Question

我的目标是获得以下输出：

1,2（等待2s ）3,4（等待2s ）5但是我得到1,2,4（等待2s）3,5。 / p>

有谁知道如何解决这个问题？

console.log("1");
console.log("2");
setTimeout(function(){console.log("3")}, 2000);
console.log("4");
setTimeout(function(){console.log("5")}, 2000);

Answer 1

因为setTimeout是异步的。 JS转到第一行，打印1，然后打印第二行，打印2，然后打印第三行 - 它是异步，所以它将在稍后执行（在放置时为2000ms），然后在第四行，打印4和第五行，在2000ms执行（作为第三行）。因此，您可以立即获得1,2和4，然后在3和5之后获得2000毫秒。要获得所需内容，请参阅以下代码：

＆＃13;

console.log("1")
console.log("2")

setTimeout(function() {
  console.log("3")
  console.log("4")
}, 2000)

setTimeout(function() {
  console.log("5")
}, 4000)

＆＃13;

Answer 2

以下是使用-keep class kotlin.reflect.jvm.internal.** { *; } -dontwarn kotlin.reflect.jvm.internal.** ..

的简单示例

async / await

Answer 3

您需要了解how JavaScript runs the code：

当您调用代码时，它会运行完成，这意味着代码会立即执行。设置超时的作用是将内部函数推送到事件循环以便稍后执行。让我评论一下发生了什么，请记住堆栈的执行模型是先进先出。执行return时，函数将从堆栈中弹出：

// a log function is pushed to stack and popped
console.log("1");

// another log function is pushed to stack and popped
console.log("2");

// setTimeout is pushed to stack and popped
// When popped it added an anonymous function to the event loop
setTimeout(function(){console.log("3")}, 2000);

// another log function is pushed to stack
console.log("4");

// settimeout is pushed to stack and popped

// When popped it added an anonymous function to the event loop
setTimeout(function(){console.log("5")}, 2000);

当没有更多功能可以继续堆叠时，JavaScript引擎开始执行事件循环。因此setTimeous的内部函数等待被推入堆栈。当超时时，它们将被推到堆叠状态。

因此您需要更改代码，如下所示：

// a log function is pushed to stack and popped
console.log("1");

// another log function is pushed to stack and popped
console.log("2");

// first settimeout is pushed to stack
// When executed inner function will be added to event loop because of setTimeout
// In 2000 miliseconds, the event loop will push this function to the stack
setTimeout(function(){
  // When executed
  // another log function will be pushed to stack and popped
  console.log("3")
  
  // another log function will be pushed to stack and popped
  console.log("4");
  
  // second setTimeout is pushed to stack, which in turn pushes inner anonymous function to event loop
  setTimeout(function(){

    // another log function will be pushed to stack and popped
    console.log("5");
      
    // After this point second setTimeout and first setTimeout will be popped in order
      
  }, 2000);
  
}, 2000);

Answer 4

＆＃13;

console.log("1");
console.log("2");
setTimeout(function(){
  console.log("3");
  console.log("4");
  }, 2000);
setTimeout(function(){console.log("5")}, 4000);

＆＃13;

Answer 5

async / await

的最佳解释之一

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Statements/async_function

同步执行和setTimeout（）

5 个答案: