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.

Answer 1

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".

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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.