datalab

bits.c

/* 
 * CS:APP Data Lab 
 * 
 * <Please put your name and userid here>
 * bei wei qiang
 * 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:
  只允许使用:
  1. Integer constants 0 through 255 (0xFF), inclusive. You are
      not allowed to use big constants such as 0xffffffff.
     整数常量 0 ~ 255
  2. Function arguments and local variables (no global variables).
     只能使用 函数参数 和 局部变量 (不能使用全局变量)
  3. Unary integer operations ! ~
     一元操作符 ! ~
  4. Binary integer operations & ^ | + << >>
     二元操作符 & ^ | + << >>
    
  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:
  不允许使用:
  1. Use any control constructs such as if, do, while, for, switch, etc.
     不能使用任何控制结构
  2. Define or use any macros.
     不能定义和使用任何宏
  3. Define any additional functions in this file.
     不能在文件里定义任何其他的函数
  4. Call any functions.
     不能调用任何函数
  5. Use any other operations, such as &&, ||, -, or ?:
     不能使用任何其他操作符, 比如: && || - ?:
  6. Use any form of casting.
     不能使用任何形式的 类型转换
  7. Use any data type other than int.  This implies that you
     cannot use arrays, structs, or unions.
     不能使用除了 int 类型外的其他数据类型.
     比如, 不能使用 array, struct, union

 
  You may assume that your machine:
  你可以假设你的机器是这样情况:
  1. Uses 2s complement, 32-bit representations of integers.
     使用补码, 32位 表示整数
  2. Performs right shifts arithmetically.
     右移 是 算术右移
  3. Has unpredictable behavior when shifting an integer by more
     than the word size.
     将整数移位超过字的大小时, 后果不可预知!

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
浮点数编码规则

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.
对于浮点数运算, 编码规则没有那么严格.
允许使用 循环 和 条件控制.
可以使用 int 和 unsigned.
可以使用任意 整数 和 unsigned 常量.


You are expressly forbidden to:
不能使用:
  1. Define or use any macros.
     不能定义和使用宏
  2. Define any additional functions in this file.
     不能定义另外的函数
  3. Call any functions.
     不能调用任何函数
  4. Use any form of casting.
     不能使用任何形式的 类型转换
  5. Use any data type other than int or unsigned.  This means that you
     cannot use arrays, structs, or unions.
     除了 int 和 unsigned 类型, 其他数据类型不能使用.
     不能使用 array, struct, union
  6. Use any floating point data types, operations, or constants.
     不能使用任何浮点数据类型, 操作 或者 常量


NOTES:
  1. Use the dlc (data lab checker) compiler (described in the handout) to 
     check the legality of your solutions.
     使用 dlc 编译器检查解决方案的合法性
  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.
     对于实现的函数, 都有最大操作符数量的使用限制.
     dlc 会检查最大操作符数量.
     '=' 不会检查.
  3. Use the btest test harness to check your functions for correctness.
     使用 btest 测试函数的正确性
  4. Use the BDD checker to formally verify your functions
     使用 BDD
  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.
     每个函数操作符的最大数量会在头部的 comment 里面给出.
     如果操作符的最大数量, 在 writeup 和 这个文件 中给出的数量不一致, 以该文件为权威.


/*
 * 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) {
  int mask = ((1 << 31) >> n) << 1;
  return (x >> n) & (~mask);
}
/*
 * 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) {

  return 2;
}
/* 
 * 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) {
  return (1 << 31) & (~0);
}
/* 
 * 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) {

  return 2;
}
/* 
 * 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) {
  /*
   * 正数: x >> n
   * 负数: (x + (1 << n) - 1) >> n
   * */
  int isNeg = x >> 31;
  int a = ((isNeg & 1) << n ) + isNeg;

  return (x + a) >> 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) {
  int sign = (x >> 31) & 1;
  return (!sign) & !!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 + ((~y) + 1);
//  int sign = (a >> 31) & 1;
//  return sign ^ (!a);
  return 2;
}
/*
 * 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) {
  return 2;
}
/* 
 * 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) {
  if ((((uf >> 23) & 0xff) == 0xff) && ((uf & 0x7fffff) != 0)) return uf;
  else return uf ^ (1 << 31);
}
/* 
 * 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) {


  return 2;
}
/* 
 * 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 sign = (uf & (1 << 31)) | ~(1 << 31);

  if ((((uf >> 23) & 0xff) == 0xff) || (uf == 0) || (uf == 0x80000000)) {
    // 如果是 无穷/NaN/+0/-0 返回参数
    return uf;
  } else if (((uf >> 23) & 0xff) == 0) {
    // 如果是 非规格化, 左移一位
    return ((uf << 1) | (1 << 31)) & sign;
  } else {
    // 如果是 规格化, 加上 1 << 23
    return uf + (1 << 23);
  }
}