How I can pass callable object to function as parameter

Question

In c++ standard library almost all algo function takes a callable object as a argument. Now I want to try this thing with my program. I opened the the headers for function like find_if or search_n() but could not understand much about how these callable object parameters are handled and off course how argument is passed to them especially for lambda object ( can bind() be used for lambdas, I dont know)

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Can any one explain me how this thing works. Thanks in advance

Answer 1

Just have a template parameter for the function-like type and take an argument of that type:

template <typename Function>
void foo(Function func) {
  // Use func like so
  func(10);
}

This is how the standard library algorithms do it. You could call it with a lambda like so:

foo([](int x) { std::cout << (x * 2) << std::endl; });

Of course, this requires that you specify in your documentation what kind of function you expect as the Function type (until we get (until we get Concepts). Should it be unary? Binary? What type arguments should it take? What should it return?

Alternatively, you can take a function of a specific type by using std::function:

int foo(std::function<int(char)> func) {
  return func('a');
}

This time, foo will only take function-like objects that take a char argument and return an int. There is one downside to this method, however, which is that the compiler is unlikely to inline any lambdas you pass as func.

The most basic way to take a function as an argument is with a function pointer:

int foo(int (*func)(char)) {
  return func('a');
}

However, this will only take function pointers (some lambdas can be converted to function pointers). It won't take functors or anything else of the sort.