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typescript/no-unsafe-return Pedantic

💭 This rule requires type information.

What it does

This rule disallows returning a value with type any from a function.

Why is this bad?

The any type in TypeScript disables type checking. When you return a value typed as any from a function, you're essentially passing the type-safety problem to the caller without providing any guarantees about what the function actually returns.

Examples

Examples of incorrect code for this rule:

ts
declare const anyValue: any;

function getString(): string {
  return anyValue; // unsafe return
}

const getNumber = (): number => anyValue; // unsafe return

function processData(): { name: string; age: number } {
  return anyValue; // unsafe return
}

Examples of correct code for this rule:

ts
declare const stringValue: string;
declare const numberValue: number;
declare const unknownValue: unknown;

function getString(): string {
  return stringValue; // safe
}

const getNumber = (): number => numberValue; // safe

function processUnknown(): unknown {
  return unknownValue; // safe - explicitly returning unknown
}

// Type guard to safely return
function safeGetString(): string | null {
  if (typeof unknownValue === "string") {
    return unknownValue; // safe after type guard
  }
  return null;
}

How to use

To enable this rule using the config file or in the CLI, you can use:

json
{
  "rules": {
    "typescript/no-unsafe-return": "error"
  }
}
bash
oxlint --type-aware --deny typescript/no-unsafe-return

References