I've always thought of a reduce function as a way to create an aggregation. For example, to add up all the numbers in a list:

[1, 2, 3].reduce((memo, i) => memo + i);
//=> 6

Because of this, I've always assumed that the first argument to the reduce function callback (memo) is mutated in each iteration of the function and so, at the end of the iterations, memo becomes the final mutated value.

Last week I learned that memo isn't being mutated, it's being replaced by the return value of the previous loop.

That means that reduce isn't simply an aggregator function, it can be used in other flexible ways:

For example:

[1, 2, 3].reduce((memo, i) => 2);
// => 2

[1, 2, 3].reduce((memo, i) => i);
// => 3

So you can apply arbitrary logic to your memo value:

[1, 2, 3].reduce((memo, i) => {
  let returnValue;
  switch (i) {
    case 1:
      returnValue = memo + i;
      break;
    case 2:
      returnValue = memo - i;
      break;
    case 3:
      returnValue = memo * i;
      break;
    default:
      returnValue = memo / i;
  }
  return returnValue;
}, 0);
// => -3

To understand this flexibility, t helps me to replace the memo moniker with previousReturnValue instead.