question

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Extract the return type of a function without calling it (using templates?)

Starting with this (provided by somebody else):

int my_function(int, int *, double);

I want to get to 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 do I get my_function_type?

NOTE: I know about BOOST_TYPEOF(), but it seems a bit scary, as in “perhaps not totally portable”?

Answer 1

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 to decltype is evaluated at compile time and thus must be resolvable. You could pass any constexpr integer to it.