typescript/no-meaningless-void-operator Correctness

✅ This rule is turned on by default.

🚧 An auto-fix is still under development.

What it does

This rule disallows the void operator when its argument is already of type void or undefined.

Why is this bad?

The void operator is useful when you want to execute an expression and force it to evaluate to undefined. However, using void on expressions that are already of type void or undefined is meaningless and adds unnecessary complexity to the code.

Examples

Examples of incorrect code for this rule:

function foo(): void {
  return;
}

void foo(); // meaningless, foo() already returns void

void undefined; // meaningless, undefined is already undefined

async function bar() {
  void (await somePromise); // meaningless if somePromise resolves to void
}

Examples of correct code for this rule:

function getValue(): number {
  return 42;
}

void getValue(); // meaningful, converts number to void

void console.log("hello"); // meaningful, console.log returns undefined but we want to be explicit

function processData() {
  // some processing
}

processData(); // no void needed since we don't care about return value

How to use

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

CLI

oxlint --deny typescript/no-meaningless-void-operator

Config (.oxlintrc.json)

{
  "rules": {
    "typescript/no-meaningless-void-operator": "error"
  }
}

References

• Rule Source