213 lab1 datalab

Lab1

第一次做213的lab 呕出一口老血 下面的代码都是运行通过的

/*  * 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 byediting 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 RULESFor the problems that require you to implent floating-point operations,the coding rules are less strict.  You are allowed to use looping andconditional 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. */#endif

以上部分是一些说明和要求
下面开始是练习题

1.bitAnd

/*  * 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))); //    ~| == &}

2.getByte

/*  * 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) {  return (x>>(n<<3))&0xff;//将一个byte单独取出来}

3.logicalShift

/*  * 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  */向右移动n bit高位补0int logicalShift(int x, int n) { int mask=~(((1<<31)>>n)<<1); //做好一个mask 把右移高位的1去掉return (x>>n)&mask; }

4.bitCount

/* * bitCount - returns count of number of 1's in word *   Examples: bitCount(5) = 2, bitCount(7) = 3 *   Legal ops: ! ~ & ^ | + << >> *   Max ops: 40 *   Rating: 4 */计算1的个数int bitCount(int x) {          int result;          //int mask1=(0x55)|(0x55<<8)|(0x55<<16)|(0x55<<24); 0101010101...         int tmp_mask1=(0x55)|(0x55<<8);          int mask1=(tmp_mask1)|(tmp_mask1<<16);          //int mask2=(0x33)|(0x33<<8)|(0x33<<16)|(0x33<<24);  001101100110011...        int tmp_mask2=(0x33)|(0x33<<8);          int mask2=(tmp_mask2)|(tmp_mask2<<16);          //int mask3=(0x0f)|(0x0f<<8)|(0x0f<<16)|(0x0f<<24);  000011110000111100001111...        int tmp_mask3=(0x0f)|(0x0f<<8);          int mask3=(tmp_mask3)|(tmp_mask3<<16);          int mask4=(0xff)|(0xff<<16);          int mask5=(0xff)|(0xff<<8);          //add every two bits          result=(x&mask1)+((x>>1)&mask1);          //add every four bits          result=(result&mask2)+((result>>2)&mask2);          //add every eight bits          //result=(result&mask3)+((result>>4)&mask3);          result=(result+(result>>4))&mask3;          //add every sixteen bits          //result=(result&mask4)+((result>>8)&mask4);          result=(result+(result>>8))&mask4;          //add every thirty two bits          //result=(result&mask5)+((result>>16)&mask5);          result=(result+(result>>16))&mask5;          return result;  }

5.bang

/*  * bang - Compute !x without using ! *   Examples: bang(3) = 0, bang(0) = 1 *   Legal ops: ~ & ^ | + << >> *   Max ops: 12 *   Rating: 4  */判断x是否为0 非0即1  x如果非0，则(x|(~x+1)的符号位为1，如果为0，反之int bang(int x) {return ~((x|(~x+1))>>31)&0x1; //返回值也是一个int}

6.tmin

/*  * tmin - return minimum two's complement integer 补码最小的整数 *   Legal ops: ! ~ & ^ | + << >> *   Max ops: 4 *   Rating: 1 */int tmin(void) {  return 1<<31;}/* 

7.fitsBits

 * 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 */如果能用n位补码表示int fitsBits(int x, int n) {  int sign = x>>31;//all 0 or all 1  return (~sign & !(x>>(n+~0)))+(sign & !((~x)>>(n+~0))) ;//除去符号位 剩下的数值部分是否能用n-1位来表示}

8.divpwr2

/*  * 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) {//要分情况讨论 正数可以直接右移 负数要考虑符号位    int s=!!(x>>31);//取x的符号位  0 or 1    int t=(1<<n)+~0;//制作一个只有低n位为1的mask    int lowx=t&x;//lowx保存x的低n位的值    return  (x>>n)+((!!lowx)&s);//}负数进行右移的时候要进行Biasing 加上2的k次方减1后在右移k位[草稿](http://img.blog.csdn.net/20160921114817987)

9.negate

/*  * negate - return -x  *   Example: negate(1) = -1. *   Legal ops: ! ~ & ^ | + << >> *   Max ops: 5 *   Rating: 2 */int negate(int x) {  return ~x+1;}

10.isPositive

/*  * isPositive - return 1 if x > 0, return 0 otherwise  *   Example: isPositive(-1) = 0. *   Legal ops: ! ~ & ^ | + << >> *   Max ops: 8 *   Rating: 3 */int isPositive(int x) {  int sign = x>>31;  return (!sign) & (!!x);//仅仅判断符号位是不够的 因为0要单独考虑 不为0且符号位为0才行}

11.isLessOrEqual

/*  * 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) {//consider overflowint sub=y+(~x+1); //比较数的大小 要考虑符号int signx = (x >> 31) & 1; int signy = (y >> 31) & 1; int subsign =!((sub >> 31) & 1);//正的话 1 负的话是0 return ((signx ^ signy) & signx) + ((signx ^ signy ^ 1) & subsign); }

12.ilog2

/* * 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){  int byte3 = x >> 24;  int byte2 = (x >> 16) & 0xff;  int byte1 = (x >> 8) &0xff;  int byte0 = x &0xff;// get the four parts 每个byte单独取出来  int i3 = !!byte3;//see if byte>0 看byte里面是不是为0  int i2 = i3 | !!byte2;  int i1 = i2 | !!byte1;  int i0 = i1 | !!byte0;  int i = i3 + i2 +i1 +i0 +~0;//注意要减一 得出的结果就是最高位在第几个byte里  int highbyte = x >> (i <<3);//get highest byte  int b7 = (highbyte >> 7) & 1;  int b6 = (highbyte >> 6) & 1;  int b5 = (highbyte >> 5) & 1;  int b4 = (highbyte >> 4) & 1;  int b3 = (highbyte >> 3) & 1;  int b2 = (highbyte >> 2) & 1;  int b1 = (highbyte >> 1) & 1;  int b0 = (highbyte) & 1;//get the number of every bit of highbyte 0 or 1 每一位都get  int k7 = b7;  int k6 = k7|b6;  int k5 = k6|b5;  int k4 = k5|b4;  int k3 = k4|b3;  int k2 = k3|b2;  int k1 = k2|b1;  int k0 = k1|b0;  int k = k7 +k6 +k5 +k4 +k3 +k2+k1+k0+~0; 找到最高位在哪里  return (i << 3) + k;//}

13.float_neg

/*  * 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 */  f变-f 这里要考虑NAN的特殊情况 判断是不是Nan exp（之间8位）部分全为1 数值部分（后23位）不全为0 //NaN for Not a Nunmber unsigned float_neg(unsigned uf) {  if(   (((uf >> 23 )& 0xff)^0xff) ||  !(uf & ((1<<23)-1))){// two evidence to define if uf is NaN  uf ^=(1<<31);}//异或 只修改符号位  return uf;//如果是Nan 直接return本身}

14.float_i2f

/*  * 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 */强行格式转换 int变floatunsigned float_i2f(int x) {   unsigned shiftLeft=0;    unsigned afterShift, tmp, flag;    unsigned absX=x;    unsigned sign=0;    //special case 考虑正数负数和0    if (x==0) return 0;    //if x < 0, sign = 1000...,abs_x = -x    if (x<0)负数的情况 取符号 取绝对值    {        sign=0x80000000;        absX=-x;    }    afterShift=absX;    //count shift_left and after_shift    while (1)    {        tmp=afterShift;        afterShift<<=1;        shiftLeft++;//一直左移 直到出现最高位的1        if (tmp & 0x80000000) break;    }    if ((afterShift & 0x01ff)>0x0100)        flag=1;//在截断数值部分低位的时候有两种情况要进行进位 第一种后面九位最高位为1 剩余八位不为0    else if ((afterShift & 0x03ff)==0x0300)        flag=1;//第二种最后九位为100000000 截断前一位也为1    else        flag=0;    return sign + (afterShift>>9) + ((159-shiftLeft)<<23) + flag;}

15.float_twice

/*  * 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) {   unsigned f = uf;    if ((f & 0x7F800000) == 0) //    {        //如果exp位全为0 符号位不变 左移        f = ((f & 0x007FFFFF) << 1) | (0x80000000 & f);    }    else if ((f & 0x7F800000) != 0x7F800000)    {//如果exp不全为1 exp部分加一        f =f + 0x00800000;    }    return f;//其余情况 无穷以及Nan直接返回本身}