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

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".

---

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.