Do c++11 lambdas capture variables they don't use?

Question

Each variable expressly named in the capture list is captured. The default capture will only capture variables that are both (a) not expressly named in the capture list and (b) used in the body of the lambda expression. If a variable is not expressly named and you don't use the variable in the lambda expression, then the variable is not captured. In your example, my_huge_vector is not captured.

Per C++11 §5.1.2[expr.prim.lambda]/11:

If a lambda-expression has an associated capture-default and its compound-statement odr-uses 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.

Your lambda expression has an associated capture default: by default, you capture variables by value using the [=].

If and only if a variable is used (in the One Definition Rule sense of the term "used") is a variable implicitly captured. Since you don't use my_huge_vector at all in the body (the "compound statement") of the lambda expression, it is not implicitly captured.

To continue with §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 &.

Since your my_huge_vector is not implicitly captured and it is not explicitly captured, it is not captured at all, by copy or by reference.

No, my_huge_vector will not be captured. [=] means all used variables are captured in the lambda.