javascript - 回调函数抛出错误

Question

我正在尝试解决这个异步问题。我解决了问题。

但是，在调用“完成”回调后，它会抛出一个错误。我不知道为什么。

问题:

I want to print the task number once "Done" is invoked. But it throws an error saying TypeError: "callback" argument must be a function

问题:

A "TaskRunner" Constructor takes one argument, "concurrency", and exposes one method "push" on its prototype. The "push" method takes one argument "task" which is a "function"

Passing a task to a push method of a TaskRunner instance should immediately execute (call/run/invoke) the task, unless the number of currently running tasks exceeds the concurrency limit.

function TaskRunner(concurrency) {
    this.count = concurrency;
    this.queue = [];
  }
  
  TaskRunner.prototype.push = function(task) {
    if (this.count === 0) {
      this.queue.push(task);
    } else {
      this.invoke(task);
    }
  }
  
  TaskRunner.prototype.invoke = function(task) {
    task.call(null, this.done.bind(this));
    this.count--;
  }
  
  TaskRunner.prototype.done = function(num) {
    console.log(`After Executing done: ${num}`)
    this.count++;
    this.execute();
  }
  
  TaskRunner.prototype.execute = function() {
    if (this.queue.length > 0) {
      var task = this.queue.shift();
      this.invoke(task);
    }
  }

  function factory(num) {
    return function exampleSimpleTask(done) {
      this.num = num;
      console.log("task", "Before " + new Date().getTime());
      setTimeout(done(num), 2000);
    }
  }


  var r = new TaskRunner(2);

r.push(factory(1));
r.push(factory(2));
r.push(factory(3));

编辑:出于某种原因，在 jsfiddle 上它运行良好，但是当我在本地运行相同的代码时它失败了。

请帮忙。

Answer 1

您将函数的结果传递给setTimeout:

setTimeout(done(num), 2000);

这将立即调用 done(num) 并且 setTimeout 将尝试调用返回的任何 done() ，因为它是一个函数。

你应该给它传递一个它可以调用的函数:

setTimeout(() => done(num), 2000);

或者[正如@JaromandaX 在评论中指出的那样] 你可以利用 setTimeOut 的选项第三个参数，它将被传递到回调函数中:

setTimeout(done, 2000, num);

这将调用函数done 并传入num