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;
}

• user751747: Thanks for your answer ..Is there any way/chance of explaing ..this point ..in terms of a class Example
• Johannes Schaub - litb: I believe that this is well-formed though: auto closure = []{}; decltype(closure) c{};. That may be changed by a defect report for C++0x rev1 though.

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;

• user751747: yes the closure obj type is created on-the-fly by the compiler..atleast is there any chance to check whether it exists or not
• Xeo: @user: What do you mean with "exists"?