Web Reference: Sep 29, 2011 · An illustration. Suppose that array contains three integers, 0, 1, 2, and that i is equal to 1. array[i]++ changes array[1] to 2, evaluates to 1 and leaves i equal to 1. array[i++] does not modify array, evaluates to 1 and changes i to 2. A suffix operators, which you are using here, evaluates to the value of the expression before it is ... We'll use that fact later. Array.apply(null, [undefined, undefined, undefined]) is equivalent to Array(undefined, undefined, undefined), which produces a three-element array and assigns undefined to each element. How can you generalize that to N elements? Consider how Array() works, which goes something like this: Jul 29, 2009 · The third way of initializing is useful when you declare an array first and then initialize it, pass an array as a function argument, or return an array. The explicit type is required.
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