Accept any function (but only functions) as a parameter

Question

Is there a way to accept any function signature as a parameter? For example:

void accepter(void (*func)(int)) {  ... }

would accept anything of type void (*)(int). I know I could use a template, such as:

template<class T>
void accepter(T func) { ... }

which would work, but then it allows non-functions to be passed in. How do I filter only for functions in the same way that a void * filters for any type of pointer?

If you try to call it, you will get a compile-time error. What are you doing that you are getting a link-time error? The only way I see would be if you're explicitly instantiating the template only for function types, but you don't seem to be doing that... — Dark Falcon, Jul 02 '14 at 15:37
Use C++11 std::function or boost::function, e.g. http://stackoverflow.com/questions/2304203/how-to-use-boost-bind-with-a-member-function — Martin Meeser, Jul 02 '14 at 15:47

score 2 · Accepted Answer · answered Jul 02 '14 at 15:44

2

Using C++11's std::function should work:

#include <functional>

template<typename T>
void test (std::function<T> func) {

You can use this, e.g.:

int foo (int);
std::sting bar (bool);

test(std::function<int(int)>(foo));
test(std::function<std::string(bool)>(bar));

answered Jul 02 '14 at 15:44

CodeClown42

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Shame there's no implicit conversion for `std::function`. This works though! – sircodesalot Jul 02 '14 at 15:48

Accept any function (but only functions) as a parameter

1 Answers1