How do you pass a function as a parameter in C?

Declaration

A prototype for a function which takes a function parameter looks like the following:

void func ( void (*f)(int) );

This states that the parameter f will be a pointer to a function which has a void return type and which takes a single int parameter. The following function (print) is an example of a function which could be passed to func as a parameter because it is the proper type:

void print ( int x ) {  printf("%d\n", x);}

Function Call

When calling a function with a function parameter, the value passed must be a pointer to a function. Use the function's name (without parentheses) for this:

func(print);

would call func, passing the print function to it.

Function Body

As with any parameter, func can now use the parameter's name in the function body to access the value of the parameter. Let's say that func will apply the function it is passed to the numbers 0-4. Consider, first, what the loop would look like to call print directly:

for ( int ctr = 0 ; ctr < 5 ; ctr++ ) {  print(ctr);}

Since func's parameter declaration says that f is the name for a pointer to the desired function, we recall first that if f is a pointer then *f is the thing that f points to (i.e. the function print in this case). As a result, just replace every occurrence of print in the loop above with *f:

void func ( void (*f)(int) ) {  for ( int ctr = 0 ; ctr < 5 ; ctr++ ) {    (*f)(ctr);  }}

Source

This question already has the answer for defining function pointers, however they can get very messy, especially if you are going to be passing them around your application. To avoid this unpleasantness I would recommend that you typedef the function pointer into something more readable. For example.

typedef void (*functiontype)();

Declares a function that returns void and takes no arguments. To create a function pointer to this type you can now do:

void dosomething() { }functiontype func = &dosomething;func();

For a function that returns an int and takes a char you would do

typedef int (*functiontype2)(char);

and to use it

int dosomethingwithchar(char a) { return 1; }functiontype2 func2 = &dosomethingwithcharint result = func2('a');

There are libraries that can help with turning function pointers into nice readable types. The boost function library is great and is well worth the effort!

boost::function<int (char a)> functiontype2;

is so much nicer than the above.

Since C++11 you can use the functional library to do this in a succinct and generic fashion. The syntax is, e.g.,

std::function<bool (int)>

where bool is the return type here of a one-argument function whose first argument is of type int.

I have included an example program below:

// g++ test.cpp --std=c++11#include <functional>double Combiner(double a, double b, std::function<double (double,double)> func){  return func(a,b);}double Add(double a, double b){  return a+b;}double Mult(double a, double b){  return a*b;}int main(){  Combiner(12,13,Add);  Combiner(12,13,Mult);}

Sometimes, though, it is more convenient to use a template function:

// g++ test.cpp --std=c++11template<class T>double Combiner(double a, double b, T func){  return func(a,b);}double Add(double a, double b){  return a+b;}double Mult(double a, double b){  return a*b;}int main(){  Combiner(12,13,Add);  Combiner(12,13,Mult);}