CS:APP3e 深入理解计算机系统_3e Datalab实验

**由于http://csapp.cs.cmu.edu/并未完全开放实验，很多附加实验做不了，一些环境也没办法搭建，更没有标准答案。做了这个实验的朋友可以和我对对答案；）**

实验内容和要求可在http://csapp.cs.cmu.edu/3e/labs.html获得。

Data Lab [Updated 5/4/16]

bits.c

/*
 * CS:APP Data Lab
 *
 * <李秋豪 Richard Li>
 *
 * 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.
 */
/*
 * 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.
You are expressly forbidden to:
  1. Define or use any macros.
  2. Define any additional functions in this file.
  3. Call any functions.
  4. Use any form of casting.
  5. Use any data type other than int or unsigned.  This means that you
     cannot use arrays, structs, or unions.
  6. Use any floating point data types, operations, or constants.
NOTES:
  1. Use the dlc (data lab checker) compiler (described in the handout) to
     check the legality of your solutions.
  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.
  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.
*/ 
/*----------------------------------------------------*/
/*
 * bitAnd - x&y using only ~ and |
 *   Example: bitAnd(6, 5) = 4
 *   Legal ops: ~ |
 *   Max ops: 8
 *   Rating: 1
 */
int bitAnd(int x, int y) {
  return ~(~x | ~y);
}
/*
 * getByte - Extract byte n from word x
 *   Bytes numbered from 0 (LSB) to 3 (MSB)
 *   Examples: getByte(0x12345678,1) = 0x56
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 6
 *   Rating: 2
 */
int getByte(int x, int n) {
  n = n << 3; /* n = n*8 */
  return (x & (0xFF<<n)) >> n;
}
/*
 * logicalShift - shift x to the right by n, using a logical shift
 *   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 mask = ~0 << n;
   return (mask & x) >> n;
}
/*
 * bitCount - returns count of number of 1's in word
 *   Examples: bitCount(5) = 2, bitCount(7) = 3
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 40
 *   Rating: 4
 */
int bitCount(int x) {
	/*这个题我一开始的思路是逐个位的将x向右移动，每次移动后将和0x01与的
	结果和sum相加。但是这样操作符的数量会超过40个。后来想到，其实没有必
	要非要每次把加法放在起始位，其他位置也是符合加法规律的，所以可以在
	多个位置同时进行加法，最后“汇总”到首位的地方*/
	/*同时要注意两个问题：1.第一次一位相加的时候可能会溢出到两位”10“，
	第二次两位相加的时候可能会溢出到三位”100“，第三次相加的时候是不会
	溢出超过四位的”1000“，所以会产生0x0y0z0a这种格式。将这四个四位相加
	即可。2.四个四位相加时不能使用x+(x>>24)因为之前x = x + (x>>8)已经
	将第三个和第四个4位相加了，最后只需要将新的第三位和第四位相加。*/
    int mask1 = (((((0x55 << 8) + 0x55) << 8) + 0x55) << 8) + 0x55;
    int mask2 = (((((0x33 << 8) + 0x33) << 8) + 0x33) << 8) + 0x33;
    int mask3 = (((((0x0f << 8) + 0x0f) << 8) + 0x0f) << 8) + 0x0f;
    x = (x & mask1) + ((x >>1) & mask1);
    x = (x & mask2) + ((x >>2) & mask2);
    x = (x & mask3) + ((x >>4) & mask3);
    x = x + (x>>8);
    x = x + (x>>16);
    return x & 0x3F;
}
/*
 * bang - Compute !x without using !
 *   Examples: bang(3) = 0, bang(0) = 1
 *   Legal ops: ~ & ^ | + << >>
 *   Max ops: 12
 *   Rating: 4
 */
int bang(int x) {
    x = x | (x >> 16);
    x = x | (x >> 8);
    x = x | (x >> 4);
    x = x | (x >> 2);
    x = x | (x >> 1);
    return x & 1;
}
/*
 * tmin - return minimum two's complement integer
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 4
 *   Rating: 1
 */
int tmin(void) {
  int x = 0x80 << 24;
  return x;
}
/*
 * fitsBits - return 1 if x can be represented as an
 *  n-bit, two's complement integer.
 *   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) {
    n = ~n + 1; /* 取得-n */
    int y = x << (32 + n) >> (32 + n); /* 将缩短后的”符号位“扩展”*/
    return !(y ^ x);
}
/*
 * divpwr2 - Compute x/(2^n), for 0 <= n <= 30
 *  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) {
    int bias = (x >> 31) & ((1 << n) + ~0); /* 如果x为正数，与运算后bias为0 */
    x = x + bias;
    return x >> n;
}
/*
 * negate - return -x
 *   Example: negate(1) = -1.
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 5
 *   Rating: 2
 */
int negate(int x) {
  return ~x + 1;
}
/*
 * isPositive - return 1 if x > 0, return 0 otherwise
 *   Example: isPositive(-1) = 0.
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 8
 *   Rating: 3
 */
int isPositive(int x) {
  return (!((x >> 31) & 1)) & (!!x);
}
/*
 * isLessOrEqual - if x <= y  then return 1, else return 0
 *   Example: isLessOrEqual(4,5) = 1.
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 24
 *   Rating: 3
 */
int isLessOrEqual(int x, int y) {
    int a = x >> 31;
    int b = y >> 31; /*正数的话为0，负数的话为-1*/
    int c = a ^ b;   /*分同异号来处理，异号的话c为-1，同号的话c为0*/
					 /*接下来对同号和异号两种情况分别处理，处理的结果不为零
					（准确讲是全f）代表x<=y，最后将两种情况或（实际只有一个不为零）*/
    int case1 = c & a; /*异号，只有当x为负数的时候返回1*/
    int case2 = ~c & ( ~((y + (~x + 1)) >> 31));/*同号，x-y的结果非负即返回1*/
    int result = case1 | case2;
    return !!result; /*由于result可能为全f，需要用!处理一下*/
}
/*
 * ilog2 - return floor(log base 2 of x), where x > 0
 *   Example: ilog2(16) = 4
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 90
 *   Rating: 4
 */
int ilog2(int x) {
    /*开始判断是在32位里的高16位还是低16位*/
    int a16 = !!(x >> 16); /*a16为1代表1在高16位，为0代表在低十六位*/
    int b16 = a16 << 4; /*1在高16位时b16 = 16*/
    /*开始判断是在16里的高8位还是低8位*/
    int a8 = !!(x >> (8 + b16));/*a8为1代表1在高8位，为0代表在低8位*/
    int b8 = a8 << 3;
    /*开始判断是在8里的高4位还是低4位*/
    int a4 = !!(x >> (4 + b8 + b16));/*a8为1代表1在高8位，为0代表在低8位*/
    int b4 = a4 << 2;
    /*开始判断是在4里的高2位还是低2位*/
    int a2 = !!(x >> (2 + b4 + b8 + b16));/*a8为1代表1在高8位，为0代表在低8位*/
    int b2 = a2 << 1;
    /*开始判断是在16里的高1位还是低1位*/
    int a1 = !!(x >> (1 + b2 + b4 + b8 + b16));/*a8为1代表1在高8位，为0代表在低8位*/
    int b1 = a1 << 0;
    return b1 + b2 + b4 + b8 + b16;
}
/*
 * float_neg - Return bit-level equivalent of expression -f for
 *   floating point argument f.
 *   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.
 *   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 mask = ~0 << 31;
    return mask ^ x;
}
/*
 * float_i2f - Return bit-level equivalent of expression (float) x
 *   Result is returned as unsigned int, but
 *   it is to be interpreted as the bit-level representation of a
 *   single-precision floating point values.
 *   Legal ops: Any integer/unsigned operations incl. ||, &&. also if, while
 *   Max ops: 30
 *   Rating: 4
 */
unsigned float_i2f(int x) {
    int position = 0; /*计算最高有效位距离第0位的位置*/
    unsigned result = 0;
    unsigned bias = 127;
    int tmp = 0;
    if (x == 0)
    {
        return result;
    }
    if (x == 0x80000000) /*由于-x = x，此种情况单独处理*/
    {
        result = 0xcf000000;
        return result;
    }
    for(int i = 16; i >= 1; i /= 2) /*计算距离*/
    {
        tmp = i + positon;
        if (x << (tmp) >> (tmp) == x)
        {
            position += i;
        }
    }
    result += (((30 - position) + bias) << 23);/*放置exp位*/
    if (x < 0)
    {
        x = -x;
        result | 0x80000000; /*放置符号位*/
    }
    tmp = 7 - position;
    if (tmp > 0)
    {
        x >>= tmp;
    }
    else
    {
        x = (x <<-tmp) & 0x007fffff; /*注意移到第24位的数字要变为0*/
    }
    return result | x; /*放置frac位*/
	/*刚好30个，O(∩_∩)O哈哈~*/
}
/*
 * float_twice - Return bit-level equivalent of expression 2*f for
 *   floating point argument f.
 *   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) {
    int tmp = 0x7f800000;
    int tmpp = uf & tmp; /*切割出exp位*/
    int tmppp = uf & 0x80000000; /*切割出符号位*/
    if (tmpp == tmp) /* infinity or NaN */
    {
        return uf;
    }
    if (tmpp == 0) /* Denormalized */
    {
        return (uf << 1) | tmppp; /*恢复符号位*/
    }
    return (uf & 0x807fffff) | (((tmpp >> 23) + 1) << 23); /* Noemalized */
}