useArraySortCompare

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JavaScript (and super languages)

Summary

Rule available since: v2.3.5
Diagnostic Category: lint/nursery/useArraySortCompare
This rule doesn’t have a fix.
The default severity of this rule is information.
This rule belongs to the following domains:
- project
Sources:
- Same as @typescript-eslint/require-array-sort-compare

How to configure

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{
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  "linter": {
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    "rules": {
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      "nursery": {
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        "useArraySortCompare": "error"
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      }
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    }
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  }
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}

Description

Require Array#sort and Array#toSorted calls to always provide a compareFunction.

When called without a compare function, Array#sort() and Array#toSorted() converts all non-undefined array elements into strings and then compares said strings based off their UTF-16 code units ECMA specification.

The result is that elements are sorted alphabetically, regardless of their type. For example, when sorting numbers, this results in a “10 before 2” order:

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[1, 2, 3, 10, 20, 30].sort(); //→ [1, 10, 2, 20, 3, 30]

This rule reports on any call to the sort methods that do not provide a compare argument.

Examples

Invalid

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const array: any[] = [];
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array.sort();

/invalid.ts:2:1 lint/nursery/useArraySortCompare ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

  ℹ Compare function missing.
  
    1 │ const array: any[] = [];
  > 2 │ array.sort();
      │ ^^^^^^^^^^^^
    3 │ 
  
  ℹ When called without a compare function, Array#sort() and Array#toSorted() converts all non-undefined array elements into strings and then compares said strings based off their UTF-16 code units.
  
  ℹ Add a compare function to prevent unexpected sorting.

Valid

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const array: any[] = [];
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array.sort((a, b) => a - b);