Why can [=] be used to modify member variables in a lambda?

Question

http://coliru.stacked-crooked.com/a/29520ad225ced72d

#include <iostream>

struct S
{
    void f()
    {
        //auto f0 = []     { ++i; }; // error: 'this' was not captured for this lambda function

        auto f1 = [this] { ++i; };    
        auto f2 = [&]    { ++i; };
        auto f3 = [=]    { ++i; };
        f1();
        f2();
        f3();
    }

  int i = 10;  
};

int main()
{
    S s;
    std::cout << "Before " << s.i << std::endl;
    s.f();
    std::cout << "After " << s.i << std::endl;
}

Before 10
After 13

Question: Why does [=] enable modification of member variables in a lambda?

max66 · Accepted Answer · 2017-03-19T01:24:22.017

19

Question> Why [=] enables modification of member variables in lambda?

Because (see cpp reference) [=] "captures all automatic variables odr-used in the body of the lambda by value and current object by reference if exists"

So [=] capture the current object by reference and can modify i (that is a member of the current object).

edited Mar 19 '17 at 01:24

answered Mar 19 '17 at 01:18

max66

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1

what the meaning of 'odr-used'? – q0987 Mar 19 '17 at 01:26
1

@q0987 http://stackoverflow.com/questions/19630570/what-does-it-mean-to-odr-use-something – Mar 19 '17 at 01:28
14

Think of it as capturing `this` by value. – Yakk - Adam Nevraumont Mar 19 '17 at 02:59
@q0987 One Definition Rule – aggsol Mar 21 '17 at 07:46

Yankes · Answer 2 · 2017-03-19T19:12:13.453

9

If you write in equivalent way, this behavior will be more clear:

auto f3 = [=]    { ++(this->i); };

You not catch i but this, it will be constant but thing that its points to will be editable as in any other pointer.

edited Mar 19 '17 at 19:12

answered Mar 19 '17 at 16:49

Yankes

1,958
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`as in any other pointer` well, not any pointer-to-`const` :P – underscore_d Mar 21 '17 at 00:32

score 6 · Answer 3 · answered Mar 19 '17 at 16:53

6

You can imagine that [=] capture this by value, then using the copied this to access this->i.

a copied this still allows you to access the object this points at.

answered Mar 19 '17 at 16:53

David Haim

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Why can [=] be used to modify member variables in a lambda?

3 Answers3