标签:java ar for 数据 on c new har as
实际中估计也没有这个需求,只是写写。
能不能直接由2进制转为16进制呢,最直接的办法是通过位移操作,也就是通常所说的 每四位二进制对应一个16进制,
假如java中有一种 二进制的基础类型 Binary bi = new Binary(11010101);这样的话 进行 bi >>> 4 位移操作 就可以很容易转换了,关于这个可以参考Integer.java中的toHexString(int i)方法.
但是没有这种类型,java中貌似没有直接转化的方法,也就只能先转为10进制,再转为16进制。
写下自己实现的比较笨的办法:
法1:和Integer源码中的进制转换在某些思路上类似
/** * 10 0010 1111 按每4个一组分开,在binarys[]中找到对应的下标idx,对应的hexs[idx]相连接。 * @param binaryStr 10 0010 1111 * @return hexStr 22F */ public static String binaryToHex(String binaryStr){ String binarys[] = new String[]{ "0000","0001","0010","0011", "0100","0101","0110","0111", "1000","1001","1010","1011", "1100","1101","1110","1111" }; String hexs[] = new String[]{ "0","1","2","3", "4","5","6","7", "8","9","A","B", "C","D","E","F" }; int n = binaryStr.length()%4; if(n != 0){ int zeroNum = 0; zeroNum = 4 - n; for(int i=0; i< zeroNum; i++){ binaryStr = "0"+binaryStr; } n = n+1; } String hexStr = ""; for(int j=0; j<binaryStr.length()/4; j++){ String temp = binaryStr.substring(j*4, (j+1)*4); int index = -1; for(int k=0;k<16;k++){ if(binarys[k].equals(temp)){ index = k; } } if(index != -1){ hexStr = hexStr + hexs[index]; } } return hexStr; }
/** * 10 0010 1111 按每4个一组分开,转化为int型数据index,然后hexs[index]相加。 * @param binaryStr 10 0010 1111 * @return hexStr 22F */ public static String binaryToHex2(String binaryStr){ String hexs[] = new String[]{ "0","1","2","3", "4","5","6","7", "8","9","A","B", "C","D","E","F" }; int n = binaryStr.length()%4; if(n != 0){ int zeroNum = 0; zeroNum = 4 - n; for(int i=0; i< zeroNum; i++){ binaryStr = "0"+binaryStr; } n = n+1; } String hexStr = ""; for(int j=0; j<binaryStr.length()/4; j++){ String temp = binaryStr.substring(j*4, (j+1)*4); int index =0;//下标 index=15,则对应输出 hexs[15] = F int i = 3; for(char c : temp.toCharArray()) { String str = new String(new char[]{c}); index = index + ((Integer.parseInt(str))<<i) ; i--; } hexStr = hexStr + hexs[index]; } return hexStr; }
标签:java ar for 数据 on c new har as
原文地址:http://blog.csdn.net/wufengui1315/article/details/39030511