Passing a function template as a function argument in C++

Question

Suppose I have a template function that takes an integer as the template argument as follows:

template <int i> void f(int x) {...}

Now I want to write another function that takes the function f as an argument. However I do not know how to achieve this. For example, consider the following wrong code (that can not be compiled):

template <template <int> typename T> void g(T fun, int i, int x) {
    if (i == 0) fun<0>(x);
    else if (i == 1) fun<1>(x);
    //...
}

I have searched Google but It seems that all related questions are to pass a standard function as a template argument which is not the case here. Hope anyone can help me. Thank you very much!

You can't do that. Why not make the template argument a regular argument? — Aykhan Hagverdili, May 10 '21 at 14:07
You can't pass a template as a runtime argument. Templates are instantiated at compile-time. — molbdnilo, May 10 '21 at 14:09
Because the constant i can make the function f much faster. For example, suppose f calculates the ith power of x, and I will overload the function f when i=0, 1 or 2. — zbh2047, May 10 '21 at 14:10
a template is nothing you can pass around at runtime. Would it be fine if the template to be instantiated in `g` is determined at compile time but not passed as parameter? — 463035818_is_not_an_ai, May 10 '21 at 14:11
at some point you need to map the runtime value to the compile time template parameter. Maybe you made `f` faster, but now `g` needs to branch — 463035818_is_not_an_ai, May 10 '21 at 14:15
i was just a little confused because you have it as template parameter and function argument. If template argument is fine, you already have the answer — 463035818_is_not_an_ai, May 10 '21 at 14:16
I see. The branching in g is performed only once and f is in fact a complicated function. This is why I will make i a constant. — zbh2047, May 10 '21 at 14:29
Re: " takes the function `f`" -- `f` is not a function. It's a template. `f<3>` is a function. — Pete Becker, May 10 '21 at 15:46

Aykhan Hagverdili · Accepted Answer · 2021-05-10T14:18:24.667

You can emulate that with a callable class:

template <int i>
struct f {
  void operator()(int x) {
    // ...
  }
};

template <template <int> typename T>
void g(int i, int x) {
  if (i == 0)
    T<0>{}(x);
  else if (i == 1)
    T<1>{}(x);
  //...
}

int main() { 
  g<f>(10, 12); 
}

Could also do the same with a named static method. But I suggest you make the template argument a regular argument so you don't need stuff like this.

Passing a function template as a function argument in C++

1 Answers1