consistent-generic-constructors

Enforce specifying generic type arguments on type annotation or constructor name of a constructor call.

When constructing a generic class, you can specify the type arguments on either the left-hand side (as a type annotation) or the right-hand side (as part of the constructor call):

// Left-hand side
const map: Map<string, number> = new Map();

// Right-hand side
const map = new Map<string, number>();

This rule ensures that type arguments appear consistently on one side of the declaration.

Attributes

• Included in configs
  • ✅ Recommended
  • 🔒 Strict
• Fixable
  • 🔧 Automated Fixer
  • 🛠 Suggestion Fixer
• 💭 Requires type information

Options

{
  "rules": {
    "@typescript-eslint/consistent-generic-constructors": [
      "error",
      "constructor"
    ]
  }
}

This rule takes a string option:

• If it's set to constructor (default), type arguments that only appear on the type annotation are disallowed.
• If it's set to type-annotation, type arguments that only appear on the constructor are disallowed.

Rule Details

The rule never reports when there are type parameters on both sides, or neither sides of the declaration. It also doesn't report if the names of the type annotation and the constructor don't match.

constructor

❌ Incorrect

const map: Map<string, number> = new Map();
const set: Set<string> = new Set();

✅ Correct

const map = new Map<string, number>();
const map: Map<string, number> = new MyMap();
const set = new Set<string>();
const set = new Set();
const set: Set<string> = new Set<string>();

type-annotation

❌ Incorrect

const map = new Map<string, number>();
const set = new Set<string>();

✅ Correct

const map: Map<string, number> = new Map();
const set: Set<string> = new Set();
const set = new Set();
const set: Set<string> = new Set<string>();

When Not To Use It

You can turn this rule off if you don't want to enforce one kind of generic constructor style over the other.