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    <p>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) <em>used</em> 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, <code>my_huge_vector</code> is not captured.</p> <p>Per C++11 §5.1.2[expr.prim.lambda]/11:</p> <blockquote> <p>If a <em>lambda-expression</em> has an associated <em>capture-default</em> and its <em>compound-statement</em> <em>odr-uses</em> <code>this</code> or a variable with automatic storage duration and the <em>odr-used</em> entity is not explicitly captured, then the <em>odr-used</em> entity is said to be implicitly captured.</p> </blockquote> <p>Your lambda expression has an associated capture default: by default, you capture variables by value using the <code>[=]</code>. </p> <p>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 <code>my_huge_vector</code> at all in the body (the "compound statement") of the lambda expression, it is not implicitly captured.</p> <p>To continue with §5.1.2/14</p> <blockquote> <p>An entity is captured by copy if </p> <ul> <li>it is implicitly captured and the <em>capture-default</em> is <code>=</code> or if </li> <li>it is explicitly captured with a capture that does not include an <code>&amp;</code>. </li> </ul> </blockquote> <p>Since your <code>my_huge_vector</code> is not implicitly captured and it is not explicitly captured, it is not captured at all, by copy or by reference.</p>
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    1. PODo c++11 lambdas capture variables they don't use?
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    1. CODoes thou hath a Holy Quote?
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    2. CO@GMan: I sure do!
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    3. COI will say, though, that the entirety of §5.1.2 is important to understand all of the details. There are a lot of technical terms defined in that section and because the definitions of the various components of lambda expressions are necessarily entangled, it is difficult to pull out short quotes that definitively say "this is X and this is why X."
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