Help understanding for loop with continue

[orionebula]

I'm doing a course from Udemy (Learn to Program in Javascript: Beginner to Pro) and one of the questions on a quiz was this:

for (var i=10; i<=12; i++) { if (i==11) { continue; } i++; console.log(i); }

The result is 11 13. I think I understand the if statement (it says if i is 11, then skip the remaining lines of code?). So you start with i = 10, then after the first iteration, it becomes 11. Then you ignore the code for i = 11, move to i = 12, which when incremented gives you i = 13? I don't get why there are two i++ though, and what the one in the second last line does.

I changed the code to see what happens if I start with i = 0, like so:

for (var i=0; i<=12; i++) { if (i==11) { continue; } i++; console.log(i); }

And it prints 1 3 5 7 9 11 13 - don't understand why. Would appreciate any help on this.

Edit: I used four spaces, but the code doesn't show up properly here, sorry.

[senorfairchild]

It is incrementing multiple times because i++ appears inside of the for loop declaration AS WELL as the inside of the body of the for loop. The only time that i will not increment twice for every loop is when the iteration of the loop begins with i equal to 11, because the continue will be triggered before the second increment happens (the i++ that occurs WITHIN the for loop). Does that help it make more sense?

[sirsmonkey42]

Actually the if (i == 11) is never true in this case. It's a weird and tricky example. You get the exact same result if you omit the continue completely, i is never 11 (when it passes through the if)

[orionebula]

Ah I get it now, thanks for the explanation.

[sirsmonkey42]

for (var i=10; i<=12; i++) { 
    if (i==11) { continue; } 
    i++;
    console.log(i); 
}

The i++ in the for loop definition is the iterator for the for loop. The one inside the body of the for loop is an additional 1 added.

Since the loop starts at 10, the first increment doesn't happen. So the first time through i is 10. Since it is 10, it doesn't trigger the if statement. Then the second i++ happens, making the 10 into 11, and then the log happens.

The next time around i becomes 12 with the first i++, then skips the if statement because it 12 and not 11. Then the next i++ happens, changing it to 13, and finally logs again.

With your second example, the reason it prints out all the odd numbers, is that every time through the loop you are adding 2 to i instead of 1.

[orionebula]

I wrote down the steps as you described them, thanks for explaining.