CSAPP深入理解计算机系统实验datalab解析       

        看完这一本《CSAPP深入理解计算机系统》自然应该将配套的实验好好做做，这是巩固知识提升运用能力的一个非常好的方法，第一个实验就是这个datalab，要求我们对位运算有很深的理解和掌握，在做这个实验的过程中，遇到了很多难题，基本上都能通过查阅资料和自我分析解决，下面我将我的答案贴出来，以便以后再做此实验的同仁能够参考一下，所有函数都通过检测，并且大部分函数都包含我自己的理解所以我都加上了注释，有网址的地方是借鉴和参考的原网址，在原内容的基础上我又增加了自己的思考与修改。由于CMU的这门计算机导论课程一直在更新，所以很多最新的内容在网络上比较乱，大家在这篇日志可以看到对该实验的最新解析。下面提供几篇可供参考的关于该实验的日志：

       http://blog.csdn.net/lwfcgz/article/details/8515188   这篇文章的实验内容比较新，少部分不同，美中不足的是注释不多。

        http://blog.sina.com.cn/s/blog_cd4677b40101ieps.html  这篇文章介绍了几个需要考虑的细节问题。

        http://codejam.diandian.com/archive   这个博客讲解比较详细，注释丰富，但实验内容不全。

       http://blog.csdn.net/kqzxcmh/article/details/11845165 这篇文章有很多其他函数的实现，可以继续深入学习。

/*  * 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/*  * 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) {  return x>>(n<<3)&0xff;}/*  * 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) {  return (x>>n)&(~((0x1<<31)>>n<<1));}/* * bitCount - returns count of number of 1's in word *   Examples: bitCount(5) = 2, bitCount(7) = 3 *   Legal ops: ! ~ & ^ | + << >> *   Max ops: 40 *   Rating: 4 */ /* **参考《CSAPP深入理解计算机系统》例3.50解答 */int bitCount(int x) {  int tmp=((0x01<<8|0x01)<<8|0x01)<<8|0x01;  int val=x&tmp;  val+=tmp&(x>>1);  val+=tmp&(x>>2);  val+=tmp&(x>>3);  val+=tmp&(x>>4);  val+=tmp&(x>>5);  val+=tmp&(x>>6);  val+=tmp&(x>>7);  val=val+(val>>16);  val=val+(val>>8);  return val&0xff;}/*  * bang - Compute !x without using ! *   Examples: bang(3) = 0, bang(0) = 1 *   Legal ops: ~ & ^ | + << >> *   Max ops: 12 *   Rating: 4  *//***If x!=0，x+(~x+1)＝2^32，the highest bit of x and (~x+1)cannot be both 0(至少有一个为1再加上进位就可以继续进位，否则不会变成2^32).*/int bang(int x) {  return (~((x|(~x+1))>>31))&0x1;}/*  * tmin - return minimum two's complement integer  *   Legal ops: ! ~ & ^ | + << >> *   Max ops: 4 *   Rating: 1 */int tmin(void) {  return 0x1<<31;}/*  * 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 *//* if all negative numbers(e.g x) are changed into ~x, all the numbers those    * can be represented as an n-bit, two's complement integer will be in range    * of 0~2^(n-1). If it was arithmatically shifted (n-1) bits to right, it'll   * be 0x00000000.    http://codejam.diandian.com/post/2011-04-16/427784*/int fitsBits(int x, int n) {  int isPositive=!(x>>31);  int shift=n+~0;    //shift=n-1;  return (isPositive&!(x>>shift))|(!isPositive&!((~x)>>shift));}/*  * 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 sign=x>>31;    int bias=sign & ((1<<n) + (~0));    return (x+bias)>>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)|(!x));}/*  * isLessOrEqual - if x <= y  then return 1, else return 0  *   Example: isLessOrEqual(4,5) = 1. *   Legal ops: ! ~ & ^ | + << >> *   Max ops: 24 *   Rating: 3 *//* Because the substraction of two numbers with different signs may cause    * overflow, we should only judge whether x is negative then return    * 1(true). Otherwise we'll judge whether (x-y) is a non-positive     ((x+(~y)+1-1)>>31,which includes the situation of x-y=0)   * number and return 1(true).    reference:http://codejam.diandian.com/post/2011-04-16/427885 */int isLessOrEqual(int x, int y) {  int signx=x>>31;  int signy=y>>31;  int IsSameSign=(!(signx^signy));  return (IsSameSign & ((x+(~y))>>31)) | ((!IsSameSign) & signx);}/* * ilog2 - return floor(log base 2 of x), where x > 0 *   Example: ilog2(16) = 4 *   Legal ops: ! ~ & ^ | + << >> *   Max ops: 90 *   Rating: 4 */  /* ilog2 is equivlance of finding the index of first 1 from left, the method is as following:   * check whether x's left half bits are all 0's, if so then throw them away,    * else right shift to reject the right half(remember the 16 bits thrown from    * right), to repeat it until only one "1" remaining and return the remembered   * number of x having been eliminated.     http://codejam.diandian.com/post/2011-04-16/428134   */int ilog2(int x) {  int shift1,shift2,shift3,shift4,shift5;  //The value of 2^n has the form:0……010……0,so we can cut it short and increment the count variable with the corresponding number.  int sign = !!(x >> 16); // whether x's left 16 bits aren't all 0's  shift1 = sign << 4;  x = x >> shift1; // if so, right shift 16 bits  sign = !!(x >> 8); // whether x's left 8 bits aren't all 0's  shift2 = sign << 3;  x = x >> shift2; // if so, right shift 8 bits   sign = !!(x >> 4); // whether x's left 4 bits aren't all 0's  shift3 = sign << 2;  x = x >> shift3; // if so, right shift 4 bits   sign = !!(x >> 2); // whether x's left 2 bits aren't all 0's  shift4 = sign << 1;  x = x >> shift4; // if so, right shift 2 bits   sign = !!(x >> 1); // whether x's left 1 bits aren't all 0's  shift5 = sign; // if so, count will be added 1  // return all the shifts whose sum means the index of first "1" from left  return shift1+shift2+shift3+shift4+shift5; }/*  * 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 result= uf ^ 0x80000000;//change the signed bit of uf  unsigned tmp=uf & 0x7fffffff;//change the signed bit of uf to 0  if( tmp > 0x7f800000 ) //When argument is NaN(exponent bits are all 1 and mantissa bits arenot all 0), return argument.                         //IEEE754 standard:1 bit signed bit,8 bits exponent,23 bits mantissa.      result=uf; return result;}/*  * 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) {//Rounding is important!  unsigned sign=0,shiftleft=0,flag=0,tmp;  unsigned absx=x;  if( x==0 ) return 0;  if( x<0 ){     sign=0x80000000;     absx=-x;  }  while(1){//Shift until the highest bit equal to 1 in order to normalize the floating-point number     tmp=absx;     absx<<=1;     shiftleft++;     if( tmp&0x80000000 ) break;  }  if( (absx & 0x01ff) > 0x0100 ) flag=1;//***Rounding 1  if( (absx & 0x03ff) == 0x0300 ) flag=1;//***Rounding 2    return sign+(absx>>9)+((159-shiftleft)<<23)+flag;}/*  * 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) {  /* Computer 2*f. If f is a NaN, then return f. */      unsigned exponent = uf & 0x7F800000;      unsigned sign = uf & 0x80000000;      if (exponent) {         if (  exponent != 0x7F800000 ) {  //if exponent bits are all 1 and mantissa bits arenot all 0,the number is NaN.          uf = uf + 0x00800000;   //Increment exponent by 1,effectively multiplying by 2.            }    }    else       uf = ( uf << 1) | sign ;  //shift one bit to left      return uf;}

运行结果如下：

Score    Rating    Errors    Function
 1           1              0             bitAnd
 2           2              0             getByte
 3           3              0             logicalShift
 4           4              0             bitCount
 4           4              0             bang
 1           1              0             tmin
 2           2              0             fitsBits
 2           2              0             divpwr2
 2           2              0             negate
 3           3              0             isPositive
 3           3              0             isLessOrEqual
 4           4              0             ilog2
 2           2              0             float_neg
 4           4              0             float_i2f
 4           4              0             float_twice
Total points: 41/41