中返回值和返回promise时，执行顺序有何不同。然后？

Question

原文是这样的：

Promise.resolve('foo')
  // 1. Receive "foo" concatenate "bar" to it and resolve that to the next then
  .then(function(string) {
    return new Promise(function(resolve, reject) {
      setTimeout(function() {
        string += 'bar';
        resolve(string);
      }, 1);
    });
  })
  // 2. receive "foobar", register a callback function to work on that string
  // and print it to the console, but not before return the unworked on
  // string to the next then
  .then(function(string) {
    setTimeout(function() {
      string += 'baz';
      console.log(string);
    }, 1)
    return string;
  })
  // 3. print helpful messages about how the code in this section will be run
  // before string is actually processed by the mocked asynchronous code in the
  // prior then block.  
  .then(function(string) {
    console.log("Last Then:  oops... didn't bother to instantiate and return " +
                "a promise in the prior then so the sequence may be a bit " +
                "surprising");

    // Note that `string` will not have the 'baz' bit of it at this point. This 
    // is because we mocked that to happen asynchronously with a setTimeout function
    console.log(string);
});

这确实令人惊讶，但最终明白了（嗯，至少我想我明白了）为什么会这样发生。

现在，我将第二个.然后更改为：

  .then(function(string) {
    setTimeout(function() {
      string += 'baz';
      console.log(string);
      return string;
    }, 1)

如果我遗漏了一些明显的东西，或者我的问题是愚蠢的，让我提前说声对不起。还有谢谢你！

Answer 1

我得到最后一个。然后在第二个之前执行。然后返回一个值。

不，您只是没有看到第二个和返回值的位置。您的return语句位于settimeout回调中，并从该函数返回（稍后），而不是从then回调返回。事实上，该函数没有任何return语句，因此默认情况下，它在执行主体后返回undefined。

你基本上在做

.then(function(string) {
  setTimeout(function() {
    string += 'baz';
    console.log(string);
    return string; // irrelevant for the promise chain
  }, 1)
  return undefined; // this is the value that the promise resolves with
//       ^^^^^^^^^     and which the next `then` callback receives
})