prefer-spread: Add prefer spread over .slice()/.concat()

[fisker]

Currently, prefer-spread only check Array.from(), how about add prefer spread over .slice()/.concat()?

Fail

const foo = array.slice();

const foo = array.concat(item);

const foo = array.concat(anotherArray);

Pass

const foo = [...array];

const foo = array.slice(1);

The concat() case, we can't know item is array or not.

Maybe with suggestions?
Suggestion 1: [...array, itemOrArray]
Suggestion 2: [...array, ...itemOrArray]

Another problem is TypedArray#slice() and String#concat(). String#slice should not a problem. Nobody use string.slice()(no arguments).

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Any other method people use to clone an array?

[sindresorhus]

The concat() case, we can't know item is array or not.

Maybe, if the user is using the TypeScript parser (#347), we could handle it?

And if not, just use suggestions.

[sindresorhus]

Any other method people use to clone an array?

Array.from([1, 2]) and Array.of([1, 2]).

[fisker]

@sindresorhus I mean add to our existing rule prefer-spread, not new rule.

[fisker]

I feel bad that I can't distinguish Array and TypedArray, so I can't make this rule work on Array#slice()

Let's close this?

[sindresorhus]

@fisker Maybe we can make it opt-in? I have personally never used or seen .slice() used on TypedArray. For me, it's more valuable to catch Array#slice(). // eslint-disable are good for edge-cases like this.

[coreyfarrell]

Are you open to an option to ignore [].concat(value)? I use this pattern to normalize arguments where a single element or an array of elements is accepted. I know I could use Array.isArray(value) ? [...value] : [value] but IMO this is less readable.

[sindresorhus]

@coreyfarrell You can just use [value].flat() for that (we will eventually enforce that #975).

[coreyfarrell]

That is a reasonable replacement, thanks!