Does [=] have all local variables copied?

Question

When I write lambda with [=] , does this mean that all my local variables will be copied to the members of the created structure, or can I assume that only those that are really used in 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?

• and 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 implantation / optimization settings.

Suppose for an argument that a vector copy constructor has side effects.

Answer 1

No. It just means that all local variables from scope are searchable inside the lambda body. Only if you refer to the name of a local local variable will this variable be captured and fixed by value.

"Remove something" abbreviates = and & is just syntactic sugar, essentially telling the compiler to "find out what I mean."

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Formal link from 5.1.2 / 11-12:

If a lambda expression has a default capture-related function, and its compound operator odr uses [...] a variable with automatic storage duration, and an object using odr is not explicitly captured, then an object using odr is called implicit capture [.. .]

An object is captured if it is latched explicitly or implicitly.

Note that "capture-default" refers to [=] and [&] . Repeat, indicating capture default, does not fix anything; only the odr-using variable is used.

Answer 2

No! (Luckily)

You can program your code to check if your compiler really does (or not). For example, gcc 4.8.0 meets the requirements.

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In accordance with what the Standard actually provides (addresses back):

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

$ 5.1.2 / 11 If a lambda expression has a default capture and its compound operator odr-uses (3.2) this or a variable with automatic storage duration, and an object using odr is not explicitly captured, then an object using odr is called implicit capture; such objects must be declared within the scope of the lambda expression.

§5.1.2 / 9 A lambda expression, the smallest enclosing area of ​​which is the block region (3.3.3), is a local lambda expression; any other lambda expression should not have a capture list in its lambda introducer. The scope for reaching a local lambda expression is a set of encompassing action areas up to the innermost encompassing function and its parameters. [Note. This scope includes any intermediate lambda expressions. -end note]