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 ... Why go through the trouble of Array.apply(null, {length: N}) instead of just Array(N)? After all, both expressions would result an an N -element array of undefined elements. The difference is that in the former expression, each element is explicitly set to undefined, whereas in the latter, each element was never set. 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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