括号中的四个“东西”的for循环是如何工作的？

本文介绍了括号中的四个“东西”的for循环是如何工作的？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 我在这里遇到了一些代码 https://stackoverflow.com/a/2922778/2176900 ，这对于（i = 0，j = nvert-1; i< nvert; j = i ++）{... b $ b 。 我不明白它在括号内有四件事情时会做什么（ i = 0，j = nvert-1，i $ b 谢谢 解决方案它没有四件事;它仍然有三个。 预循环语句 i = 0，j = nvert-1 迭代条件 i 迭代后操作 j = i ++ 预循环语句实际上应该是： var i = 0，j = nvert-1 记住我们可以声明多个变量在一个单一的声明。这就是发生在这里的事情。 I came across some code here https://stackoverflow.com/a/2922778/2176900 that started off like this:for (i = 0, j = nvert-1; i < nvert; j = i++) {...`and I don't understand what it does when it has four things within the parenthesis ("i = 0", "j = nvert-1", "i < nvert" and "j = i++"). I've obviously been trying to google this, but can't seem to find any answers. I'm sure if I just knew what the things in the parenthesis was called I'd be glad.Thanks 解决方案 It doesn't have four things; it still has three.The pre-loop statementi = 0, j = nvert-1The iteration conditioni < nvertThe post-iteration actionj = i++That pre-loop statement is actually probably supposed to read:var i = 0, j = nvert-1Remember that we can declare multiple variables in a single declaration. That's what's happening here. 这篇关于括号中的四个“东西”的for循环是如何工作的？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！