java 如何将十六进制字符串转换为 float 符点型？相互转换

java 如何将十六进制字符串转换为 float 符点型？

先上代码：

package com.weixiao.network;/** * java 如何将十六进制字符串转换为 float 符点型？相互转换 * Hex2Float * @author 微wx笑 * @date   2017年12月6日下午5:22:10 */public class Hex2Float {public static void main(String[] args) {String hexString = "46105cec";Long l = Hex2Float.parseLong(hexString, 16);Float f = Float.intBitsToFloat(l.intValue());System.out.println(hexString);System.out.println(l);System.out.println(f);System.out.println(Integer.toHexString(Float.floatToIntBits(f)));Integer i = Integer.parseInt(hexString, 16);f = Float.intBitsToFloat(i.intValue());System.out.println("");System.out.println(i);System.out.println(f);System.out.println(Integer.toHexString(Float.floatToIntBits(f)));hexString = "-c6105cec";l = Hex2Float.parseLong(hexString, 16);f = Float.intBitsToFloat(l.intValue());System.out.println("");System.out.println(hexString);System.out.println(l);System.out.println(f);System.out.println(Integer.toHexString(Float.floatToIntBits(f))); // 使用 Long 会输出：ffffffffc6105ceci = Integer.parseInt(hexString, 16); // 使用 Integer 会抛异常： java.lang.NumberFormatException: For input string: "c6105cec"f = Float.intBitsToFloat(i.intValue());System.out.println("");System.out.println(i);System.out.println(f);System.out.println(Integer.toHexString(Float.floatToIntBits(f)));}/** * 代码来自：java.lang.Long * 因为要跟踪看变量的值，所以要copy出来，或者是去附加源码，否则 eclipse 调试时查看变量的值会提示 xxx cannot be resolved to a variable * @author 微wx笑 * @date   2017年12月6日下午5:19:40 * @param s * @param radix * @return * @throws NumberFormatException */public static long parseLong(String s, int radix) throws NumberFormatException {if (s == null) {throw new NumberFormatException("null");}if (radix < Character.MIN_RADIX) {throw new NumberFormatException("radix " + radix + " less than Character.MIN_RADIX");}if (radix > Character.MAX_RADIX) {throw new NumberFormatException("radix " + radix + " greater than Character.MAX_RADIX");}long result = 0;boolean negative = false;int i = 0, len = s.length();long limit = -Long.MAX_VALUE;long multmin;int digit;if (len > 0) {char firstChar = s.charAt(0);if (firstChar < '0') { // Possible leading "+" or "-"if (firstChar == '-') {negative = true;limit = Long.MIN_VALUE;} else if (firstChar != '+')throw NumberFormatException.forInputString(s);if (len == 1) // Cannot have lone "+" or "-"throw NumberFormatException.forInputString(s);i++;}multmin = limit / radix;while (i < len) {// Accumulating negatively avoids surprises near MAX_VALUEdigit = Character.digit(s.charAt(i++), radix);if (digit < 0) {throw NumberFormatException.forInputString(s);}if (result < multmin) {throw NumberFormatException.forInputString(s);}result *= radix;if (result < limit + digit) {throw NumberFormatException.forInputString(s);}result -= digit;}} else {throw NumberFormatException.forInputString(s);}return negative ? result : -result;}}/** * 代码来自：java.lang.NumberFormatException * NumberFormatException * @author 微wx笑 * @date   2017年12月6日下午5:20:36 */class NumberFormatException extends IllegalArgumentException {/** *  */private static final long serialVersionUID = 1L;public NumberFormatException(String s) {super(s);}static NumberFormatException forInputString(String s) {return new NumberFormatException("For input string: \"" + s + "\"");}}

对应的输出如下：

46105cec11754774849239.2346105cec11754774849239.2346105cec-c6105cec-33229611324.5707135E-439efa314Exception in thread "main" java.lang.NumberFormatException: For input string: "-c6105cec"at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)at java.lang.Integer.parseInt(Integer.java:583)at com.weixiao.network.Hex2Float.main(Hex2Float.java:36)

对，你没看错，上面的代码有会抛出异常的！

Why？

你注意到将十六进制字符串转换为 float，和将 float 转换为十六进制字符串 的区别了吗？

前面是用 Long，后面是用 Integer，为什么不用相同的类型呢？

后面的异常就是解释这个的。

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