How to declare a function that accepts a lambda?

Question

I read on the internet many tutorials that explained how to use lambdas with the standard library (such as std::find), and they all were very interesting, but I couldn't find any that explained how I can use a lambda for my own functions.

For example:

int main()
{
    int test = 5;
    LambdaTest([&](int a) { test += a; });

    return EXIT_SUCCESS;
}

How should I declare LambdaTest? What's the type of its first argument? And then, how can I call the anonymous function passing to it - for example - "10" as its argument?

Answer 1

If you don't want to template everything, you can do the following:

void LambdaTest (const std::function <void (int)>& f)
{
    ...
}

Answer 2

I would like to contribute this simple but self-explanatory example. It shows how to pass "callable things" (functions, function objects, and lambdas) to a function or to an object.

// g++ -std=c++11 thisFile.cpp

#include <iostream>
#include <thread>

using namespace std;

// -----------------------------------------------------------------
class Box {
public:
  function<void(string)> theFunction; 
  bool funValid;

  Box () : funValid (false) { }

  void setFun (function<void(string)> f) {
    theFunction = f;
    funValid = true;
  }

  void callIt () {
    if ( ! funValid ) return;
    theFunction (" hello from Box ");
  }
}; // class

// -----------------------------------------------------------------
class FunClass {
public:
  string msg;
  FunClass (string m) :  msg (m) { }
  void operator() (string s) {
    cout << msg <<  s << endl; 
  }
};

// -----------------------------------------------------------------
void f (string s) {
  cout << s << endl;
} // ()

// -----------------------------------------------------------------
void call_it ( void (*pf) (string) ) {
  pf( "call_it: hello");
} // ()

// -----------------------------------------------------------------
void call_it1 ( function<void(string)> pf ) {
  pf( "call_it1: hello");
} // ()

// -----------------------------------------------------------------
int main() {

  int a = 1234;

  FunClass fc ( " christmas ");

  f("hello");

  call_it ( f );

  call_it1 ( f );

  // conversion ERROR: call_it ( [&] (string s) -> void { cout << s << a << endl; } );

  call_it1 ( [&] (string s) -> void { cout << s << a << endl; } );

  Box ca;

  ca.callIt ();

  ca.setFun (f);

  ca.callIt ();

  ca.setFun ( [&] (string s) -> void { cout << s << a << endl; } );

  ca.callIt ();

  ca.setFun (fc);

  ca.callIt ();

} // ()

Answer 3

Given that you probably also want to accept function pointers and function objects in addition to lambdas, you'll probably want to use templates to accept any argument with an operator(). This is what the std-functions like find do. It would look like this:

template<typename Func>
void LambdaTest(Func f) {
    f(10);
}

Note that this definition doesn't use any c++0x features, so it's completely backwards-compatible. It's only the call to the function using lambda expressions that's c++0x-specific.