为什么zlib.deflate(buf，callback)是异步的? [英] Why zlib.deflate(buf, callback) is async?

问题描述

当接受输入缓冲区并且知道压缩是CPU绑定的任务时，是否需要提供回调?

When accepting an input buffer, and knowing compression is a CPU bound task, does it needs to offer a callback? Is it just there to follow common practice like callback(err, result)?

文档: http://nodejs.org/api/zlib.html#zlib_zlib_deflate_buf_callback

推荐答案

压缩/解压缩是在单独的线程中执行的，而不是在主线程中执行的.这就是为什么线程完成工作时需要回调.

The compression/decompression is executed in a separate thread and not in the main thread. That is why a callback is needed for when the thread completes its work.

回调样式与整个节点和大多数第三方模块(错误优先)使用的通用签名相同.

The callback style is the same common signature used throughout node and most third-party modules (error first).

这篇关于为什么zlib.deflate(buf，callback)是异步的?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！