Is there a way to shorten the C++11 lambda signature in declaration?

c++, c++11, lambda, syntactic-sugar

Solution

Looks like you're looking for an empty lambda which does nothing, so that your `std::function` object will always be in callable state!

If so, then use this one which can be reused, for any number of parameters:

static const struct empty_lambda_t //static and const applies to the object!
{
      template<typename ...T>
      void operator()(T && ... ) const {} //does nothing

}empty_lambda {}; //declare an object which is static and const

And then use it as:

 std::function<void()>          fun1 = empty_lambda;
 std::function<void(int,int)>   fun2 = empty_lambda;
 std::function<void(whatever)>  fun3 = empty_lambda;

Hope that helps.

Problem

I want to shorten the following type of lambdas: ``` [] (SomeVeryLongTemplateType<int, float, char, std::string>, AnotherLongType) {}; ``` Since the only reason for this lambda is to initialize some class `std::function<...>` member - it doesn't capture anything, it doesn't have argument names, it returns nothing, it does nothing. If the shortening operation is expressed as a function of the number of arguments in signature, then I want this function to have the complexity O(1). Is there a way to do that?

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