c# declaring variables inside Lambda expressions

Question

The following code outputs 33 instead of 012. I don't understand why a new variable loopScopedi isn't captured in each iteration rather than capturing the same variable.

Action[] actions = new Action[3];

for (int i = 0; i < 3; i++)
{

   actions [i] = () => {int loopScopedi = i; Console.Write (loopScopedi);};
}

foreach (Action a in actions) a();     // 333

Hopwever, this code produces 012. What's the difference between the two?

Action[] actions = new Action[3];

for (int i = 0; i < 3; i++)
{
    int loopScopedi = i;
    actions [i] = () => Console.Write (loopScopedi);
}

foreach (Action a in actions) a();     // 012

Answer 1

This is called "access to a modified closure". Basically, there is only one i variable, and all three lambdas are referring to it. At the end, the one i variable has been incremented to 3, so all three actions print 3. (Note that int loopScopedi = i in the lambda only runs once you call the lambda later.)

In the second version, you are creating a new int loopScopedi for every iteration, and setting it to the current value of i, which is 0 and 1 and 2, for each iteration.

You can try imagining inlining the lambdas to see more clearly what is happening:

foreach (Action a in actions)
{
   int loopScopedi = i; // i == 3, since this is after the for loop
   Console.Write(loopScopedi); // always outputs 3
}

Versus:

foreach (Action a in actions)
{
   // normally you could not refer to loopScopedi here, but the lambda lets you
   // you have "captured" a reference to the loopScopedi variables in the lambda
   // there are three loopScopedis that each saved a different value of i
   // at the time that it was allocated
   Console.Write(loopScopedi); // outputs 0, 1, 2
}

Answer 2

Variables captured in a lambda are hoisted into a class shared between the lambda and the outside code.

In your first example, i is being hoisted once and used with both the for() and all the passed lambdas. By the time you reach Console.WriteLine, the i has reached 3 from the for() loop.

In your second example, A new loopScopedi is being hoisted for each run of the loop, so it is left unaffected by the subsequent loops.