C++ template deduction from lambda
C++ template deduction from lambda
I have a function which takes two std::function
s as arguments. The parameter of the second function has the same type as the result of the first.
std::function
I wrote a function template like this:
template<typename ResultType>
void examplFunction(std::function<ResultType()> func, std::function<void(ResultType)> func2)
auto x = func();
func2(x);
I can call it with:
void f()
examplFunction<int>(() return 1; , //
(int v) std::cout << "result is " << v << std::endl; );
Is there a way to to get rid of the <int>
at examplFunction<int>
and let the compiler deduce the type of ResultType
?
<int>
examplFunction<int>
ResultType
std::function
std::function
4 Answers
4
Do you actually need std::function
in there? std::function
is useful when you need type erasure. With templates, you can usually skip it altogether:
std::function
std::function
template<class F1, class F2>
void examplFunction(F1 func, F2 func2, decltype(func2(func()))* sfinae = nullptr)
auto x = func();
func2(x);
The sfinae
parameter makes sure the function can only be called with functions such that func2
can be called with the result of func
.
sfinae
func2
func
Note that your SFINAE is not actually corresponding to the implementation. If
func
returned a move-only type (say, std::unique_ptr
) then the SFINAE would allow it but the function body wouldn't compile. Example.– Max Langhof
Sep 7 '18 at 15:27
func
std::unique_ptr
@MaxLanghof Yes, but the exact same issue applies to the OP's original code. The SFINAE is only there to guarantee what the OP asked for.
– Angew
Sep 7 '18 at 15:29
True I guess. I'm left to wonder if this can somehow be broken with implicit conversions (and an overloaded
func2
)... But I don't see a way.– Max Langhof
Sep 7 '18 at 15:31
func2
Hmm it seems to do what it should. But the
decltype(func2(func()))* sfinae = nullptr
looks very magical to me. How can a third function argument restrict the first and second?– ErWu
Sep 7 '18 at 16:19
decltype(func2(func()))* sfinae = nullptr
@ErWu en.wikipedia.org/wiki/SFINAE Basically, if
func2(func())
would be ill-formed, it prevents the template from being instantiated.– Angew
Sep 7 '18 at 16:34
func2(func())
Yes, there is.
template<typename ResultType>
void examplFunction_impl(std::function<ResultType()> func, std::function<void(ResultType)> func2)
auto x = func();
func2(x);
template<class F1, class F2>
void examplFunction(F1&& f1, F2&& f2)
using ResultType = decltype(f1());
examplFunction_impl<ResultType>(std::forward<F1>(f1), std::forward<F2>(f2));
Demo
In this case you require that f1
be invocable with no arguments, so you can figure out the return type in the helper function. Then you call the real function while explicitly specifying that return type.
f1
You could add some SFINAE to make sure this function only participates in overload resolution when f1
can indeed be invoked like that (and if f2
can also be invoked with the return value of f1
).
f1
f2
f1
Although I have to agree with @Angew that in the given example there is no need for std::function
. That might of course be different in a real-world situation.
std::function
Also nice, thank you.
– ErWu
Sep 7 '18 at 16:29
std::function
has a templated (and otherwise unconstrained) constructor, so deducing it from simply an argument type is not that easy a deal. If those arguments still need to be std::function
s, you can skip one <int>
for the price of two std::function
s, and let deduction guides do the rest:
std::function
std::function
<int>
std::function
void f()
examplFunction(std::function(() return 1; ), //
std::function((int v) std::cout << "result is "
<< v << std::endl; ));
A fun fact is that this does not always work. For instance, the current implementation of libc++ lacks guides for std::function
, thus violating the standard.
std::function
Angew's answer is great (and should be the accepted answer), but is missing the minor detail of checking that the section function doesn't return anything. To do that you'll need to use the std::is_void
type trait, and std::enable_if:
std::is_void
template<class F1, class F2>
void examplFunction(F1 func, F2 func2, std::enable_if_t<std::is_void_v<decltype(func2(func()))>, void*> sfinae = nullptr)
auto x = func();
func2(x);
This obvious is more verbose and difficult to read if you aren't familiar with type traits and SFINAE, so it probably isn't the best way forward if you don't need to make sure F2 returns void
.
void
The OP's original code doesn't require that
func2
return void
. std::function<void(ResultType)>
is constructible-from a function that returns something; it'll just cast the return value to void
, which is to say, it'll ignore it. std::function<void()>( () return 42; )
compiles just fine.– Quuxplusone
Sep 8 '18 at 5:50
func2
void
std::function<void(ResultType)>
void
std::function<void()>( () return 42; )
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Lambdas are not
std::function
s.std::function
s are type erasure tools. Type deduction and type erasure are opposites of each other. If you are deducing the type of a type erasure tool you are almost always screwing up.– Yakk - Adam Nevraumont
Sep 7 '18 at 15:38