JXS In Depth

JXS In Depth

1、Spread Attributes

  If you already have props as an object, and you want to pass it in JSX, you can use ... as a "spread" operator to pass the whole props object. These two components are equivalent:

  

2、Functions as Children

  props.children可以是一个函数。只要render()最终返回的是jsx就行了。

  Normally, JavaScript expressions inserted in JSX will evaluate to a string, a React element, or a list of those things. However, props.children works just like any other prop in that it can pass any sort of data, not just the sorts that React knows how to render. For example, if you have a custom component, you could have it take a callback as props.children:

  

  Children passed to a custom component can be anything, as long as that component transforms them into something React can understand before rendering. This usage is not common, but it works if you want to stretch what JSX is capable of.

3、jsx中使用条件expression

  

参考:https://facebook.github.io/react/docs/jsx-in-depth.html#user-defined-components-must-be-capitalized

时间: 2024-08-26 22:41:10

JXS In Depth的相关文章

Maximum Depth of Binary Tree

这道题为简单题 题目: Given a binary tree, find its maximum depth.The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. 思路: 我是用递归做的,当然也可以用深搜和广搜,递归的话就是比较左右子树的深度然后返回 代码: 1 # Definition for a binary tre

测试不同格式下depth buffer的精度

这篇文章主要是参考MJP的"Attack of The Depth Buffer",测试不同格式下depth buffer的精度. 测试的depth buffer包含两类: 一是非线性的depth buffer,存储着perspective z(也就是最常用的,透视投影后归一化的z/w的buffer),二是线性的depth buffer,存储着view space z(这里的线性指的是在view space 中是否线性).测试的格式包括16位浮点数,32位浮点数,16位定点数,还有最常

[leetcode] 104. Maximum Depth of Binary Tree

Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. 递归遍历 左子树 和 右子树 一刷: public int maxDepth(TreeNode root) { if(root == null){ return 0; } int

leetcode笔记:Minimum Depth of Binary Tree

一. 题目描述 Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. 二. 题目分析 这道题属于二叉树的深度优先搜索,然后返回深度最小的值,可以递归(当然,也可以使用迭代)来实现.递归退出的条件是到达叶子节点或者到达空子树,使用空子树

leetcode笔记:Maximum Depth of Binary Tree

一. 题目描述 Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. 二. 题目分析 这道题和Minimum Depth of Binary Tree一题相似,这个是求最大深度的,就是对二叉树进行递归,然后将子树的最大深度进行返回,最

leetcode 104 Maximum Depth of Binary Tree二叉树求深度

Maximum Depth of Binary Tree Total Accepted: 63668 Total Submissions: 141121 My Submissions Question Solution Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the

[LeetCode] 111. Minimum Depth of Binary Tree Java

题目: Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. 题意及分析:求一棵二叉树的最低高度,即根节点到叶子节点的最短路径.遍历二叉树,然后用一个变量保存遍历到当前节点的最小路径即可,然后每次遇到叶子节点,就判断当前叶节点高度和先

[LeetCode] 104. Maximum Depth of Binary Tree Java

题目: Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. 题意及分析:找出一棵树的高度,即最深子节点.使用深度遍历的方法即可,用一个变量记录遍历到当前点的最大高度,然后当前点若有子节点,遍历到子节点,那么该点的高度+1和当前的最大

手势跟踪论文学习:Realtime and Robust Hand Tracking from Depth(三)Cost Function

iker原创.转载请标明出处:http://blog.csdn.net/ikerpeng/article/details/39050619 Realtime and Robust Hand Tracking from Depth中的Cost Function 学习 首先,我们应该知道,输入的数据是什么:3D 点云数据. 3D点云给我的感觉应该是这种 输出的是:拟合好的手模型(48球体模型). 而这里的的3D 点云数据用p表示,每个球体用Sx 表示. Ci 第i个球体的中心:D表示深度图( 区分还