swap() function for variables's values

2018-06-06 17:32:10

This question already has an answer here:

Is JavaScript a pass-by-reference or pass-by-value language? 29 answers

Your function is swapping the values internally, but the function does not return a value.

The following method returns an array of values in reverse order of what was supplied.

function swap(x, y) {
    var t = x;
    x = y;
    y = t;
    return [x, y];
}

console.log(swap(2, 3));

If you don't actually need the values to swap:

function swap (x, y)
{
  return [y, x];
}

If you do need the values to swap, but you don't want to declare another variable:

function swap (x, y)
{
  x = x + y;
  y = x - y;
  x = x - y;
  return [x, y];
}

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