JavaScript style for optional callbacks

Question

I have some functions which occasionally (not always) will receive a callback and run it. Is checking if the callback is defined/function a good style or is there a better w

Answer 1

Rather than make the callback optional, just assign a default and call it no matter what

const identity = x =>
  x

const save (..., callback = identity) {
  // ...
  return callback (...)
}

When used

save (...)              // callback has no effect
save (..., console.log) // console.log is used as callback

Such a style is called continuation-passing style. Here's a real example, combinations, that generates all possible combinations of an Array input

const identity = x =>
  x

const None =
  Symbol ()

const combinations = ([ x = None, ...rest ], callback = identity) =>
  x === None
    ? callback ([[]])
    : combinations
        ( rest
        , combs =>
            callback (combs .concat (combs .map (c => [ x, ...c ])))
        )

console.log (combinations (['A', 'B', 'C']))
// [ []
// , [ 'C' ]
// , [ 'B' ]
// , [ 'B', 'C' ]
// , [ 'A' ]
// , [ 'A', 'C' ]
// , [ 'A', 'B' ]
// , [ 'A', 'B', 'C' ]
// ]

Because combinations is defined in continuation-passing style, the above call is effectively the same

combinations (['A', 'B', 'C'], console.log)
// [ []
// , [ 'C' ]
// , [ 'B' ]
// , [ 'B', 'C' ]
// , [ 'A' ]
// , [ 'A', 'C' ]
// , [ 'A', 'B' ]
// , [ 'A', 'B', 'C' ]
// ]

We can also pass a custom continuation that does something else with the result

console.log (combinations (['A', 'B', 'C'], combs => combs.length))
// 8
// (8 total combinations)

Continuation-passing style can be used with surprisingly elegant results

const first = (x, y) =>
  x

const fibonacci = (n, callback = first) =>
  n === 0
    ? callback (0, 1)
    : fibonacci
        ( n - 1
        , (a, b) => callback (b, a + b)
        )
        
console.log (fibonacci (10)) // 55
// 55 is the 10th fibonacci number
// (0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...)