Generic type not assignable if extends type with optional field and passed object does not include the optional field

TypeScript Version: 3.9.2

Search Terms:
optional generic "not assignable"

Code

const foo = <V extends { s?: string }>(v: V) => {};

// This works:
foo({ s: "1", v: 2 });

// This doesn't work:
foo({ v: 2 }); // Argument of type '{ v: number; }' is not assignable to parameter of type '{ s?: string | undefined; }'. Object literal may only specify known properties, and 'v' does not exist in type '{ s?: string | undefined; }'.

Compare with

const bar = <V extends { }>(v: V) => {};

// Both work:
bar({ s: "1", v: 2 });
bar({ v: 2 });

Expected behavior:
Both statements in the first snippet should compile.

Actual behavior:
foo({ v: 2 }); doesn't compile for const foo = <V extends { s?: string }>(v: V) => {};

Playground Link:
https://www.typescriptlang.org/play?#code/MYewdgzgLgBAZiEMC8MA8A1GBTAHlbMAEwhgG8YIB+ALkqgCcBLMAcxgF8A+ACgDc6GAJQou5ALAAoKRwDcUhCB4UIdAEQBGNQBoYAmACZOQ+ZMXK9dIxxNSpoSLABGAQwYp0WPAWKkK3fkERZDEyOTtJAHpImAAhECgACxgAdxAGAGsaKVcGC1UYTR1LQ2NTXIt9axMgA

Related Issues:

Possible remotely related... https://github.com/microsoft/TypeScript/issues/13747

Commencez par exécuter la reproduction TypeScript 3.9.2 dans le lien Playground fourni et comparez les deux appels génériques. Suivez le comportement de la vérification des types pour les propriétés excédentaires sur une contrainte générique avec un champ optionnel. C'est terminé lorsque les deux appels du premier extrait compilent sans modifier le comportement montré pour le second extrait.