I wish to implement motion blur or perhaps gaussian blur using purely GLSL. I've created a few basic shaders and I have already a few ideas in mind. My shaders: Vertex shader: attribute vec4 a_color; attribute vec2 a_position; attribute vec2 a_texCoord0; uniform mat4 u_projTrans; varying vec4 v_color; varying vec2 v_texCoord0; void main() { v_color = a_color; v_texCoord0 = a
我希望使用纯粹的GLSL来实现运动模糊或者高斯模糊。 我已经创建了一些基本着色器，并且已经有一些想法。 我的着色器： 顶点着色器： attribute vec4 a_color; attribute vec2 a_position; attribute vec2 a_texCoord0; uniform mat4 u_projTrans; varying vec4 v_color; varying vec2 v_texCoord0; void main() { v_color = a_color; v_texCoord0 = a_texCoord0; gl_Position = u_projTrans * vec4(a_posi
im trying out java regex quantifiers Pattern p = Pattern.compile("^s(.*?)n$"); Matcher m = p.matcher("sensation"); if(m.matches()){ System.out.println("dude how is this even possible"); System.out.print(m.group() + m.start()+m.end()+"n"); }else { System.out.println("sorry dude someting wrong"); } since it was relecutant quantifier it was supposed t
即时通讯尝试了正则表达式量词 Pattern p = Pattern.compile("^s(.*?)n$"); Matcher m = p.matcher("sensation"); if(m.matches()){ System.out.println("dude how is this even possible"); System.out.print(m.group() + m.start()+m.end()+"n"); }else { System.out.println("sorry dude someting wrong"); } 因为它是重量级量词，所以它应该给出以下结果说明 但是我不知
I was reading K.Sierra and found the following sentance: The greedy quantifier does in fact read the entire source data, and then it works backward (from the right) until it finds the rightmost match. At that point, it includes everything from earlier in the source data up to and including the data that is part of the rightmost match. Now, Suppose we have a source as follows: "proj3.txt,pro
我正在读K.Sierra并发现以下错误： 贪婪的量词实际上读取整个源数据，然后它向后（从右边）工作，直到找到最右边的匹配。 在这一点上，它包含了源数据之前的所有内容以及包含最右边匹配数据的数据。 现在，假设我们有一个来源如下： "proj3.txt,proj1sched.pdf,proj1,proj2,proj1.java" 和模式： proj1([^,])* 为什么它不符合整个文本？ 贪婪它应该匹配最右边的“proj1.java”，返回的匹配应该是最右边最匹配之前的整个
I want to match roman numbers using Groovy regular expressions (I have not tried this in Java but should be the same). I found an answer in this website in which someone suggested the following regex: /M{0,4}(CM|CD|D?C{0,3})(XC|XL|L?X{0,3})(IX|IV|V?I{0,3})/ The problem is that a expression like /V?I{0,3}/ is not greedy in Groovy. So for a string like "Book number VII" the matcher /V
我想用Groovy正则表达式来匹配罗马数字（我没有在Java中尝试过，但应该是相同的）。 我在这个网站上找到了一个答案，其中有人提出了以下正则表达式： /M{0,4}(CM|CD|D?C{0,3})(XC|XL|L?X{0,3})(IX|IV|V?I{0,3})/ 问题是在Groovy中像/V?I{0,3}/这样的表达式不是贪婪的。 因此，对于像“书号VII”这样的字符串，匹配器/V?I{0,3}/返回“V”而不是“VII”，因为它是所希望的。 显然，如果我们使用模式/VI+/那么我们会得到匹配“VII”..
I have issue with following example: import java.util.regex.*; class Regex2 { public static void main(String[] args) { Pattern p = Pattern.compile(args[0]); Matcher m = p.matcher(args[1]); boolean b = false; while(b = m.find()) { System.out.print(m.start() + m.group()); } } } And the command line: java Regex2 "d*" ab34ef Can someone
我有以下例子的问题： import java.util.regex.*; class Regex2 { public static void main(String[] args) { Pattern p = Pattern.compile(args[0]); Matcher m = p.matcher(args[1]); boolean b = false; while(b = m.find()) { System.out.print(m.start() + m.group()); } } } 和命令行： java Regex2 "d*" ab34ef 有人可以解释我，为什么结果是：01234456
This question already has an answer here: Greedy vs. Reluctant vs. Possessive Quantifiers 7 answers There are three types of quantifiers: the "regular" quantifiers (*, +, ?) also called "greedy" quantifiers; the "lazy", quantifiers (*?, +?, ??); the "possessive" quantifiers (*+, ++, ?+). For instance, take this input: The answer is 42 Now, tak
这个问题在这里已经有了答案： 贪婪与不愿意与拥有量词7答案 量词有三种类型： “规则”量词（*，+，？）也被称为“贪婪”量词; “懒惰”，量词（* ?, + ?, ??）; “占有”量词（* +，++，？+）。 例如，请输入以下内容： The answer is 42 现在，采取这个正则表达式： .*(d+) 问题是，根据你使用哪个*版本， (d+)会捕获什么.* ： 如果* ，将被捕获的是2 ; 如果*? ，将被捕获的是42 ; 如果*+ ，正则表达式不匹配
This question already has an answer here: Greedy vs. Reluctant vs. Possessive Quantifiers 7 answers 这是我喜欢想到的方式 - X?? Negative bias, 0 or 1 time - preference to 0 if possible X? Neutral bias, 0 or 1 time X?+ Positive bias, 0 or 1 time - preference to 1 if possible, and if 1 won't give it up (backtrack) You need more complex patterns to see the difference. A greedy q
这个问题在这里已经有了答案： 贪婪与不愿意与拥有量词7答案 这是我喜欢想到的方式 - X?? Negative bias, 0 or 1 time - preference to 0 if possible X? Neutral bias, 0 or 1 time X?+ Positive bias, 0 or 1 time - preference to 1 if possible, and if 1 won't give it up (backtrack) 您需要更复杂的模式才能看出差异。 一个贪婪的量词首先尽可能匹配（但回溯）。 不情愿的或“非贪婪的”量词首先尽
This question already has an answer here: Greedy vs. Reluctant vs. Possessive Quantifiers 7 answers The ? here acts as a 'modifier' if I can call it like that and makes .* match the least possible match (termed 'lazy') until the next match in the pattern. In fall/2005 , the first .*? will match up to the first match in ([am/]*) , which is just before f . Hence, .*? match
这个问题在这里已经有了答案： 贪婪与不愿意与拥有量词7答案 这个? 如果我可以这样称呼它并使得.*匹配尽可能少的匹配（称为“懒惰”），直到模式中的下一场比赛为止。 在fall/2005年fall/2005 ，第一个.*? 将匹配在([am/]*)的第一个匹配，这就在f之前。 因此， .*? 匹配0字符，以便([am/]*)将匹配fall/以来([am/]*)无法比拟的了，该模式的下一部分.*匹配什么字符串中离开，这意味着2005 。 与此相反.*([am/]*).* ，你
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