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Could you please tell me the difference between capturing of [this] and [&]? (In case of a lambda is defined inside a member function of the class)

As I understood, there is no difference between both of them.

  1. [this] - [this] means capturing all the member data of the class. Capturing [this] means capturing the address of the object.

    #include <iostream>
class Foo {
    int x;
public:
    Foo() : x(10) {}

    void bar() {
        // Increment x every time we are called
        auto lam = [this](){ return ++x; };
        std::cout << lam() << std::endl;
    }
};

int main() {
    Foo foo;
    foo.bar(); // Outputs 11
    foo.bar(); // Outputs 12
}

  2. [&] - [&] means capture all variables by reference all variables mean local variable, global variable and member data of the class.

    class Foo {
    int x;
public:
    Foo() : x(10) {}

    void bar() {
        // Increment x every time we are called
        auto lam = [&](){ return ++x; };
        std::cout << lam() << std::endl;
    }
};

int main() {
    Foo foo;
    foo.bar(); // Outputs 11
    foo.bar(); // Outputs 12
}

I believe that there is no difference between [this] and [&]. Because by both capturing data is captured by reference. so the data can be changed.

  • [this] captures address of the data (reference value)
  • [&] captures all data by reference

Is my understanding is wrong? Please give me some advice.

Evg
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harrbanggg
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    Does this answer your question? [C++ lambda capture this vs capture by reference](https://stackoverflow.com/questions/33575563/c-lambda-capture-this-vs-capture-by-reference) – Louis Go Jul 30 '20 at 06:15
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    `[&]` itself doesn't capture data members at all. But when you write `[&]`, `this` is captured implicitly: `[&]` is equivalent to `[&, this]`. – Evg Jul 30 '20 at 06:22
  • @Evg Thanks. then basically if i use member data of the class in lambda expression, can I use [&] or [&, this] or [this] for the same purpose? Thanks. I will choose one of them for coding. – harrbanggg Jul 30 '20 at 06:31
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    Yes, you can use any one of them. But if you only access data members inside a lambda, `[this]` is all you need. So I suggest using `[this]` to be explicit about your intention. – Evg Jul 30 '20 at 06:34
  • @Evg Thank you for the kind explanation. thanks. – harrbanggg Jul 30 '20 at 06:35
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    For clarification: `[=]` won't capture data members by value. It will capture `this` implicitly (before C++20), *effectively* capturing data members by reference. – Evg Jul 30 '20 at 06:42

0 Answers0