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:

Inizia con la riproduzione collegata in TypeScript Playground e la funzione di test, confrontando l'inferenza quando continuation viene inizializzato a undefined invece che a un altro valore. Traccia il percorso del controllo dei tipi per l'argomento opzionale e aggiungi un test di regressione; il lavoro è completato quando summaries viene inferito come ContinuationQueryResult invece che come any.