JDK源码分析:Integer.java部分源码解析

1)声明部:

public final class Integer extends Number implements Comparable<Integer> 

  extends Number, 重写方法:

public byte byteValue() {  
    return (byte)value;  
}  
public short shortValue() {  
    return (short)value;  
}  
public int intValue() {  
    return value;  
}  
public long longValue() {  
    return (long)value;  
}  
public float floatValue() {  
    return (float)value;  
}  
public double doubleValue() {  
    return (double)value;  
} 

  implements Comparable<T>,接口实现如下:

public int compareTo(Integer anotherInteger) {  
    return compare(this.value, anotherInteger.value);  
} 
public static int compare(int x, int y) {  
    return (x < y) ? -1 : ((x == y) ? 0 : 1);  
}

  与实现该接口无关观察compareUnsigned方法:

  compareUnsigned 例子:

Integer a1 = 0b00000000000000000000000000000001;  
Integer a2 = 0b00000000000000000000000000000011;  
Integer a3 = 0b10000000000000000000000000000001;  
Integer a4 = 0b10000000000000000000000000000011;  
int r1 = Integer.compareUnsigned(a1.intValue(),a2.intValue());  
int r2 = Integer.compareUnsigned(a1.intValue(),a3.intValue());  
int r3 = Integer.compareUnsigned(a3.intValue(),a4.intValue());  
Out.println("a1,a2,a3,a4分别为:" + a1 +"," + a2 + "," + a3 + "," + a4);  
Out.println("r1,r2,r3分别为:" + r1 +"," + r2 + "," + r3);  

  result:

a1,a2,a3,a4分别为:1,3,-2147483647,-2147483645  
r1,r2,r3分别为:-1,-1,-1  

  算法分析:比较是无符号比较方法,而默认是有符号整型,所以需要特殊方法处理。先看Integer的范围:
  0B10000000000000000000000000000000
  0B10000000000000000000000000000001
  ..
  0B11111111111111111111111111111111
  …
  0B00000000000000000000000000000000
  0B00000000000000000000000000000001
  0B00000000000000000000000000000010
  …
  0B01111111111111111111111111111111

  在同一个符号下:除符号位其他位的数字越大,那么该数字越大。+MIN.VALUE就是修改符号位。不影响同一个符号位下的大小比较,
  假设a=-1,b=1按无符号考虑:a>b;
  按有符合考虑:a<b,通过修改符号位,a+MIN.VALUE(正数)>b+MIN.VALUE(负数)
  实现了无符号比较。

2)属性

@Native public static final int   MIN_VALUE = 0x80000000;  
@Native public static final int   MAX_VALUE = 0x7fffffff;  
/** 
 * The {@code Class} instance representing the primitive type {@code int}. 
 */  
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'  
};  
//十位上的数值,可以根据36,取的十位上数为3  
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',  
} ;  
//个位上的数值,可以根据36,取的个位上数为6  
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',  
} ;  
final static int [] sizeTable = { 9, 99, 999, 9999, 99999, 999999, 9999999,99999999, 999999999, Integer.MAX_VALUE };  
//避免除法操作,直接比较大小返回位数  
static int stringSize(int x) {  
    for (int i=0; ; i++)  
        if (x <= sizeTable[i])  
            return i+1;  
}  
public Integer(int value) { this.value = value;}  
@Native public static final int SIZE = 32;  
public static final int BYTES = SIZE / Byte.SIZE; 

  3)内部私有类

private static class IntegerCache {  
    static final int low = -128;  
    static final int high;  
    static final Integer cache[];  
  
    static {  
        <span style="color:#ff0000;">// high value may be configured by property</span>  
        int h = 127;  
        String integerCacheHighPropValue =  
            sun.misc.VM.getSavedProperty("java.lang.Integer.IntegerCache.high");  
        if (integerCacheHighPropValue != null) {  
            try {  
                int i = parseInt(integerCacheHighPropValue);  
                i = Math.max(i, 127);  
                // Maximum array size is Integer.MAX_VALUE  
                h = Math.min(i, Integer.MAX_VALUE - (-low) -1);  
            } catch( NumberFormatException nfe) {  
                // If the property cannot be parsed into an int, ignore it.  
            }  
        }  
        high = h;  
  
        cache = new Integer[(high - low) + 1];  
        int j = low;  
        for(int k = 0; k < cache.length; k++)  
            cache[k] = new Integer(j++);  
  
        // range [-128, 127] must be interned (JLS7 5.1.7)  
        assert IntegerCache.high >= 127;  
    }  
  
    private IntegerCache() {}  
}  

  和Short.java和Byte.java一样,都是通过静态代码块初始化缓存对象数组,不过,IntegerCache对象的high可以通过JVM的启动参数设置,缺省为127。

4)初始化方法

  构造函数:

public Integer(String s) throws NumberFormatException {  
    this.value = parseInt(s, 10);  
}  
public Integer(int value) {  
    this.value = value;  
}

   其他方法:

public static int parseInt(String s, int radix);  
public static int parseInt(String s) throws NumberFormatException;  
public static Integer valueOf(String s, int radix) throws NumberFormatException;   
public static Integer valueOf(String s) throws NumberFormatException;  
public static Integer valueOf(int i);  
public static int parseUnsignedInt(String s) throws NumberFormatException;  
public static int parseUnsignedInt(String s, int radix);

  观察“其他方法”的代码,难点主要在下面的方法:

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;//设不同进制下的极限值  
        while (i < len) {  
<span style="color:#ff0000;">           // Accumulating negatively avoids surprises near MAX_VALUE  
            //该方法返回该字符根据进制对应的数字,比如2进制1那么返回1,比如16进制F那么返回15  
            digit = Character.digit(s.charAt(i++),radix);//i++:0->1  
            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;</span>  
        }  
    } else {  
        throw NumberFormatException.forInputString(s);  
    }  
    return negative ? result : -result;  
}  

  

该代码段例子分析:

F1F(16进制)

       (1)digit 15;result *= radix -》0;result -= digit -》 -15 
                    (2)digit 1;result *= radix -》-15*16;result -= digit -》 -15*16 – 1(相当于累加) 
                    (3)digit 15;result *= radix -》-(15*16+1)*16;result -= digit -》 -(15*16+1)*16 – 15
                    (4)-( -(15*16+1)*16 – 15)= (15*15+1)*16 + 15

    public static int parseUnsignedInt(String s, int radix)
                throws NumberFormatException {
        if (s == null)  {
            throw new NumberFormatException("null");
        }

        int len = s.length();
        if (len > 0) {
            char firstChar = s.charAt(0);
            if (firstChar == '-') {
                throw new
                    NumberFormatException(String.format("Illegal leading minus sign " +
                                                       "on unsigned string %s.", s));
            } else {
                if (len <= 5 || // Integer.MAX_VALUE in Character.MAX_RADIX is 6 digits
                    (radix == 10 && len <= 9) ) { // Integer.MAX_VALUE in base 10 is 10 digits
                    return parseInt(s, radix);
                } else {
                    long ell = Long.parseLong(s, radix);
                    if ((ell & 0xffff_ffff_0000_0000L) == 0) {
                        return (int) ell;
                    } else {
                        throw new
                            NumberFormatException(String.format("String value %s exceeds " +
                                                                "range of unsigned int.", s));
                    }
                }
            }
        } else {
            throw NumberFormatException.forInputString(s);
        }
    }

  例子:

Integer a11 = 3;  
Integer a6 = Integer.parseInt("11",2);  
Integer a7 = Integer.parseUnsignedInt("80000000",16);  
Integer a9 = Integer.valueOf(3);  
Integer a10 = new Integer(3);  
boolean b1 = a6==a9;//true  
boolean b2 = a6==a11;//true  
boolean b3 = a6==a10;//false

  初始化对象进行使用parseInt或者valueOf方法,如果值在IntegerCache范围内,可以直接获取对象。

5)一些toXXXString方法:

public static String toString(int i, int radix) {  
    if (radix < Character.MIN_RADIX || radix > Character.MAX_RADIX)  
        radix = 10;  
  
    /* Use the faster version */  
    if (radix == 10) {  
        return toString(i);  
    }  
  
    char buf[] = new char[33];//二进制  
    boolean negative = (i < 0);  
    int charPos = 32;  
  
    if (!negative) {  
        i = -i;  
    }  
  
    while (i <= -radix) {  
        buf[charPos--] = digits[-(i % radix)];  
        i = i / radix;  
    }  
    buf[charPos] = digits[-i];  
  
    if (negative) {  
        buf[--charPos] = '-';  
    }  
        //从buf数组的charPos位置开始截取33-charPos字符转换为String类型  
    return new String(buf, charPos, (33 - charPos));  
}  

  分析该函数之前,我们用十进制转换为二进制来解释下算法:比如10,10%2 商5余0;5%2 商2余1;2%2 商1余0;2%1 商0余1,把余数倒序拼接1010,1010就是二进制。i++号在后面的意思是先赋值然后自身加1;++i在前面的是先自身加1后赋值;–同样。

public static String toUnsignedString(int i, int radix);  
public static String toHexString(int i);  
public static String toOctalString(int i);  
public static String toBinaryString(int i){  
    return toUnsignedString0(i, 1);  
}  
  
//前面4个方法都是调用该方法,shift:次幂  
private static String toUnsignedString0(int val, int shift) {  
    // assert shift > 0 && shift <=5 : "Illegal shift value";  
    //获取有效的二进制的位数 31位二进制  
    int mag = Integer.SIZE - Integer.numberOfLeadingZeros(val);  
    //一个算法,可以根据二进制的有效位数,求出8进制或者16进制的位数  
    //比如0100 0000 shift=3 -》 (7+(3-1))/3 = 3 八进制需要3位表示  
    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);  
}  
  
//转换为2进制,高位-》低位,直至遇到1停止,0的个数  
//比如0000 0000 0000 0000 0000 0000 0000 0001-》31  
//比如0100 0000 0000 0000 0000 0000 0000 0001-》1  
//比如0010 0000 0000 0000 0000 0000 0000 0001-》2  
public static int numberOfLeadingZeros(int i) {  
    // HD, Figure 5-6  
    if (i == 0)  
        return 32;  
    int n = 1;  
    if (i >>> 16 == 0) { n += 16; i <<= 16; }  
    if (i >>> 24 == 0) { n +=  8; i <<=  8; }  
    if (i >>> 28 == 0) { n +=  4; i <<=  4; }  
    if (i >>> 30 == 0) { n +=  2; i <<=  2; }  
    n -= i >>> 31;  
    return n;  
}  
  
static int formatUnsignedInt(int val, int shift, char[] buf, int offset, int len) {  
    int charPos = len;  
    int radix = 1 << shift;  
    int mask = radix - 1;  
    do {  
        //val&mask 可以获取末尾使用该进制表示的数值  
        //例子 假设16进制  
        //  1010 1010 1010 1000  
        //& 0000 0000 0000 1111  
        //  0000 0000 0000 1000  
        buf[offset + --charPos] = Integer.digits[val & mask];  
        val >>>= shift;  
    } while (val != 0 && charPos > 0);  
    return charPos;  
}  

  

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];  
     // 将Integer数读入到char[]数组  
    getChars(i, size, buf);  
    return new String(buf, true);  
}  
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 iteration  
    // 处理超过2的16次方的大数  
    while (i >= 65536) {  
        q = i / 100;  
        // really: r = i - (q * 100);  
        // 假设 65536 那么等于 36,根据36取的十位数和个位数上的数值         
        r = i - ((q << 6) + (q << 5) + (q << 2));  
        //655  
        i = q;  
        //个位 6  
        buf [--charPos] = DigitOnes[r];  
        //十位 3  
        buf [--charPos] = DigitTens[r];  
    }  
  
    // Fall thru to fast mode for smaller numbers  
    // assert(i <= 65536, i);  
    // 处理<2的16次方的大数  
    for (;;) {  
        q = (i * 52429) >>> (16+3);//i*52429/524288 ≈0.1000003。相当于i/10  
        //获取个位的数字  
        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;  
    }  
} 

 一个例子: 

for(int i = 0;i<65536;i++){  
    int q = (i * 52429) >>> (16+3);//相当于q = i/10;  
    int j = i/10;  
    if(q != j){  
        Out.println(false);  
    }  
} 
输出:true

  

    原文作者:knbsyoo
    原文地址: https://www.cnblogs.com/knsbyoo/p/9032563.html
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