One-liner to call a function if it's defined, else call another function

Question

I want to call a callback function if one was provided by the user, or default to a default defaultCallback function.

I've done it as follows:

function defaultCallback(x) {
  console.log('default callback ' + x)
}

function test(callback) {
  let x = 'x'

  if (callback) {
    callback(x)
  } else {
    defaultCallback(x)
  }
}

I feel there should be a more concise way of doing this?

you could set the default argument. `function test(callback = defaultCallback)` — rlemon, May 29 '18 at 14:28

Daniel Gimenez · Answer 1 · 2018-05-29T14:39:16.017

2

You can just use || operator to get callback or fallback to defaultCallback.

function test(callback) {
   (callback || defaultCallback)('x')
}

Here is a test snippet you can use to view the results in the console.

function defaultCallback(x) { console.log('Used default ' + x); }

function test(callback) {
  (callback || defaultCallback)('x')
}

test(undefined);
test((y) => console.log('Used func ' + y));

edited May 29 '18 at 14:39

answered May 29 '18 at 14:28

Daniel Gimenez

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Bergi · Accepted Answer · 2018-05-29T14:31:14.030

1

I think you're looking for the conditional operator:

function test(callback) {
  let x = 'x'

  (typeof callback == "function" ? callback : defaultCallback)(x);
}

Alternatively, if you don't want to check the type of the argument, but just assert that it's something else than undefined, you can use a default initialiser in your parameter:

function test(callback = defaultCallback) {
  let x = 'x'

  callback(x);
}

edited May 29 '18 at 14:31

answered May 29 '18 at 14:29

Bergi

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2

I think a downvote warrants an explanation in this case ?? Why downvoted? – Zak May 29 '18 at 14:31

score 0 · Answer 3 · answered May 29 '18 at 14:30

0

How about:

function test(callback) {
  let x = 'x'

  callback ? callback(x) : defaultCallBack(x);
}

answered May 29 '18 at 14:30

RicardoAgra

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And why is this downvoted? Does it not work? – RicardoAgra May 29 '18 at 14:37

One-liner to call a function if it's defined, else call another function

3 Answers3