How / Can C ++ lambdas maintains internal state?

Question

I have a basic lambda that looks like this:

auto l = [](){ int i = 0; cout << i++; } 

Calling this several times will print 0. How can I save i? Can I do this without functors?

Answer 1

  auto l = [](){ static int i = 0; // ^^^^^^ cout << i++; } 

should fix your problems.

In common functions, you cannot preserve the internal state without using the local static variable. There is no difference using lambda actually.

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If you want to count copies, you can use the usual functor class implementation as @Revolver_Ocelot .

Answer 2

Depending on what you want to do with this lambda, you might consider the following alternative:

 auto exec = [i = 0]() mutable { cout << ++i << ' '; }; exec(); // 1 exec(); // 2 auto exec2 = exec; // copy state of lambda exec2(); // 3 exec(); // 3 

Using []() { static int i = 0; cout << ++i << ' '; }; []() { static int i = 0; cout << ++i << ' '; }; instead, prints the sequence 1 2 3 4 .

Living example

Answer 3

Think of lambda as a class with operator() . How would you save the state in the class? You have a member. Captures here are equivalent.

 #include <iostream> auto l = [i=0]()mutable{ std::cout << i++; }; auto l2=l; int main(){ l(); // 0 l(); // 1 l(); // 2 l2(); // 0 l2(); // 1 l(); // 3 std::cout << '\n'; } 

Answer 4

If you want i keep its value, you have three options:

• Declare i as a global variable (Bad).
• Pass previous i value to function every time (better).

• Declare i as a static variable (best).

   auto l = []() { static int i = 0; cout << i++ << endl; }; l(); l(); l(); 

  This will give as output:

   0 1 2