為什么我可以等待承諾解決執行，而不是承諾拒絕執行？

Question

我一直在玩承諾，我發現了一種奇怪的行為。 為什么在我等待resolve調用時會等待這個承諾，而在我等待reject調用時卻沒有等待？

 (async () => { let resolve, reject; const promise = new Promise((_resolve, _reject) => { resolve = _resolve; reject = _reject; }); promise.then(_ => console.log('Promise resolved')); await resolve(); // Instead of "resolve(); await promise;", which would be more natural console.log('Thread still running'); })()

 (async () => { let resolve, reject; const promise = new Promise((_resolve, _reject) => { resolve = _resolve; reject = _reject; }); await reject(); // Instead of "reject(); await promise;", which would be more natural console.log('Thread still running'); })()

Answer 1

您只能有用地await承諾。

resolve返回undefined 。 reject返回undefined 。 這些都不是承諾，所以await沒有實際效果。

 (async() => { let resolve, reject; const promise = new Promise((_resolve, _reject) => { resolve = _resolve; reject = _reject; }); promise.then(_ => console.log('Promise 1 resolved')); const resolve_return_value = resolve(); console.log({ resolve_return_value }); await resolve_return_value; console.log('Thread 1 still running'); })(); (async() => { let resolve, reject; const promise = new Promise((_resolve, _reject) => { resolve = _resolve; reject = _reject; }); const reject_return_value = reject(); console.log({ reject_return_value }); await reject_return_value; console.log('Thread 2 still running'); })();

當您將一個函數傳遞給Promise構造函數時，它會被賦予兩個參數，這些參數是允許您解析或拒絕該承諾的函數。

不要將它們與Promise.reject()和Promise.resolve()混淆，它們返回立即被拒絕或解決的新承諾。