How can I pass a lambda (c++11) into a templated function?

Question

I'm playing around with lambda functions in gcc 4.6.2, and would like to implement a templated "map" function like this:

template<typename A, typename B> std::vector<B> map(const std::vector<A>& orig, const std::function<B(A)> f) {
  std::vector<B> rv;
  rv.resize(orig.size());
  std::transform(begin(orig), end(orig), begin(rv), f);
  return rv;
}

This doesn't work, because the test code:

int main(int argc, char **argv) {
  std::vector<int> list;
  list.push_back(10);
  list.push_back(20);
  list.push_back(50);

  std::vector<int> transformed = map(list, [](int x) -> int { return x + 1; });
  std::for_each(begin(transformed), end(transformed), [](int x) { printf("-> %d\n", x); });
  return 0;
}

gives this error:

test.cpp:49:80: error: no matching function for call to ‘map(std::vector<int>&, main(int, char**)::<lambda(int)>)’
test.cpp:49:80: note: candidate is:
test.cpp:6:49: note: template<class A, class B> std::vector<B> map(const std::vector<A>&, std::function<B(A)>)

If I remove the templating, and use a vector directly, it compiles fine:

std::vector<int> map(const std::vector<int>& orig, const std::function<int(int)> f) {
  std::vector<int> rv;
  rv.resize(orig.size());
  std::transform(begin(orig), end(orig), begin(rv), f);
  return rv;
}

so it must be a problem with the way I'm defining the template.

Has anyone run into this before? I know lambdas are incredibly new.

Answer 1

You don't need to use std::function. Just make the predicate parameter a template value. For example,

template<typename A, typename B> std::vector<B> map(const std::vector<A>& orig, B f) {

std::function<> is more useful as a member value type or for defining non-templated code.

Answer 2

I'm tackling with lambdas too and i noticed that you can declare a function pointer in a function definition's parameter list and when you make a call to that function you can pass a lambda expression as an argument if it matches the function prototype of course.

#include <iostream>

#include <vector>

#include <algorithm>

#include <iterator>



template <typename T,typename C>

struct map  {

   typedef C (*F)(const T&);

   std::vector<C> rv;

   map () {}

   map (const std::vector<T>& o,F f)  {

      rv.resize(o.size());

      std::transform (o.begin(),o.end(),rv.begin(),f);

    }

   ~map () {}

   operator std::vector<C> () const  {

      return rv;

    }

 };



int main ()  {

   std::vector<int> asd(5,12);

   std::vector<char> transformed=map<int,char>(asd,[](const int& x)->char {return x+1;});

   std::copy (transformed.begin(),transformed.end(),std::ostream_iterator<int>(std::cout," "));

 }

Answer 3

The problem is that the compiler can't figure out what to use for B. In order to determine that type it wants to use the function<> you pass in for f, but you don't pass an std::function<> directly. You pass in something you expect to be used to construct a function<>. And in order to do that implicit construction it needs to know the type of argument. So you've got this circular dependency where the type of argument depends on what you pass in, but what gets passed in depends on the type of argument.

You can break this circular dependency by specifying the template parameters, such as map_<int,int>(list, [](int x) -> char { return x + 1; });

(although I see the functor actually returns a char, not an int, so if the type deduction worked for you here you'd be getting back a vector<char> which cannot be converted to a vector<int> when you assign the result to transformed)

However as has been pointed out, generally templates take functors as just a plain template type:

template<typename A,typename Func>
auto map_(const std::vector<A>& orig, Func f) -> std::vector<decltype(f(A()))> {
    std::vector<decltype(f(A()))> rv;
    /*...*/
}

(we use the trailing return type because we need to use the expression f in the return type, which isn't available unless the return type comes afterwards.)

This allows the template to deduce the functor type directly and avoids any type conversions and best allows for optimization.

It's also customary to use iterators as arguments on these sorts of functions, in which case your function is just a wrapper around std::transform, so you can just use that directly. I'm not sure there's a whole lot of value in a special version that deals with vectors specifically.