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! (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.

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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 ]