CSAPP data Lab

注意，本文代码出于节省括号避免繁杂的考虑，对运算符优先级利用得比较充分，比如 1>>n+1 等价于 1>>(n+1)，所以代码里写了1>>n+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);}

思路

• 德摩根定律

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) {    int bias = n<<3;    return (x>>bias)&0xFF;}

思路

• 移位到最低的1byte然后用0xFF提取

logicalShift

/*  * logicalShift - shift x to the right by n, using a logical shift    i *   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 (1<<32+~n<<1)+~0 & (x>>n);  //equal to ((1<<31-n<<1)-1)&(x>>n);  //负号优先级高于移位}

思路

• 因为不能用-，所以用取反加一代替取负

• 构造低32-nbit的1来提取移位后的数值

• 因为移位量不能小于0或大于等于32，所以对于n可能是0而导致移位量是32的情况，先移位31位，再移位1位

  小技巧，如果n移位k，k∈[0, 32]，则可以n>>(k-!!k)>>!!k

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 */int bitCount(int x) {    int mark1 = 0x55;    int mark2 = 0x33;    int mark3 = 0x0F;    mark1 |= mark1<<8;    mark1 |= mark1<<16;    mark2 |= mark2<<8;    mark2 |= mark2<<16;    mark3 |= mark3<<8;    mark3 |= mark3<<16;    x = (x>>1&mark1)+(x&mark1); //every two bits; clear record;    x = (x>>2&mark2)+(x&mark2); //every four bits; clear record;    x = (x>>4&mark3)+(x&mark3); //every eight bits; clear record;    x = (x>>8)+x;   //every 16 bits; record in the low 8 bits;    x = (x>>16)+x;  //every 32 bits; record in the low 8 bits;    return x&0xFF;}

思路

• 构造0x55555555，提取每两位中的low bit。通过移位及0x55555555，提取每两位中的高位。然后相加，使得结果中，每两位的二进制值就是该两位的bit数目
• 同样的思路，提取每四位的low bit、high bit，然后相加
• 因为32==100000(二级制)，也就是只需要5位就可以记录有多少bit数，所以不需要每次都构造常数屏蔽高位的值，直接移位相加然后取低8bit就可以得到最终结果

bang

/* * 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>>1;    x |= x>>2;    x |= x>>4;    x |= x>>8;    x |= x>>16;    return ~x&0x1;}

思路

• 如果非0，位模式从最高位的1到最低位都填充为1，
• 如果为0，则位模式还是保持全0

tmin

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

fitBits

/*  * 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 !(x>>n+~0)|!((x>>n+~0)+1);  //equal to !(x>>n-1) | !((x>>n-1)+1)}

思路

• 算术移n-1位，如果是负数，且可以用n bits的补码表示，则得到-1。如果是正数，则得到0。

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) {    int t = x>>31;    return (x+(t&1<<n)+(~(t&1)+1))>>n;  //equal to (x+(t&1<<n)-(t&1))>>n;  //note that & 的优先级低于<<}

思路

• 直接移位是round down，无论是负数还是正数
• 所以要实现round to zero , C表达式为x<0 ? x+(pow(2,n)-1)>>n : x>>n

negate

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

思路

• 直接取反再加1

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) {    return ~(x>>31)&!!x;}

思路

• 符号位判断，并且非0

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) {    return !!(x>>31&~(y>>31)) | !(~(x>>31)&(y>>31))&(x+~y+1>>31) | !(x^y);    //equal to  !!(x>>31&~(y>>31)) | !(~(x>>31)&(y>>31))&(x-y>>31) | !(x^y)}

思路

• x<0&&y>0 | !(x>0&&y<0)&&(x-y>0) | x==y

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 mark1 = 0x55;    int mark2 = 0x33;    int mark3 = 0x0F;    mark1 |= mark1<<8;    mark1 |= mark1<<16;    mark2 |= mark2<<8;    mark2 |= mark2<<16;    mark3 |= mark3<<8;    mark3 |= mark3<<16;    x |= x>>1;    x |= x>>2;    x |= x>>4;    x |= x>>8;    x |= x>>16;    x >>= 1;    x = (x>>1&mark1)+(x&mark1); //every two bits; clear record;    x = (x>>2&mark2)+(x&mark2); //every four bits; clear record;    x = (x>>4&mark3)+(x&mark3); //every eight bits; clear record;    x = (x>>8)+x;   //every 16 bits; record in the low 8 bits;    x = (x>>16)+x;  //every 32 bits; record in the low 8 bits;    return x&0xFF;}

思路

• 先构造从最高的1到最低位均为1的二进制，然后类似bitCount

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 */unsigned float_neg(unsigned uf) {    unsigned t = uf&0x7FFFFFFF;    if(t^0x7F800000 && (t>>23)+1>>8)        return uf;    else         return uf^0x80000000;}

思路

• 判别是否是NaN。先判断尾数是否全0，然后用(t>>23)+1>>8判断exp是否全1

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 */unsigned float_i2f(int x) {    unsigned shiftLeft=0;    unsigned afterShift, tmp, flag;    unsigned absX=x;    unsigned sign=0;    //special case    if (x==0) return 0;    //if x < 0, sign = 1000...,abs_x = -x    if (x<0)    {        sign=0x80000000;        absX=-x;    }    afterShift=absX;    //count shift_left and after_shift    while (1)    {        tmp=afterShift;        afterShift<<=1;        shiftLeft++;        if (tmp & 0x80000000) break;    }    if ((afterShift & 0x01ff)>0x0100)        flag=1;    else if ((afterShift & 0x03ff)==0x0300)        flag=1;    else        flag=0;    return sign + (afterShift>>9) + ((159-shiftLeft)<<23) + flag;}//from http://www.cnblogs.com/tenlee/p/4951639.html

思路

• 分情况处理0、负数、正数

• 要处理舍人

  • 向接近的舍入
  • 如果处于中间，向偶数舍入

• 舍入时，如果尾数加一，exp有可能需要进位，这时候直接加一效果一样，可以导致exp进位，不需要特殊处理。如果exp等于0xFE，那么进位就变成了inf，也是合法的

float_twict

/*  * 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 t = uf&0x7FFFFFFF;    unsigned temp = t&0x7F800000;    unsigned temp2 = uf&0xFF800000;    int expFull = !(temp^0x7F800000);    if(t^0x7F800000 && expFull)        return uf;    if(expFull){        return temp2;    }    if(!(t&0x7F800000)){        unsigned k = (uf&0x7FFFFF);        return temp2+(k<<1);    }    return (temp>>23)+1<<23 | uf&0x807FFFFF;}

思路

• 分情况处理三种IEEE754的情况
• 需要注意exp全0时，乘以二就是尾数乘以二，如果发生进位需要exp进位，不需要特殊处理（第三个if），因为进位直接导致exp加一，这就足够了