Creating functor from lambda expression - c++

Question

I would like to know if it is possible to create an actual functor object from a lambda expression. I don't think so, but if not, why?
To illustrate, given the code below, which sorts points using various policies for x and y coordinates:
#include <vector>
#include <functional>
#include <algorithm>
#include <iostream>
struct Point
{
Point(int x, int y) : x(x), y(y) {}
int x, y;
};
template <class XOrder, class YOrder>
struct SortXY :
std::binary_function<const Point&, const Point&, bool>
{
bool operator()(const Point& lhs, const Point& rhs) const
{
if (XOrder()(lhs.x, rhs.x))
return true;
else if (XOrder()(rhs.x, lhs.x))
return false;
else
return YOrder()(lhs.y, rhs.y);
}
};
struct Ascending { bool operator()(int l, int r) const { return l<r; } };
struct Descending { bool operator()(int l, int r) const { return l>r; } };
int main()
{
// fill vector with data
std::vector<Point> pts;
pts.push_back(Point(10, 20));
pts.push_back(Point(20, 5));
pts.push_back(Point( 5, 0));
pts.push_back(Point(10, 30));
// sort array
std::sort(pts.begin(), pts.end(), SortXY<Descending, Ascending>());
// dump content
std::for_each(pts.begin(), pts.end(),
[](const Point& p)
{
std::cout << p.x << "," << p.y << "\n";
});
}
The expression std::sort(pts.begin(), pts.end(), SortXY<Descending, Ascending>()); sorts according to descending x values, and then to ascending y values. It's easily understandable, and I'm not sure I really want to make use of lambda expressions here.
But if I wanted to replace Ascending / Descending by lambda expressions, how would you do it? The following isn't valid:
std::sort(pts.begin(), pts.end(), SortXY<
[](int l, int r) { return l>r; },
[](int l, int r) { return l<r; }
>());

Answer 1

This problem arises because SortXY only takes types, whereas lambdas are objects. You need to re-write it so that it takes objects, not just types. This is basic use of functional objects- see how std::for_each doesn't take a type, it takes an object.

Answer 2

I have posted a similar question w.r.t. lambda functors within classes.
Check this out, perhaps it helps:
Lambda expression as member functors in a class

Answer 3

I had a similar problem: It was required to provide in some cases a "raw"-function pointer and in other a functor. So I came up with a "workaround" like this:
template<class T>
class Selector
{
public:
Selector(int (*theSelector)(T& l, T& r))
: selector(theSelector) {}
virtual int operator()(T& l, T& r) {
return selector(l, r);
}
int (*getRawSelector() const)(T&, T&) {
return this->selector;
}
private:
int(*selector)(T& l, T& r);
};
Assuming you have two very simple functions taking --- as described --- either a functor or a raw function pointer like this:
int
findMinWithFunctor(int* array, int size, Selector<int> selector)
{
if (array && size > 0) {
int min = array[0];
for (int i = 0; i < size; i++) {
if (selector(array[i], min) < 0) {
min = array[i];
}
}
return min;
}
return -1;
}
int
findMinWithFunctionPointer(int* array, int size, int(*selector)(int&, int&))
{
if (array && size > 0) {
int min = array[0];
for (int i = 0; i < size; i++) {
if (selector(array[i], min) < 0) {
min = array[i];
}
}
return min;
}
return -1;
}
Then you would call this functions like this:
int numbers[3] = { 4, 2, 99 };
cout << "The min with functor is:" << findMinWithFunctor(numbers, 3, Selector<int>([](int& l, int& r) -> int {return (l > r ? 1 : (r > l ? -1 : 0)); })) << endl;
// or with the plain version
cout << "The min with raw fn-pointer is:" << findMinWithFunctionPointer(numbers, 3, Selector<int>([](int& l, int& r) -> int {return (l > r ? 1 : (r > l ? -1 : 0)); }).getRawSelector()) << endl;
Of course in this example there is no real benefit passing the int's as reference...it's just an example :-)
Improvements:
You can also modify the Selector class to be more concise like this:
template<class T>
class Selector
{
public:
typedef int(*selector_fn)(T& l, T& r);
Selector(selector_fn theSelector)
: selector(theSelector) {}
virtual int operator()(T& l, T& r) {
return selector(l, r);
}
selector_fn getRawSelector() {
return this->selector;
}
private:
selector_fn selector;
};
Here we are taking advantage of a simple typedef in order to define the function pointer once and use only it's name rather then writing the declaration over and over.