使用联合重载时,是否可以获取函数参数的上下文类型?
declare function f(...args: [x: 'a', y: (a: string) => void] | [x: 'b', y: (a: boolean) => void]): void;
// Parameter 'x' implicitly has an 'any' type.(7006)
f('a', x => {});
declare function g(x: 'a', y: (a: string) => void): void;
declare function g(x: 'b', y: (a: boolean) => void): void;
// Ok.
g('a', x => {});
游乐场链接 https://www.typescriptlang.org/play?#code/CYUwxgNghgTiAEAzArgOzAFwJYHtVIAoA6E2AcwGcAueAbQA8aByKJgGngE8aCoaKMMLKjIBKeAF4AfPABuOLMAC68AD51G8JgCN2XHn3jacOCCCipx0uQuWia8xQG4AUC4D07+AAVYUALYgGCAwWvRM8Fj+AA4QWGBYGBCc8AAWUBTwFloWnBEYnNEgRAQA7AAM5QBsoi6IBCx69JIyAN4AvqKuLqCQsAgo6Nh48GQEmo0c3PC8-ILCYi02ivbLwK690HBIaJi4+GMTulMGNMam5pZLjsCrN92e8ADyANZELmOT8M3WHV1AA
尤其是工作结束后相关参数 https://devblogs.microsoft.com/typescript/announcing-typescript-4-6/#dependent-parameters-cfa,我认为这会起作用吗?是否有一个开放的问题来跟踪这个问题?
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