Should disallow using the result of calling a void-returning function

TypeScript Version: 4.1.0-dev.20201008

Search Terms: disallow function call void return usage warn error

Code

function doesntReturn() {
    // no return statement
}

const foo: string = doesntReturn() + 'foo'; // should be an error

Expected behavior:

The expression doesntReturn() + 'foo' should be an error because it uses the value of a void-returning function call. As far as I can tell, there is literally no good reason to use the return value of a function with no return statements like this.

This is similar to the current behavior of disallowing void-returning expressions inside control flow (issue #26262). It is also similar to the reasoning behind the warning that occurs when using a function value (instead of calling it) in an if condition.

Actual behavior:

No error message is given.

Playground Link: here

Related Issues: #26262

リンクされた Playground の例から始めて、不足している診断を再現し、その後、同様の control-flow の動作について関連する issue #26262 を読んでください。void を返す関数呼び出しの結果を使用する式がエラーを報告し、既存の例が引き続き期待どおりに動作すれば完了です。