c++ - Understanding how Lambda closure type has deleted default constructor - Stack Overflow#
Excerpt#
From 5.1.2
[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 const...
From 5.1.2
[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 implicitlydeclared 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 ]
I'm reading through C++ Primer 14.8.1 which explains lambda expressions being translated by the compiler in to an unnamed object of an unnamed class. How is it I can I define objects of lambda functions which do not contain a lambda capture, if the default constructor is deleted?
auto g = [](){};
Is this not conceptually the same as...
class lambdaClass{
public:
lambdaClass() = delete;
lambdaClass& operator=(const lambdaClass&) = delete;
void operator()(){ }
//copy/move constructor and destructor implicitly defined
};
auto g = lambdaClass(); //would be an error since default is deleted.
If there was a capture then a constructor other than the default constructor would be defined and it would be okay to initialise objects of such (as long as a parameter was passed). But if there is no capture and the default constructor is deleted, it doesn't seem consistent conceptually that a lambda class object can be created.
Edit: Hmm, maybe the conception that the lambda class creates constructors depending on its lambda captures is unfounded despite this being how it's described in C++ Primer (I can find no quote of it in the standard), because the following code does not work even though I would expect it to conceptually:
int sz = 2;
auto a = [sz](){ return sz;};
decltype(a) b(10); //compiler error
decltype(a) b = a; //all good though