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：ffffffffc6105cec
i = 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_VALUE
digit = 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：
46105cec 1175477484 9239.23 46105cec 1175477484 9239.23 46105cec -c6105cec -3322961132 4.5707135E-4 39efa314 Exception 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。
