Why array.forEach(() => { array.pop() }) would not empty the array

Modifying an array while iterating over it is generally a bad idea. In fact, in Java, trying to do so would cause an exception to be thrown. But let's convert the forEach into an old-school for loop, and maybe you'll see the issue.

for (let i = 0; i < a.length; ++i) {
    a.pop();
}

Is it clearer now what's going on? Each iteration you're shortening the length of the array by 1 when you pop the last element off. So the loop will end after iterating over half the elements -- because by then, it will have REMOVED half the elements, too, causing the value of i to be more than the current length of the array.

The same thing is happening when you use forEach: you're shortening the array with each iteration when you pop, causing the loop to terminate after only half the elements have been iterated. In other words, the iterator variable will move forward past the end of the array as the array shrinks.

Check out this quote from here: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/forEach

If the values of existing elements of the array are changed, the value passed to callback will be the value at the time forEach() visits them; elements that are deleted before being visited are not visited.

You're iterating from the beginning and removing the last element each iteration. This means you're moving forward 1, and reducing the length by 1, each iteration. Hence why you're ending up with floor(initialLength / 2) iterations. You're modifying the same array that you're forEaching, which as stated above means that you will not have the callback invoked for those pop'd elements.