【施工完成】CSAPP data lab

Posted by 111qqz on Thursday, February 28, 2019

TOC

CSAPP第二章的内容以前组成原理基本都学过…所以就简单翻了翻。

对应的lab是用位运算实现各种有的没的…

题目基本都很tricky…

除了用到一些常规的位运算性质,还用到了一些奇怪的条件:

  * ~0x7FFFFFFF = 0x7FFFFFFF + 1
  * 0xFFFFFFFF +1 =  0x00000000
  * 

0 == ~0+1

唯一让我觉得比较有趣的是how many bits这道题

题目要求是给一个32-bit signed int,问最少用多少位能得到它的补码表示。

考虑正数,显然,高位的连续的多个0是不必要的,只需要一个符号位的0即可。

那么对于负数,**高位的连续的多个1也是不必要的。 **原因是,-2^k + 2^(k-1) =  -2^(k-1),也就是说,去掉两个连续的1中高位的那个,数值没有改变。

我们可以将正数和负数统一来看,都是找到最高位的0和1的交界。

这可以通过和相邻的位置求异或,找到最高位的1的方式来实现。

接下来就是如何找一个数的最高位的1的位置了。

方法是构造一个单调的函数f,假设最高位位置为a,那么f((a,32))=0,f([0,a])=1.

然后在函数f上二分。

全部问题的代码如下,思路写在注释里了。还有3个涉及浮点数的问题之后补。

/* 
 * bitXor - x^y using only ~ and & 
 *   Example: bitXor(4, 5) = 1
 *   Legal ops: ~ &
 *   Max ops: 14
 *   Rating: 1
 */
/* 0 0 -> 0
   0 1 -> 1
   1 0 - > 1
   1 1 - > 0
*/
// ~(~a & ~b)&~(a&b)
int bitXor(int x, int y) {
  int ans = ~(~x & ~y)&(~(x&y));
  // printf("%d\n",ans);
  return ans;
}
/* 
 * tmin - return minimum two's complement integer 
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 4
 *   Rating: 1
 */
/*
  0X10000000
*/
int tmin(void) {

  int ans = 0x1 << 31;
  return ans;


}
//2
/* 
 * isTmax - returns 1 if x is the maximum, two's complement number,
 *     and 0 otherwise 
 *   Legal ops: ! ~ & ^ | +
 *   Max ops: 10
 *   Rating: 1
 *   0x7FFFFFFF
 */
int isTmax(int x) {
  /*
  大体思路首先是根据,如果x是最大值0x7FFFFFFF,那么~x和x+1(自然溢出)应该相等。
  不能用等号,但是我们可以用异或。x==y 等价于  !(x^y). 因此有了后半段!(x+1)^(~x)
  但是满足这个条件的还有-1,也就是0xFFFFFFFF,因此我们需要排除掉-1.
  还是用异或的性质,这回是0异或者任何数都等于其本身。
  因此如果x为-1,那么前后两部分都为1,结果为0.
  如果x为TMAX,那么前面为0,后面为1,结果为1.
  如果x为其他任何数,前后结果都应为0. 结果为0。
  */
  return (!(x+1))^!((x+1)^(~x));                                                                                                                                   
}
/* 
 * allOddBits - return 1 if all odd-numbered bits in word set to 1
 *   where bits are numbered from 0 (least significant) to 31 (most significant)
 *   Examples allOddBits(0xFFFFFFFD) = 0, allOddBits(0xAAAAAAAA) = 1
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 12
 *   Rating: 2
 */
// 理解错误。误以为是要求当前长度x的所有奇数位上都是1.
// 实际上要求和x的长度无关,而是要求[0,31]中,所有奇数位上都是1.
int allOddBits(int x) {
  int half_mask = (0xAA<<8) | 0xAA;
  int mask = (half_mask<<16) + half_mask;
  // printf("mask:X x:x x\n",mask,x,x&mask);
  return  !((x&mask)^mask);  
  
}
/* 
 * negate - return -x 
 *   Example: negate(1) = -1.
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 5
 *   Rating: 2
 */
int negate(int x) {
  return ~x + 1;
}
//3
/* 
 * isAsciiDigit - return 1 if 0x30 <= x <= 0x39 (ASCII codes for characters '0' to '9')
 *   Example: isAsciiDigit(0x35) = 1.
 *            isAsciiDigit(0x3a) = 0.
 *            isAsciiDigit(0x05) = 0.
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 15
 *   Rating: 3
 */
/* 把0-9的二进制写出来,发现0-7占满了3bit的二进制的8种组合。
   因此考虑只判断8和9两种4bit的情况
   构造mask,不在意的bit的位置放0,在意的bit位置放1.
*/
int isAsciiDigit(int x) {
  int mask = 0x0E;
  int ones = x&mask;
  int ones_3 = ones >> 3;
  int tens = x>>4;
  // printf("x: x tens: x ones:x\n",x,tens,ones);
  int ones_ok = (!(ones^0x8)) | (!ones_3);
  int tens_ok = !(tens^0x3);
  return ones_ok & tens_ok;

}
/* 
 * conditional - same as x ? y : z 
 *   Example: conditional(2,4,5) = 4
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 16
 *   Rating: 3
 */
/*
  关键思路是0xFFFFFFFF和0x00000000之间差了1.
  而这两个数一个是全部位置都取的mask,一个是全部位置都不取的mask.
*/
int conditional(int x, int y, int z) {
  return   z^(!x + ~0 )&(y^z);

}
/* 
 * isLessOrEqual - if x <= y  then return 1, else return 0 
 *   Example: isLessOrEqual(4,5) = 1.
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 24
 *   Rating: 3
 */
/*
  大体思路是,符号位相同和符号位不同分别考虑
  符号位相同:  考虑差的符号位。
  符号位不同: 当x<0,y>=0时结果为1.
*/
int isLessOrEqual(int x, int y) {
  int minus = y + (~x+1);
  int s_x = (x>>31)&1;
  int s_y = (y>>31)&1;
  int s_minus = (minus>>31) & 1;
  return (s_x&(!s_y))| (!(s_x^s_y)&!s_minus); 
}
//4
/* 
 * logicalNeg - implement the ! operator, using all of 
 *              the legal operators except !
 *   Examples: logicalNeg(3) = 0, logicalNeg(0) = 1
 *   Legal ops: ~ & ^ | + << >>
 *   Max ops: 12
 *   Rating: 4 
 */
/* 
  0 == ~0+1
  -2147483648 = ~-2147483648+1
  满足 x == ~x+1
  重点是x和~x+1的符号位相同,如果都是0那么x=0,如果都是1那么x=-214783648`
*/
int logicalNeg(int x) {
  int s1 = (x>>31)&1;
  int s2 = ((~x+1)>>31)&1;
  // printf("s1: %d s2:%d  %d  %d\n",s1,s2,s1|s2,~(s1|s2));
  //  1 + negate(0) -> 1
  //  1 + neagate(1) -> 0
  return 1+(1+~(s1|s2));
}
/* howManyBits - return the minimum number of bits required to represent x in
 *             two's complement
 *  Examples: howManyBits(12) = 5
 *            howManyBits(298) = 10
 *            howManyBits(-5) = 4
 *            howManyBits(0)  = 1
 *            howManyBits(-1) = 1 ??? should be 2?
 *            howManyBits(0x80000000) = 32
 *  Legal ops: ! ~ & ^ | + << >>
 *  Max ops: 90
 *  Rating: 4
 */
/*
  思路似乎可以转化成判断一个数(可正可负)的最高位的1的位置。
  判断最高位1用二分的办法。
  构造一个单调的函数,假设最高位位置为a,那么f((a,32))=0,f([0,a])=1.
  被 howManyBits(-1)==1 困扰了好久,实际上就是0x1,只有一位,改位就是符号位的情况。 
*/
int howManyBits(int x) {
  int n = 0 ;
  x^=(x<<1);
  n +=  (!!( x & ((~0) << (n + 16)) )) << 4;   // 看高16位是否为0,是的话区间为[0,16),否的话为[16,32)
  // printf("n:%d\n",n);
  // printf("%d\n",!!(x & ((~0) << (n + 16))));
  n +=  (!!( x & ((~0) << (n + 8)) )) << 3;
  // printf("n:%d\n",n);
  n +=  (!!( x & ((~0) << (n + 4)) )) << 2;
  // printf("n:%d\n",n);
  n +=  (!!( x & ((~0) << (n + 2)) )) << 1;
  // printf("n:%d\n",n);
  n +=  (!!( x & ((~0) << (n + 1)) ));
  // printf("n:%d\n",n);

  // int s = (x>>31)&1;
  // int ret = n+1+((1^s)&(!!x));
  // // printf("x:%d ret:%d\n",x,ret);
  
  return n+1;
}

补上三个涉及浮点数的问题…比较无聊,按照IEEE754操作即可.

//float
/* 
 * floatScale2 - 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 floatScale2(unsigned uf)
{
  int exp_ = (uf & 0x7f800000) >> 23;
  int s_ = uf & 0x80000000;
  if (exp_ == 0)
    return (uf << 1) | s_;
  if (exp_ == 255)
    return uf;
  ++exp_;
  if (exp_ == 255)
    return 0x7f800000 | s_;
  return (uf & 0x807fffff) | (exp_ << 23);
}
/* 
 * floatFloat2Int - Return bit-level equivalent of expression (int) f
 *   for floating point argument f.
 *   Argument is passed as unsigned int, but
 *   it is to be interpreted as the bit-level representation of a
 *   single-precision floating point value.
 *   Anything out of range (including NaN and infinity) should return
 *   0x80000000u.
 *   Legal ops: Any integer/unsigned operations incl. ||, &&. also if, while
 *   Max ops: 30
 *   Rating: 4
 */
int floatFloat2Int(unsigned uf)
{
  int s_ = uf >> 31;
  int exp_ = ((uf & 0x7f800000) >> 23) - 127;
  int frac_ = (uf & 0x007fffff) | 0x00800000;
  if (!(uf & 0x7fffffff))
    return 0;

  if (exp_ > 31)
    return 0x80000000;
  if (exp_ < 0)
    return 0;

  if (exp_ > 23)
    frac_ <<= (exp_ - 23);
  else
    frac_ >>= (23 - exp_);

  if (!((frac_ >> 31) ^ s_))
    return frac_;
  else if (frac_ >> 31)
    return 0x80000000;
  else
    return ~frac_ + 1;
}
/* 
 * floatPower2 - Return bit-level equivalent of the expression 2.0^x
 *   (2.0 raised to the power x) for any 32-bit integer x.
 *
 *   The unsigned value that is returned should have the identical bit
 *   representation as the single-precision floating-point number 2.0^x.
 *   If the result is too small to be represented as a denorm, return
 *   0. If too large, return +INF.
 * 
 *   Legal ops: Any integer/unsigned operations incl. ||, &&. Also if, while 
 *   Max ops: 30 
 *   Rating: 4
 */
unsigned floatPower2(int x)
{
  int exp = x + 127;
  if (exp <= 0)
    return 0;
  if (exp >= 255)
    return 0x7f800000;
  return exp << 23;
}

「真诚赞赏,手留余香」

111qqz的小窝

真诚赞赏,手留余香

使用微信扫描二维码完成支付


comments powered by Disqus