is there a way to store a generic templated function pointer? - c++

Question

The Goal:
decide during runtime which templated function to use and then use it later without needing the type information.
A Partial Solution:
for functions where the parameter itself is not templated we can do:
int (*func_ptr)(void*) = &my_templated_func<type_a,type_b>;
this line of code can be modified for use in an if statement with different types for type_a and type_b thus giving us a templated function whose types are determined during runtime:
int (*func_ptr)(void*) = NULL;
if (/* case 1*/)
func_ptr = &my_templated_func<int, float>;
else
func_ptr = &my_templated_func<float, float>;
The Remaining Problem:
How do I do this when the parameter is a templated pointer?
for example, this is something along the lines of what I would like to do:
int (*func_ptr)(templated_struct<type_a,type_b>*); // This won't work cause I don't know type_a or type_b yet
if (/* case 1 */) {
func_ptr = &my_templated_func<int,float>;
arg = calloc(sizeof(templated_struct<int,float>, 1);
}
else {
func_ptr = &my_templated_func<float,float>;
arg = calloc(sizeof(templated_struct<float,float>, 1);
}
func_ptr(arg);
except I would like type_a, and type_b to be determined during runtime. I see to parts to the problem.
What is the function pointers type?
How do I call this function?
I think I have the answer for (2): simply cast the parameter to void* and the template function should do an implicit cast using the function definition (lease correct me if this won't work as I think it will).
(1) is where I am getting stuck since the function pointer must include the parameter types. This is different from the partial solution because for the function pointer definition we were able to "ignore" the template aspect of the function since all we really need is the address of the function.
Alternatively there might be a much better way to accomplish my goal and if so I am all ears.
Thanks to the answer by #Jeffrey I was able to come up with this short example of what I am trying to accomplish:
template <typename A, typename B>
struct args_st {
A argA;
B argB;
}
template<typename A, typename B>
void f(struct args_st<A,B> *args) {}
template<typename A, typename B>
void g(struct args_st<A,B> *args) {}
int someFunction() {
void *args;
// someType needs to know that an args_st struct is going to be passed
// in but doesn't need to know the type of A or B those are compiled
// into the function and with this code, A and B are guaranteed to match
// between the function and argument.
someType func_ptr;
if (/* some runtime condition */) {
args = calloc(sizeof(struct args_st<int,float>), 1);
f((struct args_st<int,float> *) args); // this works
func_ptr = &g<int,float>; // func_ptr should know that it takes an argument of struct args_st<int,float>
}
else {
args = calloc(sizeof(struct args_st<float,float>), 1);
f((struct args_st<float,float> *) args); // this also works
func_ptr = &g<float,float>; // func_ptr should know that it takes an argument of struct args_st<float,float>
}
/* other code that does stuff with args */
// note that I could do another if statement here to decide which
// version of g to use (like I did for f) I am just trying to figure out
// a way to avoid that because the if statement could have a lot of
// different cases similarly I would like to be able to just write one
// line of code that calls f because that could eliminate many lines of
// (sort of) duplicate code
func_ptr(args);
return 0; // Arbitrary value
}

Answer 1

Can't you use a std::function, and use lambdas to capture everything you need? It doesn't appear that your functions take parameters, so this would work.
ie
std::function<void()> callIt;
if(/*case 1*/)
{
callIt = [](){ myTemplatedFunction<int, int>(); }
}
else
{
callIt = []() {myTemplatedFunction<float, float>(); }
}
callIt();

Answer 2

If I understand correctly, What you want to do boils down to:
template<typename T>
void f(T)
{
}
int somewhere()
{
someType func_ptr;
int arg = 0;
if (/* something known at runtime */)
{
func_ptr = &f<float>;
}
else
{
func_ptr = &f<int>;
}
func_ptr(arg);
}
You cannot do that in C++. C++ is statically typed, the template types are all resolved at compile time. If a construct allowed you to do this, the compiler could not know which templates must be instanciated with which types.
The alternatives are:
inheritance for runtime polymorphism
C-style void* everywhere if you want to deal yourself with the underlying types
Edit:
Reading the edited question:
func_ptr should know that it takes an argument of struct args_st<float,float>
func_ptr should know that it takes an argument of struct args_st<int,float>
Those are incompatible. The way this is done in C++ is by typing func_ptr accordingly to the types it takes. It cannot be both/all/any.
If there existed a type for func_ptr so that it could take arguments of arbitrary types, then you could pass it around between functions and compilation units and your language would suddenly not be statically typed. You'd end up with Python ;-p

Answer 3

Maybe you want something like this:
#include <iostream>
template <typename T>
void foo(const T& t) {
std::cout << "foo";
}
template <typename T>
void bar(const T& t) {
std::cout << "bar";
}
template <typename T>
using f_ptr = void (*)(const T&);
int main() {
f_ptr<int> a = &bar<int>;
f_ptr<double> b = &foo<double>;
a(1);
b(4.2);
}
Functions taking different parameters are of different type, hence you cannot have a f_ptr<int> point to bar<double>. Otherwise, functions you get from instantiating a function template can be stored in function pointers just like other functions, eg you can have a f_ptr<int> holding either &foo<int> or &bar<int>.

Answer 4

Disclaimer: I have already provided an answer that directly addresses the question. In this answer, I would like to side-step the question and render it moot.
As a rule of thumb, the following code structure is an inferior design in most procedural languages (not just C++).
if ( conditionA ) {
// Do task 1A
}
else {
// Do task 1B
}
// Do common tasks
if ( conditionA ) {
// Do task 2A
}
else {
// Do task 2B
}
You seem to have recognized the drawbacks in this design, as you are trying to eliminate the need for a second if-else in someFunction(). However, your solution is not as clean as it could be.
It is usually better (for code readability and maintainability) to move the common tasks to a separate function, rather than trying to do everything in one function. This gives a code structure more like the following, where the common tasks have been moved to the function foo().
if ( conditionA ) {
// Do task 1A
foo( /* arguments might be needed */ );
// Do task 2A
}
else {
// Do task 1B
foo( /* arguments might be needed */ );
// Do task 2B
}
As a demonstration of the utility of this rule of thumb, let's apply it to someFunction(). ... and eliminate the need for dynamic memory allocation ... and a bit of cleanup ... unfortunately, addressing that nasty void* is out-of-scope ... I'll leave it up to the reader to evaluate the end result. The one feature I will point out is that there is no longer a reason to consider storing a "generic templated function pointer", rendering the asked question moot.
// Ideally, the parameter's type would not be `void*`.
// I leave that for a future refinement.
void foo(void * args) {
/* other code that does stuff with args */
}
int someFunction(bool condition) {
if (/* some runtime condition */) {
args_st<int,float> args;
foo(&args);
f(&args); // Next step: pass by reference instead of passing a pointer
}
else {
args_st<float,float> args;
foo(&args);
f(&args); // Next step: pass by reference instead of passing a pointer
}
return 0;
}

Answer 5

Your choice of manual memory management and over-use of the keyword struct suggests you come from a C background and have not yet really converted to C++ programming. As a result, there are many areas for improvement, and you might find that your current approach should be tossed. However, that is a future step. There is a learning process involved, and incremental improvements to your current code is one way to get there.
First, I'd like to get rid of the C-style memory management. Most of the time, using calloc in C++ code is wrong. Let's replace the raw pointer with a smart pointer. A shared_ptr looks like it will help the process along.
// Instead of a raw pointer to void, use a smart pointer to void.
std::shared_ptr<void> args;
// Use C++ memory management, not calloc.
args = std::make_shared<args_st<int,float>>();
// or
args = std::make_shared<args_st<float,float>>();
This is still not great, as it still uses a pointer to void, which is rarely needed in C++ code unless interfacing with a library written in C. It is, though, an improvement. One side effect of using a pointer to void is the need for casts to get back to the original type. This should be avoided. I can address this in your code by defining correctly-typed variables inside the if statement. The args variable will still be used to hold your pointer once the correctly-typed variables go out of scope.
More improvements along this vein can come later.
The key improvement I would make is to use the functional std::function instead of a function pointer. A std::function is a generalization of a function pointer, able to do more albeit with more overhead. The overhead is warranted here in the interest of robust code.
An advantage of std::function is that the parameter to g() does not need to be known by the code that invokes the std::function. The old style of doing this was std::bind, but lambdas provide a more readable approach. Not only do you not have to worry about the type of args when it comes time to call your function, you don't even need to worry about args.
int someFunction() {
// Use a smart pointer so you do not have to worry about releasing the memory.
std::shared_ptr<void> args;
// Use a functional as a more convenient alternative to a function pointer.
// Note the lack of parameters (nothing inside the parentheses).
std::function<void()> func;
if ( /* some runtime condition */ ) {
// Start with a pointer to something other than void.
auto real_args = std::make_shared<args_st<int,float>>();
// An immediate function call:
f(real_args.get());
// Choosing a function to be called later:
// Note that this captures a pointer to the data, not a copy of the data.
// Hence changes to the data will be reflected when this is invoked.
func = [real_args]() { g(real_args.get()); };
// It's only here, as real_args is about to go out of scope, where
// we lose the type information.
args = real_args;
}
else {
// Similar to the above, so I'll reduce the commentary.
auto real_args = std::make_shared<args_st<float,float>>();
func = [real_args]() { g(real_args.get()); };
args = real_args;
}
/* other code that does stuff with args */
/* This code is probably poor C++ style, but that can be addressed later. */
// Invoke the function.
func();
return 0;
}
Your next step probably should be to do some reading on these features so you understand what this code does. Then you should be in a better position to leverage the power of C++.