Write a function that accepts a lambda expression as argument - c++

Question

I have a method like this
template<typename T, typename U>
map<T,U> mapMapValues(map<T,U> old, T (f)(T,U))
{
map<T,U> new;
for(auto it = old.begin(); it != old.end(); ++it)
{
new[it->first] = f(it->first,it->second);
}
return new;
}
and the idea is that you'd call it like this
BOOST_AUTO_TEST_CASE(MapMapValues_basic)
{
map<int,int> test;
test[1] = 1;
map<int,int> transformedMap = VlcFunctional::mapMapValues(test,
[&](int key, int value) -> int
{
return key + 1;
}
);
}
However I get the error: no instance of function template "VlcFunctional::mapMapValues" matches the argument list argument types are: (std::map, std::allocator>>, __lambda1)
Any idea what I'm doing wrong? Visual Studio 2008 and Intel C++ compiler 11.1

Answer 1

Your function is expecting a function pointer, not a lambda.
In C++, there are, in general, 3 types of "callable objects".
Function pointers.
Function objects.
Lambda functions.
If you want to be able to use all of these in your function interface, then you could use std::function:
template<typename T, typename U>
map<T,U> mapMapValues(map<T,U> old, std::function<T(T, U)> f)
{
...
}
This will allow the function to be called using any of the three types of callable objects above. However, the price for this convenience is a small amount of overhead on invokations on the function (usually a null pointer check, then a call through a function pointer). This means that the function is almost certainly not inlined (except maybe with advanced WPO/LTO).
Alternatively, you could add an additional template parameter to take an arbitrary type for the second parameter. This will be more efficient, but you lose type-safety on the function used, and could lead to more code bloat.
template<typename T, typename U, typename F>
map<T,U> mapMapValues(map<T,U> old, F f)

Answer 2

Your parameter type declaration T (f)(T,U) is of type 'free function taking a T and a U and returning a T'. You can't pass it a lambda, a function object, or anything except an actual function with that signature.
You could solve this by changing the type of the parameter to std::function<T(T,U)> like this:
template<typename T, typename U>
map<T,U> mapMapValues(map<T,U> old, std::function<T(T,U)>)
{
}
Alternately, you could declare the function type as a template argument like this:
template<typename T, typename U, typename Fn>
map<T,U> mapMapValues(map<T,U> old, Fn fn)
{
fn(...);
}

Answer 3

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 4

Lambda expressions with empty capture list should decay to function pointers, according to n3052. However it seems that this feature is not implemented in VC++ and only partially in g++, see my SO question.

Answer 5

Here is some example of how to pass a function as parameter
class YourClass
{
void YourClass::callback(void(*fptr)(int p1, int p2))
{
if(fptr != NULL)
fptr(p1, p2);
}
};
void dummyfunction(int p1, int p2)
{
cout << "inside dummyfunction " << endl;
}
YourClass yc;
// using a dummyfunction as callback
yc.callback(&dummyfunction);
// using a lambda as callback
yc.callback( [&](int p1, int p2) { cout << "inside lambda callback function" << endl; } );
// using a static member function
yc.callback( &aClass::memberfunction );