Can the 'type' of a lambda expression be expressed?

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Thinking of lambda expressions as 'syntactic sugar' for callable objects, can the unnamed underlying type be expressed?

An example: 

struct gt {
    bool operator() (int l, int r) {
        return l > r;
    }
} ;

Now, [](int l, int r) { return l > r; } is an elegant replacement for the above code (plus the necessary creation of callable objects of gt), but is there a way to express gt (the type) itself?

A simple usage: 

std::set<int, gt> s1;  // A reversed-order std::set
// Is there a way to do the same using a lambda?
std::set<int, some-magic-here-maybe([](int l, int r) { return l > r; }) > s2;
c++11 lambda stl c++
4 Answers

In Microsoft Visual Studio at least (I haven't tried this with other compilers), and if you don't capture anything, the type seems to be a regular function pointer:

std::string (*myFunctionPointer)(int x) = [] (int x) {
  char buffer[10];
  return std::string("Test ") + itoa(x, buffer, 10);
};
std::string testOutput = myFunctionPointer(123);
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