no-useless-spread should report when passing spreading parameters to new Set

[zanminkian]

1️⃣ Explain here what's wrong

function getNames() {
  return ['foo', 'bar', 'foo'];
}

const uniqueNames = [...new Set(...getNames())]; // I mistakenly spreading the `getNames()`
console.log(uniqueNames) // Print [ 'f', 'o' ]. But what I expect is ['foo', 'bar']

no-useless-spread should report error on the code new Set(...getNames()). But actually not.

Because new Set() accept one or zero parameter, we don't need to spread the parameters.

Example

const foo = ['abc'];
const set = new Set(...foo); // BAD!
const set = new Set(foo[0]); // OK

2️⃣ Specify which rule is buggy here and in the title
no-useless-spread

3️⃣ More information
TypeScript should report error when compiling the code above. Unfortunately, this problem havn't been fixed yet. Set microsoft/TypeScript#59390 and microsoft/TypeScript#48575

[fisker]

Because new Set() accept one or zero parameter, we don't need to spread the parameters.

This doesn't mean it can't spread the parameters

const foo = [[1,2]]

new Set(...foo)

[zanminkian]

This doesn't mean it can't spread the parameters

const foo = [[1,2]]

new Set(...foo)

It can, but it's unnecessary.

const foo = [[1,2]];
new Set(foo[0]);

Only the first item will be used in new Set(...foo). So Just use new Set(foo[0]).

Passing spreading parameters to new Set is a bad code style.

• It's confused.
• Not type safe enough (due to TypeScript bug).

[fregante]

There's a slight difference I think. no-useless-spread suggests that you "don't need it", while here it leads to a different result altogether, probably not what you wanted in the first place.

However I think it should still report it because the spread only works correctly when foo.length <= 1.

const foo = []; // Empty
new Set(...foo) == new Set() == new Set(undefined); // Ok, works just as well

const foo = [[1]]; // One
new Set(...foo) == new Set(foo[0]); // Ok, works just as well

const foo = [[1], [2]]; // Two
new Set(...foo) == new Set(foo[0], foo[1]); // ❌ Item 2 lost

The only caution would be to not autofix it in this case.

[zanminkian]

const foo = [[1], [2]]; // Two
new Set(...foo) == new Set(foo[0], foo[1]); // ❌ Item 2 lost

The only caution would be to not autofix it in this case.

const foo = [[1], [2]]; // Two
new Set(...foo) == new Set(foo[0], foo[1]) == new Set(foo[0]); // Ok, works just as well

Auto fixing is safe. Just confidently change all new Set(...foo) to new Set(foo[0]).

[fregante]

That only works if you know the type of foo. If it’s an iterator other than Array that breaks it.

[zanminkian]

My fault.

const map = new Map([['a','b']]);
new Set(...map); // Set(2) { 'a', 'b' }
new Set(map[0]); // Set(0) {}