Any type is inferred when function input parameter is set to some value

TypeScript Version: 3.8.2

Search Terms: any type is inferred

Code

interface ContinuationQueryResult<T> {
    documents: T[];
    continuation?: string;
}

function abc(continuation?: string): Promise<ContinuationQueryResult<string>> {
    return <any>undefined;
}

async function test() {
    // When not explicitly set to undefined type is inferred correctly.
    let continuation: string | undefined = undefined;

    do {
        const summaries = await abc(continuation);

        continuation = summaries.continuation;
    } while (continuation)
}

Expected behavior:
Type of summaries variable should be ContinuationQueryResult<string>.

Actual behavior:
summaries variable is inferred as any.

Playground Link: https://www.typescriptlang.org/play/#code/JYOwLgpgTgZghgYwgAgMIHtygK5zMTARW2gE8AlCAZ2wBswAeAFQD5kBvAWAChk-kAJugTYAthHBUAXMiYBtALoBuHv2QJM+ELnyYA-DKpgooAOYruAXx48Y2EAl0hkcAEYIAFBqza8BEAbIRiYgpgCUMgAKUOiiwFQQDBg+Ov7EZJQ09AzBZixsXLz8UBBg2FDODHAgpCz2AhAwoBACFtbcPHBUpA7Idg5OyJBGHmEcqvwA9JPIAOoAFhLIIOhgyBAAHgAOtMAIwGC0pEGlQ+jI9Y3NAkOkWyjxyKAw0CU3GlAljkcAdBN8tFO3i0qUwhmMZmQAB8LiAGk0QC1kABeWHw64Wf6Cc6FNRqbxGIJiURwEzUFEuADucAOLncXk0OD8mDCmKKePUjN8g1RNFEJLJVB+wKZTgsaksyEp82AgOQDJSzJAYR4liAA

Related Issues:

リンクされた TypeScript Playground の再現コードとテスト関数から始め、continuation を undefined で初期化した場合と別の値で初期化した場合の推論を比較してください。オプション引数の型チェック経路を追跡し、回帰テストを追加してください。summaries が any ではなく ContinuationQueryResult として推論されれば完了です。