Question

How to use lambda expression as a template parameter? E.g. as a comparison class initializing a std::set.

The following solution should work, as lambda expression merely creates an anonymous struct, which should be appropriate as a template parameter. However, a lot of errors are spawned.

Code example:

struct A {int x; int y;};
std::set <A, [](const A lhs, const A &rhs) ->bool {
    return lhs.x < rhs.x;
    } > SetOfA;

Error output (I am using g++ 4.5.1 compiler and –std=c++0x compilation flag):

error: ‘lhs’ cannot appear in a constant-expression
error: ‘.’ cannot appear in a constant-expression
error: ‘rhs’ cannot appear in a constant-expression
error: ‘.’ cannot appear in a constant-expression
At global scope:
error: template argument 2 is invalid

Is that the expected behavior or a bug in GCC?

EDIT

As someone pointed out, I’m using lambda expressions incorrectly as they return an instance of the anonymous struct they are referring to.

However, fixing that error does not solve the problem. I get lambda-expression in unevaluated context error for the following code:

struct A {int x; int y;};
typedef decltype ([](const A lhs, const A &rhs) ->bool {
    return lhs.x < rhs.x;
    }) Comp;
std::set <A, Comp > SetOfA;

Answer 1

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);