still learning to javascript. How does this below code work?
when main(params) is called does it mean it under-the-hood calls dowork(callbackXYZ, options, params) ?
const { dowork } = require('some_lib');
async function callbackXYZ(src, tgt, params) => {
// ...logic
}
const main = dowork(callbackXYZ, options);
await main(params);
Here's a simplified code example. It removes the async stuff, since it's not needed to understand.
// 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);
}
}So dowork receives your callbackXYZ and the opts and returns a function that you can invoke, passing in the params.
When you invoke that returned function, and pass it the params, that returned function invokes your original callback, and passes it data both from the options and the params.