Is there a way to pass a function template into a function?

Question

I have some template code that takes a function as an argument. I'd like to extend it to use a function template as an argument.

The existing code is:

int add1( int i ) { return i+1; }

template <typename Func, typename Integer>
Integer call_func( Func f, Integer i ) { return f(i); }

// Usage is
auto four = call_func( add1, 3 );

It works fine. If I add a version of add1 that takes a template parameter, I get:

template <typename Integer>
Integer add1_t( Integer i ) { return i+1; }

// Usage is
auto four = call_func( add1_t<int>, 3 );

What I would really like to do, however, is to not specify the full type of the function, only the template, as in:

auto four = call_func( add1_t, 3 );

and let call_func apply the template argument. Is there a way to do that?

My best guess is:

template <template <typename> typename Func, typename Integer>
Integer call_func( Func<Integer> f, Integer i ) { return f(i); }

but the type of f cannot be deduced when I do that. Even if I wanted to supply an explicit "type" at the call-site, I'm not sure what that would be.

Does this answer your question? [What are some uses of template template parameters?](https://stackoverflow.com/questions/213761/what-are-some-uses-of-template-template-parameters) — Nelfeal, Sep 23 '22 at 18:59
Afaik this cannot be done exactly this way, but you could easily write code that allows for the exact same syntax: `struct Func { template constexpr T operator()(T i) const { return i + 1;}}; constexpr Func add1; template constexpr decltype(auto) call_func(T&& t, U&& u) { return std::forward(t)(std::forward(u)); }; auto four = call_func(add1, 3);` — fabian, Sep 23 '22 at 19:06
@Nelfeal, it doesn't really answer the question directly. The examples in that article involve template classes, not template functions. The problem I'm getting is that Func is an unresolved overloaded function, which isn't a type. The other comment is essentially to wrap the function in a struct. I can certainly go that way - I was just hoping to avoid the syntax of wrapping functions in structs. — user1806566, Sep 23 '22 at 20:25

score 0 · Accepted Answer · answered Sep 23 '22 at 21:14

From your comment:

I was just hoping to avoid the syntax of wrapping functions in structs

Making functors (basically wrapping functions in classes) is a pretty common solution to things that involve passing functions around.

Here is one way to do that in your case:

template <typename Integer>
struct add1_t {
    Integer operator()( Integer i ) { return i+1; }
};

template <template <typename> class F, typename Integer>
Integer call_func( Integer i ) { return F<Integer>{}(i); }

int main() {
    auto four = call_func<add1_t>( 3 );
}

Demo

score 0 · Answer 2 · answered Sep 24 '22 at 00:01

You cannot pass template function. but you can pass functor with template operator().

Lambda does that job easily, keep your original call_func, and change call to:

auto four = call_func([](auto arg){ return add1_t(arg)], 3 );

MACRO can also help:

#define OVERLOAD_SET(FUNC_NAME) [](auto&&... args){ return FUNC_NAME(decltype(args)(args)...); }
// Can be improved with noexcept, and SFINAE friendly

auto four = call_func(OVERLOAD_SET(add1_t), 3 );

Is there a way to pass a function template into a function?

2 Answers2