从 char 数组 parseInt 实现
代码来源于 https://svn.apache.org/repos/asf/harmony/enhanced/classlib/archive/java6/modules/luni/src/main/java/java/lang/Integer.java
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public static int parseInt(char[] chars) {return parseInt(chars, 0, chars.length);}public static int parseInt(char[] chars, int offset, int length) {return parseInt(chars, offset, length, 10);}public static int parseInt(char[] chars, int offset, int length, int radix)throws NumberFormatException {if (chars == null || radix < Character.MIN_RADIX|| radix > Character.MAX_RADIX) {throw new NumberFormatException();}if (length == 0) {throw new NumberFormatException("Unable to parse");}int i = 0;boolean negative = (chars[i + offset] == '-');if (negative && ++i == length)throw new NumberFormatException("Unable to parse");if (negative) {offset++;length--;}return parseInt(chars, offset, length, radix, negative);}/** * 这个方法把要解析的字符数组映射到了负数的范围内。因为java int 的取值范围是 -2^32 ~ 2^31，负数的取值范围比正数要大 * @param chars 字符数组 * @param offset 有效的字符起始 idx * @param len 有效字符长度 * @param radix 基数，比如 10 进制，2 进制啥的 * @param negative 这个数是不是负数 * @return * @throws NumberFormatException */public static int parseInt(char[] chars, int offset, int len, int radix,boolean negative) throws NumberFormatException {//这里的 max 取值是负数，下面解析出来的数都是负数。这里的 max 是绝对值最大，而不是实际值最大int max = Integer.MIN_VALUE / radix;int result = 0;for (int i = 0; i < len; i++) {//解析出来的第一个数字int digit = Character.digit(chars[i + offset], radix);if (digit == -1) {throw new NumberFormatException("Unable to parse");}if (max > result) {throw new NumberFormatException("Unable to parse");}//映射到负数的范围内，解析出来的数字要被减去。next 的值要比已经保存的值（next） 小int next = result * radix - digit;if (next > result) {throw new NumberFormatException("Unable to parse");}result = next;}if (!negative) {result = -result;if (result < 0) {throw new NumberFormatException("Unable to parse");}}return result;}??