了解 javascript function 調用

Question

仍在學習 javascript。 下面的代碼是如何工作的？ 當調用main(params)是否意味着它在后台調用dowork(callbackXYZ, options, params) ？

const { dowork } = require('some_lib');

async function callbackXYZ(src, tgt, params) => {
    // ...logic
}
const main = dowork(callbackXYZ, options);
await main(params);

Answer 1

這是一個簡化的代碼示例。 它刪除了異步的東西，因為它不需要理解。

 // The callback you provide to `dowork` // The internals of `dowork` will give invoke it with the given args function callbackXYZ(src, tgt, params) { console.log("src is:", src); console.log("tgt is:", tgt); console.log("params is:", params); } // The options you provide to `dowork` var options = {src: "foo", tgt: "bar"}; // Invoking `dowork` receives your callback and options, and returns // a new function that has access to both of these arguments. const main = dowork(callbackXYZ, options); // The params you provide to the returned function const params = {my: "params"}; // Invoking the returned function main(params); // Here's the `dowork`. It receives your callback and options, and it // returns a new function that expects you to pass it params. // So that returned function has reference to the callback, options // and params. // The body of the returned function, simply invokes your callback and // passes it data from the options and params you provided function dowork(callback, opts) { return function(params) { callback(opts.src, opts.tgt, params); } }

因此， dowork接收您的callbackXYZ和opts並返回您可以調用的 function ，傳入params 。

當您調用返回的 function 並將其傳遞給params時，返回的 function 將調用您的原始回調，並將來自options和params的數據傳遞給它。

Answer 2

不！

• dowork是 function ，它返回另一個 function
• 將另一個 function 分配給const main
• 然后調用這個main（返回promise的函數）

您也可以在不聲明任何內容的情況下運行此代碼：

...
await dowork(callbackXYZ, options)(params);