What does the word capture mean in the context of lambdas?

Question

Can somebody please provide some insights on this? Is the lambda capturing external variables, or is the outside world capturing values produced by the lambdas? What does it mean for a certain variable to be captured?

Answer 1

The lambda is capturing an outside variable.

A lambda is a syntax for creating a class. Capturing a variable means that variable is passed to the constructor for that class.

A lambda can specify whether it's passed by reference or by value. For example:

[&] { x += 1; }       // capture by reference
[=] { return x + 1; } // capture by value

The first produces a class roughly like this:

class foo { 
    int &x;
public:
    foo(int &x) : x(x) {}
    void operator()() const { x += 1; }
};

The second produces a class something like this:

class bar {
    int x;
public:
    bar(int x) : x(x) {}
    int operator()() const { return x + 1; }
};

As with most uses of references, capturing by reference can create a dangling reference if the closure (the object of the class created by the lambda expression) out-lives the object that was captured.

Answer 2

Jerry Coffin gave you detailed response,I agree that lambda is syntax for creating a class.There are bunch of options about the way variables are captured and how.List of options:

[]  Capture nothing (or, a scorched earth strategy?)
[&] Capture any referenced variable by reference
[=] Capture any referenced variable by making a copy
[=, &foo]   Capture any referenced variable by making a copy, but capture variable foo by reference
[bar]   Capture bar by making a copy; don't copy anything else
[this]  Capture the this pointer of the enclosing class

Answer 3

Lambda captures variables which it otherwise will not have access to. One can specify how A lambda should capture a variable .i.e value,reference . This has been explained really well here In lambda functions syntax, what purpose does a 'capture list' serve?