getting type from lambda c++11

[EamonKerdoll]

Hi all,
Is there any way of creating a variable which type is the result of some lambda expression.
something like this doesn't work:

decltype([](){return 1;}()) a;

Is there a workaround?

[viliml]

Yes, and no, depending on what you really mean. If you want the type of the lambda itself, then no. Buf, if you just want the return type, it's easy!

auto x=[](){return 1;};
decltype(x()) a;

[EamonKerdoll]

Thanks!!
Do you know what is the rational behind not allowing decltype to accept the result of a lambda calculation in the first place?

[o1vk4iN6]

Why not just

decltype( 1 ) a;

Also seems a bit counter productive and you'll need two copies of the same lambda function.

[Cubbi]

Do you know what is the rational behind not allowing decltype to accept the result of a lambda calculation in the first place?

In general, it's impossible to determine the type of a lambda expression without evaluating it in the correct context. So, lamdbas are forbidden in unevaluated contexts, and decltype is one of such contexts.

[viliml]

Yes! The lambda is just a way of evluating a closure, and since you are EVALUATING it to get the closure, it can't be done in an UNEVALUATED context!