深入理解计算机系统（CSAPP）实验二 datalab-handout

实验的目的是 填写 bits.c里面的函数，使其按照规定的要求（比如只能使用有限且规定的操作符和数据类型，不能使用控制语句等等）实现函数的功能。

同时 dlc文件是用来检测 bits.c 里面的函数是否  是按照要求编写的，有没有使用非法的数据类型等。 使用方法：./dlc bits.c

检测成功后，使用 btest 测试 每一个函数功能方面是否正确无误。使用方法：./btest，如果某个函数错误，会显示错误的数据，以及正确的数据。

完整的bits.c如下。也是参考网上各路大神的。

/* 
 * CS:APP Data Lab 
 * 
 * <Please put your name and userid here>
 * 
 * bits.c - Source file with your solutions to the Lab.
 *          This is the file you will hand in to your instructor.
 *
 * WARNING: Do not include the <stdio.h> header; it confuses the dlc
 * compiler. You can still use printf for debugging without including
 * <stdio.h>, although you might get a compiler warning. In general,
 * it's not good practice to ignore compiler warnings, but in this
 * case it's OK.  
 */

#if 0
/*
 * Instructions to Students:
 *
 * STEP 1: Read the following instructions carefully.
 */

You will provide your solution to the Data Lab by
editing the collection of functions in this source file.

INTEGER CODING RULES:
 
  Replace the "return" statement in each function with one
  or more lines of C code that implements the function. Your code 
  must conform to the following style:
 
  int Funct(arg1, arg2, ...) {
      /* brief description of how your implementation works */
      int var1 = Expr1;
      ...
      int varM = ExprM;

      varJ = ExprJ;
      ...
      varN = ExprN;
      return ExprR;
  }

  Each "Expr" is an expression using ONLY the following:
  1. Integer constants 0 through 255 (0xFF), inclusive. You are
      not allowed to use big constants such as 0xffffffff.
  2. Function arguments and local variables (no global variables).
  3. Unary integer operations ! ~
  4. Binary integer operations & ^ | + << >>
    
  Some of the problems restrict the set of allowed operators even further.
  Each "Expr" may consist of multiple operators. You are not restricted to
  one operator per line.

  You are expressly forbidden to:
  1. Use any control constructs such as if, do, while, for, switch, etc.
  2. Define or use any macros.
  3. Define any additional functions in this file.
  4. Call any functions.
  5. Use any other operations, such as &&, ||, -, or ?:
  6. Use any form of casting.
  7. Use any data type other than int.  This implies that you
     cannot use arrays, structs, or unions.

 
  You may assume that your machine:
  1. Uses 2s complement, 32-bit representations of integers.
  2. Performs right shifts arithmetically.
  3. Has unpredictable behavior when shifting an integer by more
     than the word size.

EXAMPLES OF ACCEPTABLE CODING STYLE:
  /*
   * pow2plus1 - returns 2^x + 1, where 0 <= x <= 31
   */
  int pow2plus1(int x) {
     /* exploit ability of shifts to compute powers of 2 */
     return (1 << x) + 1;
  }

  /*
   * pow2plus4 - returns 2^x + 4, where 0 <= x <= 31
   */
  int pow2plus4(int x) {
     /* exploit ability of shifts to compute powers of 2 */
     int result = (1 << x);
     result += 4;
     return result;
  }

FLOATING POINT CODING RULES

For the problems that require you to implent floating-point operations,
the coding rules are less strict.  You are allowed to use looping and
conditional control.  You are allowed to use both ints and unsigneds.
You can use arbitrary integer and unsigned constants.
可以使用循环和控制语句，使用int和unsigned

You are expressly forbidden to:
  1. Define or use any macros. 定义或者使用宏
  2. Define any additional functions in this file.定义附加函数
  3. Call any functions. 调用函数
  4. Use any form of casting. 使用转换
  5. Use any data type other than int or unsigned.  This means that you
     cannot use arrays, structs, or unions.
     使用 数组，结构体，共用体
  6. Use any floating point data types, operations, or constants.
  使用任意浮点数类型，操作，常亮。


NOTES:
  1. Use the dlc (data lab checker) compiler (described in the handout) to 
     check the legality of your solutions. 
     使用dlc检查合法性
  2. Each function has a maximum number of operators (! ~ & ^ | + << >>)
     that you are allowed to use for your implementation of the function. 
     The max operator count is checked by dlc. Note that '=' is not 
     counted; you may use as many of these as you want without penalty.
     每个函数都有所使用的操作符有限，‘=’不算。
  3. Use the btest test harness to check your functions for correctness.
  4. Use the BDD checker to formally verify your functions
  5. The maximum number of ops for each function is given in the
     header comment for each function. If there are any inconsistencies 
     between the maximum ops in the writeup and in this file, consider
     this file the authoritative source.

/*
 * STEP 2: Modify the following functions according the coding rules.
 * 
 *   IMPORTANT. TO AVOID GRADING SURPRISES:
 *   1. Use the dlc compiler to check that your solutions conform
 *      to the coding rules.
 *   2. Use the BDD checker to formally verify that your solutions produce 
 *      the correct answers.
 */


#endif
/* 
 * bitAnd - x&y using only ~ and | 
 *   Example: bitAnd(6, 5) = 4
 *   Legal ops: ~ |
 *   Max ops: 8
 *   Rating: 1
 *   方法：运用摩根定律
 */
int bitAnd(int x, int y) {
  return ~((~x)|(~y)); // ~~, ~| = &
}
/* 
 * getByte - Extract byte n from word x
 *   从 字 x 中 提取第n个字节
 *   Bytes numbered from 0 (LSB) to 3 (MSB)
 *   最低有效位为0，依次，最高有效位是 3
 *   Examples: getByte(0x12345678,1) = 0x56
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 6
 *   Rating: 2
 *   方法：
 */
int getByte(int x, int n) {
	int tmp = x >> ((n) << 3); //向右移到头 0xff123456,算术右移
	tmp = tmp & 0xFF; //保留最右一个字节。
	return tmp;;

}
/* 
 * logicalShift - shift x to the right by n, using a logical shift
 *   逻辑位移，逻辑右移
 *   Can assume that 0 <= n <= 31
 *   Examples: logicalShift(0x87654321,4) = 0x08765432
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 20
 *   Rating: 3 
 */
int logicalShift(int x, int n) {
	int tmp = ~(1 << 31); //0x7f ff ff ff
	tmp = ((tmp >> n) << 1) + 1; //因为n >= 0，实现tmp >> (n-1)的功能
	tmp = tmp & (x >> n);	//
  	return tmp;
}
/*
 * bitCount - returns count of number of 1's in word
 *   返回 二进制数中 1 的个数
 *   Examples: bitCount(5) = 2, bitCount(7) = 3
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 40
 *   Rating: 4
 *   参考网上解法。采用二分法，先计算x每两位中1的个数，并用对应的两队来存储
 *   这个个数。然后计算每4位1的个数，在用对应的4位进行存储。依次类推。
 *   最后整合得到16位中1的个数，即x中的1的个数。
 */
int bitCount(int x) {
	int count;  
    int tmpMask1 = (0x55)|(0x55<<8);  
    int mask1 = (tmpMask1)|(tmpMask1<<16);  
    int tmpMask2 = (0x33)|(0x33<<8);  
    int mask2 = (tmpMask2)|(tmpMask2<<16);  
    int tmpMask3 = (0x0f)|(0x0f<<8);  
    int mask3 = (tmpMask3)|(tmpMask3<<16);  
    int mask4 = (0xff)|(0xff<<16);  
    int mask5 = (0xff)|(0xff<<8);  
    count = (x&mask1)+((x>>1)&mask1);  
    count = (count&mask2)+((count>>2)&mask2);  
    count = (count + (count >> 4)) & mask3;  
    count = (count + (count >> 8)) & mask4;  
    count = (count + (count >> 16)) & mask5;  
    return count;  
}
/* 
 * bang - Compute !x without using !
 *   0为1,1为0
 *   Examples: bang(3) = 0, bang(0) = 1
 *   Legal ops: ~ & ^ | + << >>
 *   Max ops: 12
 *   Rating: 4 
 */
int bang(int x) {
  	int tmp = ~x + 1;// tmp = -x;
	tmp = x | tmp; // 非0最高位一定是1，因为正数和负数 或	
	tmp = tmp >> 31; //非0为0xff ff ff ff，0为0x 00 00 00 00
	return (tmp+1);
}
/* 
 * tmin - return minimum two's complement integer
 *   int最小的数 
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 4
 *   Rating: 1
 */
int tmin(void) {
	return 1<<31; //很容易理解 
}
/* 
 * fitsBits - return 1 if x can be represented as an 
 *  n-bit, two's complement integer.
 *   如果x可以表示为n位二进制补码形式
 *   1 <= n <= 32
 *   Examples: fitsBits(5,3) = 0, fitsBits(-4,3) = 1
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 15
 *   Rating: 2
 */
int fitsBits(int x, int n) {
    int shiftNumber= 32 + (~n + 1);// 32 - n
 	return !(x^((x<<shiftNumber)>>shiftNumber));
 	/*先左移32-n位，在右移32-n位，即保留最后n位数。在与x异或
 	若两者相同表示x可被表示为一个n位整数，！0为1
 	*/
}
/* 
 * divpwr2 - Compute x/(2^n), for 0 <= n <= 30
 *  计算 x / (2^n)
 *  Round toward zero
 *   Examples: divpwr2(15,1) = 7, divpwr2(-33,4) = -2
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 15
 *   Rating: 2
 */
int divpwr2(int x, int n) {
	//全0或者全1
    int signx = x >> 31;     //
    int mask = (1 << n) + (~0);//得到2^n - 1
    int bias = signx & mask;//如果x是正数，则bias为0，即不用加，直接移位
    //如果x为负数，加上偏置量之后在移位
    return (x + bias) >> n;
}
/* 
 * negate - return -x 
 *   Example: negate(1) = -1.
 *   求相反数
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 5
 *   Rating: 2
 */
int negate(int x) {
  return (~x) + 1; // 取反加1
}
/* 
 * isPositive - return 1 if x > 0, return 0 otherwise 
 *   x > 0 返回1， x <= 0返回0
 *   Example: isPositive(-1) = 0.
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 8
 *   Rating: 3
 */
int isPositive(int x) {
    return !((x >> 31) | (!x)); //看符号位即可
}
/* 
 * isLessOrEqual - if x <= y  then return 1, else return 0 
 *   Example: isLessOrEqual(4,5) = 1.
 *   x - y <= 0返回1，  > 0返回1
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 24
 *   Rating: 3
 */
int isLessOrEqual(int x, int y) {
	int singx = (x >> 31) & 1;
	int singy = (y >> 31) & 1;	//比较符号位 1 0 = 1,  0 1 = 0;
	int sing = (singx ^ singy) & singx; //保证singx和singy异号
	int tmp = x + ((~y) + 1); // x - y, 同号情况下,异号情况下会越界 0 0 = , 1 1 = 
	tmp = ((tmp>>31)&1) & (!(singx ^ singy));// 保证singx 和 singy 同号
	//int t = (!(x ^ y)); //判断相等
	//printf("sing  =%d, tmp = %d\n", sing, tmp);
	return (sing | tmp | ((!(x ^ y)))); //
}
/*
 * ilog2 - return floor(log base 2 of x), where x > 0
 *   Example: ilog2(16) = 4  即得到由多少位二进制表示即可。
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 90
 *   Rating: 4
 *   方法：参考。二分法。先右移16位后若大于0即得到（10000)2=16
 *	 否则得到0，判断最高位是否为0，若不为0，则包含2的16次方。即得到最高位的log  *	 数，同理其他
 */
int ilog2(int x) {
	int bitsNumber=0;
	bitsNumber=(!!(x>>16))<<4;//
	bitsNumber=bitsNumber+((!!(x>>(bitsNumber+8)))<<3);
	bitsNumber=bitsNumber+((!!(x>>(bitsNumber+4)))<<2);
	bitsNumber=bitsNumber+((!!(x>>(bitsNumber+2)))<<1);
	bitsNumber=bitsNumber+(!!(x>>(bitsNumber+1)));
	//for non zero bitsNumber, it should add 0
	//for zero bitsNumber, it should subtract 1
	bitsNumber=bitsNumber+(!!bitsNumber)+(~0)+(!(1^x));
	//当x为0时，还需要减一才能得到正确值。
	return bitsNumber;
}
/* 
 * float_neg - Return bit-level equivalent of expression -f for
 *   floating point argument f. 返回和浮点数参数-f相等的二进制
 *   Both the argument and result are passed as unsigned int's, but
 *   they are to be interpreted as the bit-level representations of
 *   single-precision floating point values.
 *   参数和返回结果都是无符号整数，但是可以解释成单精度浮点数的二进制表示
 *   When argument is NaN, return argument.
 *   Legal ops: Any integer/unsigned operations incl. ||, &&. also if, while
 *   Max ops: 10
 *   Rating: 2
 */
unsigned float_neg(unsigned uf) {
	unsigned result;
	unsigned tmp;
  	result=uf ^ 0x80000000; //将符号位改反
  	tmp=uf & (0x7fffffff);
  	if(tmp > 0x7f800000)//此时是NaN
        result = uf;
	return result;
}
/* 
 * float_i2f - Return bit-level equivalent of expression (float) x
 *  返回int x的浮点数的二进制形式。
 *   Result is returned as unsigned int, but
 *   it is to be interpreted as the bit-level representation of a
 *   single-precision floating point values.
 *   返回的是unsigned型但是表示的时二进制单精度形式
 *   Legal ops: Any integer/unsigned operations incl. ||, &&. also if, while
 *   Max ops: 30
 *   Rating: 4
 */
unsigned float_i2f(int x) {
	unsigned shiftLeft=0;
	unsigned afterShift, tmp, flag;
	unsigned absX=x;
	unsigned sign=0;
	//special case
	if (x==0) return 0;
	//if x < 0, sign = 1000...,abs_x = -x
	if (x<0)
	{
		sign=0x80000000;
		absX=-x;
	}
	afterShift=absX;
	//count shift_left and after_shift
	while (1)
	{
		tmp=afterShift;
	    afterShift<<=1;
	    shiftLeft++;
	    if (tmp & 0x80000000) break;
	}
	if ((afterShift & 0x01ff)>0x0100)
	    flag=1;
	else if ((afterShift & 0x03ff)==0x0300)
	    flag=1;
	else
	    flag=0;

	return sign + (afterShift>>9) + ((159-shiftLeft)<<23) + flag;
}
/* 
 * float_twice - Return bit-level equivalent of expression 2*f for
 *   floating point argument f.
 ×   返回 以unsinged表示的浮点数二进制的二倍的二进制unsigned型
 *   Both the argument and result are passed as unsigned int's, but
 *   they are to be interpreted as the bit-level representation of
 *   single-precision floating point values.
 *   When argument is NaN, return argument
 *   Legal ops: Any integer/unsigned operations incl. ||, &&. also if, while
 *   Max ops: 30
 *   Rating: 4
 */
unsigned float_twice(unsigned uf) {
	unsigned f = uf;
    if ((f & 0x7F800000) == 0) // 
 	{
        //左移一位
        f = ((f & 0x007FFFFF) << 1) | (0x80000000 & f);
    }
    else if ((f & 0x7F800000) != 0x7F800000)
    {
        f =f + 0x00800000;
	}
	return f;
}