Lambda Expressions

Question

A point from ISO draft n3290 section 5.1.2 paragraph, point 19:

The closure type associated with a lambda-expression has a deleted (8.4.3) default constructor and a deleted copy assignment operator. It has an implicitly-declared copy constructor (12.8) and may have an implicitly declared move constructor (12.8). [ Note: The copy/move constructor is implicitly defined in the same way as any other implicitly declared copy/move constructor would be implicitly defined. —end note ]

Can any one please ....tell some example for this point to understand?

Is there any chance/way to check the Closure object(type)?

Answer 1

The closure type associated with a lambda-expression has a deleted (8.4.3) default constructor

int main() {
    auto closure = [](){};
    typedef decltype(closure) ClosureType;

    ClosureType closure2;   // <-- not allowed

    return 0;
}

and a deleted copy assignment operator. It has an implicitly-declared copy constructor (12.8) and may have an implicitly declared move constructor (12.8).

#include <utility>

int main() {
    auto closure = [](){};
    typedef decltype(closure) ClosureType;

    ClosureType closure2 = closure;   // <-- copy constructor
    ClosureType closure3 = std::move(closure);  // <-- move constructor
    closure2 = closure3;              // <-- copy assignment (not allowed)

    return 0;
}

Answer 2

struct LambdaExample{
  // deleted operations = not allowed
  LambdaExample() = delete;
  LambdaExample& operator=(LambdaExample const&) = delete;

  // generated by the compiler:
  LambdaExample(LambdaExample const& other);
  LambdaExample(LambdaExample&& other);

  // anything else a lambda needs
};

For your second question, if you mean that you can look into the implementation, then nope, not possible. It's created on-the-fly by the compiler. If you mean to get the type of the lambda, sure:

auto l = [](){};
typedef decltype(l) closure_type;