翻转句子的TDD议论：）

翻转句子的TDD讨论：）
http://www.iteye.com/topic/122472?page=1
需求:把字符串"Tdd is a software devolopment technology" 按照单词反转为
"technology devolopment software a is Tdd"
是看了这个帖子后一直有困惑。直接促成了我找了Kent Beck大的Test-Driven Development
By Example来看。自此进入TDD的世界。其实这题目和书后面的那个fibonacci的解法一样。
关键在于：“分解多项式” 且听我忽悠..

import junit.framework.TestCase;public class NewStringReverserTest extends TestCase{//"a b" - "b a"public void testSimpleReverse(){assertEquals("b a", reverse("a b"));//从最简单的情况开始} private String reverse(String string) {return "b a";//最“直接”的实现，绿了吧- -！}}

public class NewStringReverserTest extends TestCase{...private String reverse(String string) { return "b"+" "+"a";//拆，为什么这么拆？ 往下看}}

public class NewStringReverserTest extends TestCase{...private String reverse(String string) {                //要拆得有意义，注意描述= =                //其实是一个一个拆的，每步都执行test，边绿边拆...小步快跑 String splitToken = " ";String preWord = "a";String postWord = "b";                //下面其实是业务逻辑了，看后面的postWord被放在了前面——所谓的交换return postWord + splitToken + preWord;}}

public class NewStringReverserTest extends TestCase{...private String reverse(String inputString) {               String splitToken = " ";                //实现切词逻辑String preWord =    inputString.substring(0,inputString.indexOf(splitToken));String postWord = inputString.substring(inputString.indexOf(splitToken)+1);               return postWord + splitedToken + preWord;}}

ok 目前为止似乎无法继续了。那么继续添加test描述需求："a b c"-"c b a"

public class NewStringReverserTest extends TestCase{//"a b" - "b a"public void testSimpleReverse(){assertEquals("b a", reverse("a b"));                assertEquals("c b a", reverse("a b c"));//红了吧？- -}        ...}

Expected：c b a Actual:b c a
好吧 分析下：bc 反了，bc看做一个整体，再“交换”一次就ok了,即: reverse("b c")+"a"

        private String reverse(String inputString) {               ...          return reverse(postWord) + splitToken + preWord;}

还是会红，debug发现“切到剩一个词”了就不能切了嘛= =

        private String reverse(String inputString) {                int tokenIndex = inputString.indexOf(splitToken);if (tokenIndex == -1)return inputString;String prePart = inputString.substring(0, tokenIndex);String postPart = inputString.substring(tokenIndex + 1);               //描述为prePart，postPart 更符合语义，也是一步重构吧        return reverse(postPart)+ splitToken + prePart}

这下绿了，想想其实prePart其实是reverse(prePart)
另外把那个需求的test加上 看看符合需求否？

       public class NewStringReverserTest extends TestCase {// "a b" - "b a"public void testSimpleReverse() {assertEquals("b a", reverse("a b"));assertEquals("c b a", reverse("a b c"));                assertEquals("Tdd is a software devolopment technology",  reverse("technology devolopment software a is Tdd"));}private String reverse(String inputString) {String splitToken = " ";int tokenIndex = inputString.indexOf(splitToken);if (tokenIndex == -1)return inputString;String prePart = inputString.substring(0, tokenIndex);String postPart = inputString.substring(tokenIndex + 1);return reverse(postPart) + splitToken + reverse(prePart);}public void testReverseWithSingleWords() {assertEquals("a", reverse("a"));}public void testReverseWithSplitTokenIn() {assertEquals(" ", reverse(" "));assertEquals("   ", reverse("   "));assertEquals("   a", reverse("a   "));assertEquals("b a", reverse("a b"));assertEquals("c b a ", reverse(" a b c"));assertEquals("c b a", reverse("a b c"));}}

最后总结："a b c"到"c b a"
原来是a b c->[b c] a->[c b] a 1 楼 woods 2008-05-07   一直在想最后一步”prePart 其实是reverse(prePart) ”没啥必要= =

今天突然发现，其实这对某些情况很有意义:
reverse(postPart) + splitToken + reverse(prePart);

这不描述了这个问题的实质嘛- -#
完全可以多线程分成多个part来处理，最后再合成嘛- -#
这对提高效率，处理超大文件超有意义- -#

TDD的“副作用”? :-) 2 楼 woods 2008-05-07   My google SVN
# Project:mouse-action 
# http://code.google.com/p/mouse-action/

# Non-members may check out a read-only working copy anonymously over HTTP.
svn checkout http://mouse-action.googlecode.com/svn/trunk/ mouse-action-read-only

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