用Groovy实现判断两个int数值大小（不用比较运算符，考虑溢出）

?刚读了老紫竹的blog:? 不用比较运算符，判断int型的a，b两数的大小,考虑溢出问题.

自己试着实现了一下，用的是个苯法子 -- 把数值部分转换成0或1，再跟sign-bit整合。 谁还有更好的算法? 不妨share一下。

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我是用Groovy实现的，基本上调用的都是Java的API，在JVM上同样是BYTECODE，就算是Java的*等价*代码吧。如大家要求的话，我可以把它转成纯Java code。

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PS: 我的实现在int compare3()，老紫竹原算法的groovy实现是int compare2()，int compare1()是不考虑溢出的简单实现。

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int compare(int a,int b) { //return compare1(a,b) //naive implementation //return compare2(a,b) //老紫竹's implementation return compare3(a,b) //newmike's implementation } int compare1(int a,int b) //naive implementation without overflow concern { return (int)(a-b) } int compare2(int a,int b) //老紫竹's implementation - with overflow concern but can NOT differentiate '>'(0) from '=='(0) { return (int)((a.longValue()-b.longValue())>>>63) } int compare3(int a,int b) //newmike's implementation - with overflow concern AND -1 as <; 0 as ==; +1 as >; { long diff = a.longValue() - b.longValue() showLong 'diff',diff long sign = diff>>>63<<63 showLong 'sign',sign long val = 0L long mask = 1L for(i in 0..<32) //use looping for shorten code,can be replaced by 32 repeated statement groups without implicit compare operators! { val |= (diff & mask) val <<= 1 mask <<= 1 } showLong 'val[1]',val val = val << 31 >>> 1 showLong 'val[2]',val val |= sign showLong 'val[3]',val val >>= 62 return (int)val } // show debugging info debugging=false def showLong(String msg,long l) { if(debugging) println String.format(' --- %-6s : %#-18X',msg,l) } // test cases for compare operations def formatStr = '%1$-5s %2$#-12X = %2$+d%n' printf formatStr,'<0',compare(1,2) printf formatStr,'=0',compare(2,'>0',1) println() printf formatStr,3) printf formatStr,compare(3,compare(Integer.MIN_VALUE,Integer.MAX_VALUE) printf formatStr,Integer.MIN_VALUE) printf formatStr,compare(Integer.MAX_VALUE,Integer.MIN_VALUE)

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