110

Are there any standard ways of marking a function argument as unused in JavaScript, analogous to starting a method argument with an underscore in Ruby?

Andrew Grimm
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6 Answers6

82

Just so we have an example to work from, this is fairly common with jQuery's $.each where you're writing code that doesn't need the index, just the value, in the iteration callback and you're using this (which jQuery also sets to the value) for something else:

$.each(objectOrArrayLikeThing, (_, value) => {
    // Use `value` here
});

(Yes, $.each passes arguments to the callback backward compared to the JavaScript standard forEach.)

Using _ is the closest I've seen to a standard way to do that, yes, but I've also seen lots of others — giving it a name reflective of its purpose anyway (index), calling it unused, etc.

If you need to ignore more than one parameter, you can't repeat the same identifier (it's disallowed in strict mode, which should be everyone's default and is the default in modules and class constructs), so you have do things like _0 and _1 or _ and __, etc.

T.J. Crowder
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    you can't do function (_, _, value) though – Hanming Zeng Oct 04 '18 at 21:23
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    @HanmingZeng - Indeed not. :-( You have to do things like `(_0, _1, value)` or (in my case) `(unused0, unused1, value)`. I've also seen `(_, __, value)` but...yikes. :-) There was a strawman proposal to allow blank parameter names once upon a time but it never went anywhere. – T.J. Crowder Oct 05 '18 at 06:56
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    In Chromium it seems to work just fine to write a function declaration like `function (_, _, a) { … }` (of course inside the function the variable `_` only contains the second passed argument) but there's no error and the function is created (and works) just fine. – zrajm Aug 12 '19 at 12:27
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    @zrajm - Only in loose mode. In strict mode, duplicate parameter names are disallowed. (It's best to always use strict mode except in rare situations where you can't. It's also the default for modules and the content of `class` constructs.) – T.J. Crowder Aug 12 '19 at 12:31
59

Using destructuring assignment, one can do:

function f(...[, , third]) {
  console.log(third);
}

f(1, 2, 3);
Joseph Marinier
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29

With browsers supporting destructuring one can do:

function ({}, {}, value) {
  // console.log(value)
}

Which is kind of neat in that it avoids the problem of multiple arguments having the same name and also won't create problems with libraries that assign methods to _ (lodash, underscore, etc.).

One problem with this approach is that unused arguments of type undefined or null will throw.

For undefined one solution is to use default parameters:

function ({}={}, {}={}, value) {
  // console.log(value)
}

Sadly no such easily applied solution for null.

Fredrik Carlsson
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18

I would recommend this syntax:

function(_index, value) {...}

to not to shadow lodash variable and still have description of argument in case if it will be used.

VS Code is also highlight these names properly and these unused args won't be deleted after autofix code smells

MarkosyanArtur
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2

How about using the function arguments object? 

function third() { const [,,thirdArg] = arguments;
  return thirdArg;
}
console.log(third(1,2,3));
Doug Coburn
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  • I like the answer by @Joseph Marinier even better https://stackoverflow.com/questions/32197927/standard-conventions-for-indicating-a-function-argument-is-unused-in-javascript/58738236#58738236 – Doug Coburn Nov 18 '19 at 14:54
1

Another approach: For unused parameters to be ignore from unused errors using:

function(_1, _2, toBeUseParam) { ... }
Ahsan Farooq
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