Why lambda takes variable's initial value

Question

Question is why program print id:0, id:1, id:2, 42 instead of id:42, id:43, id:44, 45.

int main()
{

    int id = 0;
    auto f = [id] () mutable {
        std::cout << "id: " << id << std::endl;
        ++id; // OK
    };

    id = 42;
    f();
    f();
    f();
    std::cout << id << std::endl;
    return 0;
}

Are you asking why it is specified to do this?

juanchopanza
– juanchopanza

2018-09-13 06:39:50 +00:00
Commented Sep 13, 2018 at 6:39 — juanchopanza
– juanchopanza, Commented Sep 13, 2018 at 6:39

songyuanyao · Accepted Answer · 2018-09-13 06:48:04Z

8

Because id is captured by value, i.e. it gets copied. And when the lambda is declared (i.e. when capture happens) id has the value of 0.

You might want to change it to capture-by-reference.

auto f = [&id] () {
//        ^
    std::cout << "id: " << id << std::endl;
    ++id;
};

BTW: For this case mutable becomes superfluous.

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edited Sep 13, 2018 at 6:48

answered Sep 13, 2018 at 6:41

songyuanyao

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