Node.js:异步回调执行这是Zalgo吗?

【字号: 日期:2024-04-16浏览:39作者:雯心
(adsbygoogle = window.adsbygoogle || []).push({}); 如何解决Node.js:异步回调执行这是Zalgo吗??

首先,让我解释一下代码的工作原理-请参阅添加的代码中的注释:

// first you define function A and nothing happens:function A(callback) { console.log(’A’); callback();}// then you define function B and nothing happens: function B() { console.log(’B’)}// then you define function C and nothing happens:function C() { console.log(’C’);}// Now you call function A with argument C:A(C);// when A runs it prints ’A’ and calls C before it returns// Now the C runs, prints C and returns - back to A// A Now has nothing more to do and returns// Now the execution continues and B can be run:B();// B runs and prints ’B’

这与任何语言(如Java,C等)完全一样。

现在,第二个示例:

// first you define function A and nothing happens:function A(callback) { console.log(’A’); process.nextTick(() => { callback(); });}// then you define function B and nothing happens:function B() { console.log(’B’)}// then you define function C and nothing happens:function C() { console.log(’C’);}// Then you run A with C passed as an argument:A(C);// A prints ’A’ and schedules running an anonymous function:// () => { callback(); }// on the next tick of the event loop, before I/O events are handled// but after the current code stack is unrolled// then it returns// And then B is run:B();// B prints ’B’ and returns// Now nothing else is left to do so the next tick of the event loop starts// There’s one function to run, scheduled earlier so it runs.// This function runs the `callback()` which was `C`// so C starts, prints ’C’ and returns// The anonymous function has nothing else to do and returns// There is no more things on the event loop so the program exits更新资料

感谢Bergi解释了Zalgo的答案是什么。现在,我更好地了解了您的担忧。

这就是我们所说的zalgo吗?谁能提供给我一个实时的例子,这将导致重大故障?

我看过很多这样的代码:

function x(argument, callback) { if (!argument) {return callback(new Error(’Bad argument’)); } runSomeAsyncFunction(argument, (error, result) => {if (error) { return callback(new Error(’Error in async function’));}callback({data: result}); });}

现在,x()如果存在错误的参数,则可以在返回之前立即运行回调,否则可以在x()返回之后运行回调。此代码非常常见。为了测试这些论点,人们可能会争辩说应该抛出一个异常,但让我们暂时忽略一下,可能会有一些更好的操作错误示例被立即知道-这只是一个简单的示例。

现在,如果它是这样写的:

function x(argument, callback) { if (!argument) {return process.nextTick(callback, new Error(’Bad argument’)); } runSomeAsyncFunction(argument, (error, result) => {if (error) { return callback(new Error(’Error in async function’));}callback({data: result}); });}

这样可以保证回调不会在x()返回之前被调用。

现在,这是否可能导致“重大故障”完全取决于其使用方式。如果运行以下命令:

let a;x(’arg’, (err, result) => { // assume that ’a’ has a value console.log(a.value);});// assign a value to ’a’ here:a = {value: 10};

那么它将有时版本崩溃x()不process.nextTick,绝不会用的版本崩溃x()用process.nextTick()。

解决方法

使用 节点-6.0 执行以下操作

function A(callback) { console.log(’A’); callback();}function B() { console.log(’B’)}function C() { console.log(’C’);}A(C);B();// Result is A,C,B i expected that A,B,C

但是将上面的示例更改为使用process.nextTick()打印A,B,C

function A(callback) { console.log(’A’); process.nextTick(() => { callback(); });}function B() { console.log(’B’)}function C() { console.log(’C’);}A(C);B();

这就是我们所说的zalgo吗?谁能提供给我一个实时的例子,这将导致重大故障?

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