public final class Integer extends Number implements Comparable<Integer>public final 类，继承关系.
属性：
/*** A constant holding the minimum value an {@code int} can* have, -2<sup>31</sup>.*/@Native public static final int MIN_VALUE = 0x80000000;/*** A constant holding the maximum value an {@code int} can* have, 2<sup>31</sup>-1.*/@Native public static final int MAX_VALUE = 0x7fffffff;/*** The {@code Class} instance representing the primitive type* {@code int}.** @since JDK1.1*/@SuppressWarnings("unchecked")public static final Class<Integer> TYPE = (Class<Integer>) Class.getPrimitiveClass("int");/*** All possible chars for representing a number as a String*/final static char[] digits = {'0' , '1' , '2' , '3' , '4' , '5' ,'6' , '7' , '8' , '9' , 'a' , 'b' ,'c' , 'd' , 'e' , 'f' , 'g' , 'h' ,'i' , 'j' , 'k' , 'l' , 'm' , 'n' ,'o' , 'p' , 'q' , 'r' , 's' , 't' ,'u' , 'v' , 'w' , 'x' , 'y' , 'z'};toUnsignedString方法，根据进制int转对应字符串，转换代码分析在学习随笔（1）中学习了。
public static String toUnsignedString(int i, int radix) {return Long.toUnsignedString(toUnsignedLong(i), radix);}public static String toHexString(int i) {return toUnsignedString0(i, 4);}public static String toOctalString(int i) {return toUnsignedString0(i, 3);}public static String toBinaryString(int i) {return toUnsignedString0(i, 1);}全是static方法，类直接调用。
这个方法在学习随笔(1) 中学习了，不细说了
private static String toUnsignedString0(int val, int shift) {// assert shift > 0 && shift <=5 : "Illegal shift value";int mag = Integer.SIZE - Integer.numberOfLeadingZeros(val);int chars = Math.max(((mag + (shift - 1)) / shift), 1);char[] buf = new char[chars];formatUnsignedInt(val, shift, buf, 0, chars);// Use special constructor which takes over "buf".return new String(buf, true);}int转字符串
public static String toString(int i) {if (i == Integer.MIN_VALUE)return "-2147483648";int size = (i < 0) ? stringSize(-i) + 1 : stringSize(i);char[] buf = new char[size];getChars(i, size, buf);return new String(buf, true);}临界值判断，申请缓存，getChars方法为具体逻辑函数：
static void getChars(int i, int index, char[] buf) {int q, r;int charPos = index;char sign = 0;if (i < 0) {sign = '-';i = -i;}// Generate two digits per iterationwhile (i >= 65536) {q = i / 100;// really: r = i - (q * 100);r = i - ((q << 6) + (q << 5) + (q << 2));i = q;buf [--charPos] = DigitOnes[r];buf [--charPos] = DigitTens[r];}// Fall thru to fast mode for smaller numbers// assert(i <= 65536, i);for (;;) {q = (i * 52429) >>> (16+3);r = i - ((q << 3) + (q << 1)); // r = i-(q*10) ...buf [--charPos] = digits [r];i = q;if (i == 0) break;}if (sign != 0) {buf [--charPos] = sign;}}先看这段函数
// Generate two digits per iterationwhile (i >= 65536) {q = i / 100;// really: r = i - (q * 100);r = i - ((q << 6) + (q << 5) + (q << 2));i = q;buf [--charPos] = DigitOnes[r];buf [--charPos] = DigitTens[r];}实质是每次取i的最后两位数存在r中，然后i/100，q = i/100,然后(q << 6) + (q << 5) + (q << 2)就等于q*100，再r = i - q*100,r就是最后两位数，然后直接索引对应的个位和十位数组，从后到前存在buf数组中：
final static char [] DigitTens = {'0', '0', '0', '0', '0', '0', '0', '0', '0', '0','1', '1', '1', '1', '1', '1', '1', '1', '1', '1','2', '2', '2', '2', '2', '2', '2', '2', '2', '2','3', '3', '3', '3', '3', '3', '3', '3', '3', '3','4', '4', '4', '4', '4', '4', '4', '4', '4', '4','5', '5', '5', '5', '5', '5', '5', '5', '5', '5','6', '6', '6', '6', '6', '6', '6', '6', '6', '6','7', '7', '7', '7', '7', '7', '7', '7', '7', '7','8', '8', '8', '8', '8', '8', '8', '8', '8', '8','9', '9', '9', '9', '9', '9', '9', '9', '9', '9',} ;final static char [] DigitOnes = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9','0', '1', '2', '3', '4', '5', '6', '7', '8', '9','0', '1', '2', '3', '4', '5', '6', '7', '8', '9','0', '1', '2', '3', '4', '5', '6', '7', '8', '9','0', '1', '2', '3', '4', '5', '6', '7', '8', '9','0', '1', '2', '3', '4', '5', '6', '7', '8', '9','0', '1', '2', '3', '4', '5', '6', '7', '8', '9','0', '1', '2', '3', '4', '5', '6', '7', '8', '9','0', '1', '2', '3', '4', '5', '6', '7', '8', '9','0', '1', '2', '3', '4', '5', '6', '7', '8', '9',} ;当i小于等于65536时，用更快的转换方法：
for (;;) {q = (i * 52429) >>> (16+3);r = i - ((q << 3) + (q << 1)); // r = i-(q*10) ...buf [--charPos] = digits [r];i = q;if (i == 0) break;}他这个做法和上面的差不多，r为每次取的数字的最后一位，因为2的19次方为524288，比较下52429这个数字，就不难明白了。而且(q << 3) + (q << 1)就是q*10，和上面的一样吧。
再看看计算int位数的函数：
final static int [] sizeTable = { 9, 99, 999, 9999, 99999, 999999, 9999999,99999999, 999999999, Integer.MAX_VALUE };// Requires positive xstatic int stringSize(int x) {for (int i=0; ; i++)if (x <= sizeTable[i])return i+1;}明显的空间换时间啊，时间复杂度o(n).
再看看字符串转int函数：
public static int parseInt(String s, int radix)throws NumberFormatException{/** WARNING: This method may be invoked early during VM initialization* before IntegerCache is initialized. Care must be taken to not use* the valueOf method.*/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");}int result = 0;boolean negative = false;int i = 0, len = s.length();int limit = -Integer.MAX_VALUE;int multmin;int digit;if (len > 0) {char firstChar = s.charAt(0);if (firstChar < '0') { // Possible leading "+" or "-"if (firstChar == '-') {negative = true;limit = Integer.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;}首先，这个判断第一个char符号位正负性和合法性的代码很灵性啊：
if (firstChar < '0') { // Possible leading "+" or "-"if (firstChar == '-') {negative = true;limit = Integer.MIN_VALUE;} else if (firstChar != '+')throw NumberFormatException.forInputString(s);if (len == 1) // Cannot have lone "+" or "-"throw NumberFormatException.forInputString(s);i++;}首先if (firstChar < ‘0’)，因为+和-号的ascII码都小于’0’，这样就直接判断是非数字字符了。multmin为转换后的数字的下界，然后依次取每位字符转换为int，然后保存在result里。注意他默认保存的是负数，所以最后的返回值为return negative ? result : -result;
对应字符串转10进制数字函数：
public static int parseInt(String s) throws NumberFormatException {return parseInt(s,10);}先看到这，留点明天接着看~