Does [=] imply that all local variables will be copied?

Question

When I write a lambda with [=], does it mean that all my local variables will be copied into members of the created struct or can I assume that only those will that are actually used in the lambda? For example:

void f()
{
    vector<int> v(10000);
    const int n = 5;
    const int DivByNCnt = count_if(istream_iterator<int>(cin), istream_iterator<int>(), 
          [=](int i)
          {
             return i % n == 0;
          });
}

Which of the following, if any, is true?

both n and v will be copied
n will be copied, v will not
n will be copied, v may or may not be copied depending on the implmenentation/optimization settings.

Suppose for the argument's sake that vector's copy constructor has side effects.

Without knowing anything about the semantics of c++11, anything except only `n` being captured would be completely crazy with respect to the semantics of every other language with first-class anonymous functions and proper closures. — Gian, Mar 25 '13 at 10:22

score 63 · Accepted Answer · edited Jun 20 '20 at 09:12

No. It just means that all local variables from the ambient scope are available for lookup inside the body of the lambda. Only if you refer to a name of an ambient local variable will that variable be captured, and it'll be captured by value.

The "capture anything" shorthands = and & are just syntactic sugar, essentially, telling the compiler to "figure out what I mean".

A formal reference from 5.1.2/11-12:

If a lambda-expression has an associated capture-default and its compound-statement odr-uses [...] a variable with automatic storage duration and the odr-used entity is not explicitly captured, then the odr-used entity is said to be implicitly captured [...]

An entity is captured if it is captured explicitly or implicitly.

Note that "capture-default" refers to [=] and [&]. To repeat, specifying a capture-default doesn't capture anything; only odr-using a variable does.

I am in no doubt that you are correct, but have you got any references? — Alex Chamberlain, Mar 25 '13 at 10:24
@AlexChamberlain: See 5.1.2/11-12, about "implicit and explicit capturing". It's *using the variables inside the lambda body* that does the capturing. — Kerrek SB, Mar 25 '13 at 10:29

score 22 · Answer 2 · edited Jun 20 '20 at 09:12

No! (thankfully)

You can instrument your code to check whether your compiler actually does it (or not). For example gcc 4.8.0 appears to be compliant.

As to what the Standard actually mandates (working backward):

§5.1.2/14 An entity is captured by copy if it is implicitly captured and the capture-default is = or if it is explicitly captured with a capture that does not include an &. For each entity captured by copy, an unnamed nonstatic data member is declared in the closure type.

$5.1.2/11 If a lambda-expression has an associated capture-default and its compound-statement odr-uses (3.2) this or a variable with automatic storage duration and the odr-used entity is not explicitly captured, then the odr-used entity is said to be implicitly captured; such entities shall be declared within the reaching scope of the lambda expression.

§5.1.2/9 A lambda-expression whose smallest enclosing scope is a block scope (3.3.3) is a local lambda expression; any other lambda-expression shall not have a capture-list in its lambda-introducer. The reaching scope of a local lambda expression is the set of enclosing scopes up to and including the innermost enclosing function and its parameters. [ Note: This reaching scope includes any intervening lambda-expressions. —end note ]

MSVC 10.0 does not copy v, but I wanted to know if there's a guaranteed that v will not be copied! — Armen Tsirunyan, Mar 25 '13 at 10:27
@ArmenTsirunyan: I added the corresponding quotes of the Standard. Only the captured members are supposed to be "hosted" in the closure type according to §5.1.2/14. — Matthieu M., Mar 25 '13 at 10:42

Does [=] imply that all local variables will be copied?

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