I'm trying to create a delayable call object. Something along the lines of (pseudo-code):
template <class FN>
struct delayable_call
{
return-type-of-FN call(); // <-- I'd like to use result_of here.
template<class ArgTypes...>
delayable_call(FN* pFn, ArgTypes... args);
FN* fn;
args-saving-struct;
};
I tried using result_of::type for the return type of call, but get errors during instantiation of the template because apparently the argument types need to be specified separately.
Instantiation:
int foo(bool, double); // function prototype.
delayable_call<int(bool, double)> delayable_foo(foo, false, 3.14); // instantiation
The error messages and documentation I've read about result_of seem to indicate that the argument types must also be specified. So instead of result_of<FN>::type
, I'd need to specify result_of<FN(bool, double)>::type
. This does actually fix the compilation problem I'm having, but breaks the generality of the template.
So, how can I use result_of with a template parameter when the template parameter represents the function signature?
template <class FN> struct delayable_call;
template<class R, class...Args> delayable_call<R(Args...)>{
typedef R(*)(Args...) pFN;
replace your delayable_call
with a specialization, and you will extrace both R
and Args...
. You need Args...
anyhow to store the parameters.
However, a library-strength delayable call will end up using type erasure. The easiest way is a simple std::function<R()>
where you shove a lambda into it:
int foo(double);
double x = 7;
std::function<int()> delayed_foo = [x]{ return foo(x); }
and capture by value unless you really, really mean it to capture by reference.
You could deduce R
via:
template<typename Fn, typename... Args>
std::function< typename std::result_of<Fn(Args...)>::type()>
make_delayed_call( Fn&& fn, Args&&... args ) {
return [=]{ return fn(std::move(args)...); }
}
which should deduce your R
from the callable object and the arguments. This captures everything by copy -- capture by move requires either more boilerplate, or C++14.