CSAPP lab1 datalab-handout

这是一个关于机器级的整数、浮点数表示和位运算的实验。要求用给定的操作符、尽可能少的操作数去实现对应的函数功能。

完整的实验包：链接: https://pan.baidu.com/s/1xUBi3XDlidPQFNexbjXoLw 密码: 2333

以下是全部函数的代码：

 /*****************************计算log2(x)向下取整*******************************/
 int ilog2(int x) {
     int bit_16, bit_8, bit_4, bit_2, bit_1;
     bit_16 = (!!(x >> )) << ;                //如果x >> 16非零，则至少有16位
     x = x >> bit_16;
     bit_8 = (!!(x >> )) << ;
     x = x >> bit_8;
     bit_4 = (!!(x >> )) << ;
     x = x >> bit_4;
     bit_2 = (!!(x >> )) << ;
     x = x >> bit_2;
     bit_1 = x >> ;      //还剩两位时，直接判断首位。bit_1 == 1,剩两位；bit_1 == 0,剩一位
     return bit_16 + bit_8 + bit_4 + bit_2 + bit_1;
　　　　　//实际是（bit_1 + 1）-1；由于向下舍入，总位数减一
 }

 /****************************表达x所需要的最少位数******************************/
 int howManyBits(int x) {
     int bit_16, bit_8, bit_4, bit_2, bit_1, result;
     int k = x >> ;
     int temp = x ^ k;                      
　　　　　//x为正，temp = x;x为负，temp = ~x
     int isZero = (!!(temp << )) >> ; 
　　　　 //x = 0或x = -1时，temp = 0,isZero = 0...0；否则isZero = 1...1
     bit_16 = (!!(temp >> )) << ;
     temp = temp >> bit_16;
     bit_8 = (!!(temp << )) << ;
     temp = temp >> bit_8;
     bit_4 = (!!(temp << )) << ;
     temp = temp >> bit_4;
     bit_2 = (!!(temp << )) << ;
     temp = temp >> bit_2;
     bit_1 = temp >> ;
     result = bit_16 + bit_8 + bit_4 + bit_2 + bit_1 + ;
　　　　　　//真正位数为bit_16 + ... + bit_1 + 1,再加符号位一位
     return (!isZero) | (result & isZero);
 }

 /************************************逻辑右移*************************************/
 int logicshift(int x, int n)
 {
     int temp = ~( << );
     temp = ((temp >> n) << ) + ;                //生成掩码0...01...1(前面为n个0)
     return (x >> n) & temp;
 }

 /*************************类似于c语言中的x ? y : z**********************************/
 int conditional(int x, int y,int z)
 {
     int temp = (~(!x)) + ;                      //要在return中完成，必须生成x,y的掩码
     return (temp & z) | ((~temp) & y);           //当x = 0时，temp = 1...1;当x != 0时，temp = 0...0
 }

 /*********************************** x/ 2^n *****************************************/
 int divpwr2(int x, int n)
 {
     int temp = ( << n) + (~);                  //temp为baising(偏置),1...1(共n个1)
     return (x + ((x >> ) & temp)) >> n;        //只有负数才要加偏置，所以temp要与符号位相与
 }

 /****************************** x < y ? ***********************************/
 int isLessOrEqual(int x, int y)
 {
     int signx = x >> ;
     int signy = y >> ;
     int signEqual = (!(signx ^ signy) & ((x + (~y)) >> ));//符号位不同时，做差
     int signDiffer = signx & (!signy);                      //符号位相同，直接比较符号位
     return signEqual | signDiffer;
 }

 /**********************操作数更小的版本**************************/
 int isLessOrEqual_2(int x, int y)
 {
     int not_y = ~y;
     return ((((x + not_y) & (x ^ not_y)) | (x & not_y)) >> ) & ;
     //        x-y-1<0 <----------x,y不同号------>x为负，y为正，才为正
 }

 /****************************不用负号得到-x*********************************/
 int negate(int x)
 {
     return ~x + ;                //按位取反，末位加一
 }

 /**********************返回最小的补码***************************************/
 int tmin(void)
 {
     return  << ;
 }

 /*************************只用~ 和 | 实现x&y*****************************/
 int bitAnd(int x, int y)
 {
     return ~(~x | y);            //摩根律
 }

 /**************************从字x中取出第n个字节*********************************/
 int getByte(int x, int n)
 {
     return (x >> (n << )) & 0xff;        //是从0开始数的
 }

 /*********************************计算x中1的数目*********************************/
 int bitCount(int x)
 {
     int result;
     int tmpmark1 = 0x55 + (0x55 << );                //最大0xff
     int mark1 = tmpmark1 + (tmpmark1 << );
     int tmpmark2 = 0x33 + (0x33 <<);
     int mark2 = tmpmark2 + (tmpmark2 << );
     int tmpmark3 = 0x0f + (0x0f << );
     int mark3 = tmpmark3 + (tmpmark3 << );
     int mark4 = 0xff + (0xff << );
     int mark5 = 0xff + (0xff << );                  //以上生成5个掩码

     result = (x & mark1) + ((x >> ) & mark1);
     result = (result & mark2) + ((result >> ) & mark2);   //这两个由于进位问题，不能先加再与
     result = (result + (result >> )) & mark3;              //分治
     result = (result + (result >> )) & mark4;
     result = (result + (result >> )) & mark5;
     return result;
 }

 /***************************计算uf/2*********************************/
 unsigned float_half(unsigned uf)
 {
     unsigned s = uf & 0x80000000;
     unsigned exp = uf & 0x7f800000;
     int lsb = ((uf & ) == );                //判断frac最后两位是否为11
     if (exp == 0x7f800000)
         return uf;
     else if (exp <= 0x800000)
         return s | (((uf ^ s) + lsb) >> );  //uf^s将符号位置零，uf^s = frac + exp末位
     else
         return uf - 0x800000;                 //整体思路就是模拟
 }

 /****************************计算（float）x***********************************/
 int float_f2i(unsigned uf)
 {
     int abs;
     int sign = uf >> ;
     int exp = (uf >> ) & 0xff;
     int frac = uf & 0x007fffff;
     if (exp < 0x7f)        return ;
     if (exp > )        return 0x80000000;                //Tmax = 2^31 -1

     abs = ((frac >> ) + ) << (exp - );            //模拟
     if (sign)
         return -abs;
     else
         return abs;
 }

 /****************************计算2*f***************************************/
 unsigned float_twice(unsigned uf)
 {
     int result;
     int exp = uf & 0x7f800000;
     int frac = uf & 0x7fffff;
     if (exp == 0x7f800000)
         return uf;
     else if (exp == )
         frac = frac << ;              //frac也可用uf代替，因为此时frac = uf
     else
         exp = exp + 0x800000;
     result = (uf & 0x80000000) | exp | frac;
     return result;
 }