C++ – Understanding how Lambda closure type has deleted default constructor

cconstructordefaultlambda

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

Best Answer

The relationship between a closure to lambda is similar to object to class.

The C++11 standard says that the closure! type has no default constructor, and that is correct because it doesn't say it has no constructor.

The lambda is used to create a closure. But your quoted parapgraph will change for C++14.

ClosureType() = delete;                     // (until C++14)
ClosureType(const ClosureType& ) = default; // (since C++14)
ClosureType(ClosureType&& ) = default;      // (since C++14)

Closure types are not DefaultConstructible. Closure types have a deleted (until C++14) no (since C++14) default constructor. The copy constructor and the move constructor are implicitly-declared (until C++14) declared as defaulted (since C++14) and may be implicitly-defined according to the usual rules for copy constructors and move constructors.

http://en.cppreference.com/w/cpp/language/lambda

Related Solutions

C++ – Why don’t C++ compilers define operator== and operator!=

The argument that if the compiler can provide a default copy constructor, it should be able to provide a similar default operator==() makes a certain amount of sense. I think that the reason for the decision not to provide a compiler-generated default for this operator can be guessed by what Stroustrup said about the default copy constructor in "The Design and Evolution of C++" (Section 11.4.1 - Control of Copying):

I personally consider it unfortunate that copy operations are defined by default and I prohibit copying of objects of many of my classes. However, C++ inherited its default assignment and copy constructors from C, and they are frequently used.

So instead of "why doesn't C++ have a default operator==()?", the question should have been "why does C++ have a default assignment and copy constructor?", with the answer being those items were included reluctantly by Stroustrup for backwards compatibility with C (probably the cause of most of C++'s warts, but also probably the primary reason for C++'s popularity).

For my own purposes, in my IDE the snippet I use for new classes contains declarations for a private assignment operator and copy constructor so that when I gen up a new class I get no default assignment and copy operations - I have to explicitly remove the declaration of those operations from the private: section if I want the compiler to be able to generate them for me.

The difference between a ‘closure’ and a ‘lambda’

A lambda is just an anonymous function - a function defined with no name. In some languages, such as Scheme, they are equivalent to named functions. In fact, the function definition is re-written as binding a lambda to a variable internally. In other languages, like Python, there are some (rather needless) distinctions between them, but they behave the same way otherwise.

A closure is any function which closes over the environment in which it was defined. This means that it can access variables not in its parameter list. Examples:

def func(): return h
def anotherfunc(h):
   return func()

This will cause an error, because func does not close over the environment in anotherfunc - h is undefined. func only closes over the global environment. This will work:

def anotherfunc(h):
    def func(): return h
    return func()

Because here, func is defined in anotherfunc, and in python 2.3 and greater (or some number like this) when they almost got closures correct (mutation still doesn't work), this means that it closes over anotherfunc's environment and can access variables inside of it. In Python 3.1+, mutation works too when using the nonlocal keyword.

Another important point - func will continue to close over anotherfunc's environment even when it's no longer being evaluated in anotherfunc. This code will also work:

def anotherfunc(h):
    def func(): return h
    return func

print anotherfunc(10)()

This will print 10.

This, as you notice, has nothing to do with lambdas - they are two different (although related) concepts.

Best Answer

Related Solutions

C++ – Why don’t C++ compilers define operator== and operator!=

The difference between a ‘closure’ and a ‘lambda’

Related Topic