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prefer-nullish-coalescing

Enforce using the nullish coalescing operator instead of logical chaining.

🔒

Extending "plugin:@typescript-eslint/strict" in an ESLint configuration enables this rule.

💡

Some problems reported by this rule are manually fixable by editor suggestions.

💭

This rule requires type information to run.

The ?? nullish coalescing runtime operator allows providing a default value when dealing with null or undefined. Because the nullish coalescing operator only coalesces when the original value is null or undefined, it is much safer than relying upon logical OR operator chaining ||, which coalesces on any falsy value.

This rule reports when an || operator can be safely replaced with a ??.

.eslintrc.cjs
module.exports = {
  "rules": {
    "@typescript-eslint/prefer-nullish-coalescing": "warn"
  }
};

Options​

This rule accepts an options object with the following properties:

interface Options {
  ignoreConditionalTests?: boolean;
  ignoreTernaryTests?: boolean;
  ignoreMixedLogicalExpressions?: boolean;
}

const defaultOptions: Options = [
  {
    ignoreConditionalTests: true,
    ignoreTernaryTests: true,
    ignoreMixedLogicalExpressions: true,
  },
];

ignoreTernaryTests​

Setting this option to true (the default) will cause the rule to ignore any ternary expressions that could be simplified by using the nullish coalescing operator.

Incorrect code for ignoreTernaryTests: false, and correct code for ignoreTernaryTests: true:

const foo: any = 'bar';
foo !== undefined && foo !== null ? foo : 'a string';
foo === undefined || foo === null ? 'a string' : foo;
foo == undefined ? 'a string' : foo;
foo == null ? 'a string' : foo;

const foo: string | undefined = 'bar';
foo !== undefined ? foo : 'a string';
foo === undefined ? 'a string' : foo;

const foo: string | null = 'bar';
foo !== null ? foo : 'a string';
foo === null ? 'a string' : foo;

Correct code for ignoreTernaryTests: false:

const foo: any = 'bar';
foo ?? 'a string';
foo ?? 'a string';
foo ?? 'a string';
foo ?? 'a string';

const foo: string | undefined = 'bar';
foo ?? 'a string';
foo ?? 'a string';

const foo: string | null = 'bar';
foo ?? 'a string';
foo ?? 'a string';

ignoreConditionalTests​

Setting this option to true (the default) will cause the rule to ignore any cases that are located within a conditional test.

Generally expressions within conditional tests intentionally use the falsy fallthrough behavior of the logical or operator, meaning that fixing the operator to the nullish coalesce operator could cause bugs.

If you're looking to enforce stricter conditional tests, you should consider using the strict-boolean-expressions rule.

Incorrect code for ignoreConditionalTests: false, and correct code for ignoreConditionalTests: true:

declare const a: string | null;
declare const b: string | null;

if (a || b) {
}
while (a || b) {}
do {} while (a || b);
for (let i = 0; a || b; i += 1) {}
a || b ? true : false;

Correct code for ignoreConditionalTests: false:

declare const a: string | null;
declare const b: string | null;

if (a ?? b) {
}
while (a ?? b) {}
do {} while (a ?? b);
for (let i = 0; a ?? b; i += 1) {}
a ?? b ? true : false;

ignoreMixedLogicalExpressions​

Setting this option to true (the default) will cause the rule to ignore any logical or expressions that are part of a mixed logical expression (with &&).

Generally expressions within mixed logical expressions intentionally use the falsy fallthrough behavior of the logical or operator, meaning that fixing the operator to the nullish coalesce operator could cause bugs.

If you're looking to enforce stricter conditional tests, you should consider using the strict-boolean-expressions rule.

Incorrect code for ignoreMixedLogicalExpressions: false, and correct code for ignoreMixedLogicalExpressions: true:

declare const a: string | null;
declare const b: string | null;
declare const c: string | null;
declare const d: string | null;

a || (b && c);
(a && b) || c || d;
a || (b && c) || d;
a || (b && c && d);

Correct code for ignoreMixedLogicalExpressions: false:

declare const a: string | null;
declare const b: string | null;
declare const c: string | null;
declare const d: string | null;

a ?? (b && c);
(a && b) ?? c ?? d;
a ?? (b && c) ?? d;
a ?? (b && c && d);

NOTE: Errors for this specific case will be presented as suggestions (see below), instead of fixes. This is because it is not always safe to automatically convert || to ?? within a mixed logical expression, as we cannot tell the intended precedence of the operator. Note that by design, ?? requires parentheses when used with && or || in the same expression.

When Not To Use It​

If you are not using TypeScript 3.7 (or greater), then you will not be able to use this rule, as the operator is not supported.

Further Reading​

Resources​