How to use a lambda expression as a template parameter?

Question

The 2nd template parameter of std::set expects a type, not an expression, so it is just you are using it wrongly.

You could create the set like this:

auto comp = [](const A& lhs, const A& rhs) -> bool { return lhs.x < rhs.x; };
auto SetOfA = std::set <A, decltype(comp)> (comp);

For comparators used this way, you're still better off with a non-0x approach:

struct A { int x; int y; };

struct cmp_by_x {
  bool operator()(A const &a, A const &b) {
    return a.x < b.x;
  }
};

std::set<A, cmp_by_x> set_of_a;

However, in 0x you can make cmp_by_x a local type (i.e. define it inside a function) when that is more convenient, which is forbidden by current C++.

Also, your comparison treats A(x=1, y=1) and A(x=1, y=2) as equivalent. If that's not desired, you need to include the other values that contribute to uniqueness:

struct cmp_by_x {
  bool operator()(A const &a, A const &b) {
    return a.x < b.x || (a.x == b.x && a.y < b.y);
  }
};

Not sure if this is what you're asking, but the signature of a lambda which returns RetType and accepts InType will be:

std::function<RetType(InType)>

(Make sure to #include <functional>)

You can shorten that by using a typedef, but I'm not sure you can use decltype to avoid figuring out the actual type (since lambdas apparently can't be used in that context.)

So your typedef should be:

typedef std::function<bool(const A &lhs, const A &rhs)> Comp

or

using Comp = std::function<bool(const A &lhs, const A &rhs)>;