Here you can find the source of longToHex(long num)
num
to a hexadecimal number.
Parameter | Description |
---|---|
num | The number to convert |
public static String longToHex(long num)
//package com.java2s; /* This file is part of Math4J. * Math4J is free software; you can redistribute it and/or modify * it under the terms of the GNU General Public License as published by * the Free Software Foundation; either version 2 of the License, or * (at your option) any later version.//from ww w . ja v a 2s . c o m * * Math4J is distributed in the hope that it will be useful, * but WITHOUT ANY WARRANTY; without even the implied warranty of * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the * GNU General Public License for more details. * * You should have received a copy of the GNU General Public License * along with Math4J; if not, write to the Free Software * Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301, USA * * Copyright 2005 Anthony Magee */ public class Main { /** * A quick method to convert <code>num</code> to a hexadecimal number. A * hexadecimal number contains the digits zero through nine and the first * five letters of the english alphabet (A, B, C, D, E, F). * * @param num The number to convert */ public static String longToHex(long num) { /* * StringBuffer returnBuf = new StringBuffer(17); * * long power = 0; while(num > (long) Math.pow(16, power)) { power++; } * * for(long i = power - 1; i >= 0L; i--) { if((long) Math.floor(num / * (long) Math.pow(16, i)) == 10L) returnBuf.append("A"); else if((long) * Math.floor(num / (long) Math.pow(16, i)) == 11L) * returnBuf.append("B"); else if((long) Math.floor(num / (long) * Math.pow(16, i)) == 12L) returnBuf.append("C"); else if((long) * Math.floor(num / (long) Math.pow(16, i)) == 13L) * returnBuf.append("D"); else if((long) Math.floor(num / (long) * Math.pow(16, i)) == 14L) returnBuf.append("E"); else if((long) * Math.floor(num / (long) Math.pow(16, i)) == 15L) * returnBuf.append("F"); else returnBuf.append(Long.toString((long) * Math.floor(num / (long) Math.pow(16, i)))); * * num %= (long) Math.pow(16, i); } */ return (num < 16L) ? "0" + Long.toString(num) : Long.toString(num, 16); } }