环境准备
最好准备一个纯净的Linux系统,可以通过vmware创建虚拟机,或者使用docker
构建一个centos
或者 ubuntu
系统。
主机我是windows系统,本机上安装vscode,通过vscode安装remote ssh插件,连接Linux虚拟机,vscode上直接可以打开远程虚拟机中工作区,进行软件开发,比较方便。主要还是csapp的实验需要Linux环境。关于vscode的安装和配置,以及ssh插件安装和配置,大家可以网上搜一下,有很多这方面的教程,如果大家有环境弄不明白的,也欢迎私信我
如下是我搭建的实验环境
实验资料下载
http://csapp.cs.cmu.edu/3e/labs.html
另外,有兴趣做书中每章节后的homework的,书后homework答案可以参考
https://dreamanddead.github.io/CSAPP-3e-Solutions/chapter2/2.63/
实验开始
实验开始前,先阅读下README文档,我们主要的工作就是编辑bits.c源文件中的各个函数,每次修改完源文件,要通过make btest进行编译,然后通过./dlc -e bits.c来检查源文件的合法性,最后通过运行./btest 检测我们写的函数功能是否通过
如下图,如果函数功能通过,会显示得分,不通过,会给出每个函数的错误信息
下面是我的代码实现,具体的思路我写在注释里了(一定要按照函数上面的注释要求实现,限定了你的操作和次数)
1.bitXor
/*
* bitXor - x^y using only ~ and &
* Example: bitXor(4, 5) = 1
* Legal ops: ~ &
* Max ops: 14
* Rating: 1
*/
int bitXor(int x, int y) {
//a^b=(x&~y)|(y&~x)
//a|b=~(~a&~b)
int a=x&~y;
int b=y&~x;
int c=~(~a&~b);
return c;
}
2.tmin
/*
* tmin - return minimum two's complement integer
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 4
* Rating: 1
*/
int tmin(void) {
//1000...
return 1<<31;
}
3.isTmax
/*
* isTmax - returns 1 if x is the maximum, two's complement number,
* and 0 otherwise
* Legal ops: ! ~ & ^ | +
* Max ops: 10
* Rating: 1
*/
int isTmax(int x) {
int a=!(~x);//排除-1的干扰,若x=-1,此时a=1,经过下面return的逻辑取反,返回0
return !((~(x+1)^x) | a);
}
4.allOddBits
/*
* 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
*/
int allOddBits(int x) {
//mask=0xAAAAAAAA;//构造一个掩码奇数位为1,偶数位为0,如果参数为真,和掩码and操作会等于掩码值
//因为只能使用小于等于255的常数,掩码构造需要一个小技巧
int a=0xAA<<8; //0xAA00
int c=a|0xAA; //0xAAAA
int mask=c<<16|c; //0xAAAAAAAA
return !((x & mask) ^ mask);
}
5.negate
/*
* negate - return -x
* Example: negate(1) = -1.
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 5
* Rating: 2
*/
int negate(int x) {
//补码非,等于位编码取反再加1
return ~x+1;
}
6.isAsciiDigit
/*
* 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
*/
int isAsciiDigit(int x) {
//a-b=a+(-b)=a+(~b+1)
//x-0x30>=0 && 0x39-x>=0即可
int flag1=!((x+(~0x30+1))>>31 & 1);//取符号位的值
int flag2=!((0x39+(~x+1))>>31 & 1);
return flag1 & flag2;
}
7.conditional
/*
* conditional - same as x ? y : z
* Example: conditional(2,4,5) = 4
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 16
* Rating: 3
*/
int conditional(int x, int y, int z) {
//先逻辑取反后,只有两种情况0或者1,再左移31位获取符号位,然后算数右移扩展成32位掩码
int mask=((!x) << 31) >> 31;//1111111... or 0000000...
return (mask & z) | (~mask & y);//两者取其一
}
8. 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) {
//考虑到会溢出(两个数异号的情况下),不能直接通过y-x进行判断
//分两种情况,两者同号的情况下,不会溢出,进行相减,此时xor为0,判断相减后的符号(!xor & !signDiff)
int signx = x>>31 & 1;
int signy = y>>31 & 1;
int xor = signx ^ signy;
int diff = y+(~x + 1);
int signDiff = diff>>31 & 1;
return (!xor & !signDiff) | (xor & signx);//两者异号的情况,xor为1,因为判断x<=y,x若为负数,signx为1,y则为正;x为正,signx为0,y为负
//所以,xor & signx,即可判断两者异号的情况下,x<=y成立的条件
}
9.logicalNeg
/*
* 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
*/
int logicalNeg(int x) {
// 补码运算,我们知道除了0(其实还有1000…0,但是这种情况没有例外)的补码是自己本身,其他数字的补码都是其相反数,符号位改变。
// 那么如果用原数x与(~x + 0x1)进行按位或,就可以得到符号位为1,对其进行算数右移即可得到全1的位模式,而0的补码是其本身,经过这套操作,结果还是全0的位模式。
// (10000.0的补码也是其本身,但因为其本身符号位就是1,所以 x | (~x + 0x1))后也是全1的位模式,所以这种情况不是例外)。
int check= x | (~x+1);
return (check>>31)+1;
}
10.howManyBits
/* 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
* howManyBits(0x80000000) = 32
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 90
* Rating: 4
*/
int howManyBits(int x) {
int b16,b8,b4,b2,b1,b0;
int flag=x>>31;
x=(flag&~x)|(~flag&x); //x为非正数则不变 ,x 为负数 则相当于按位取反,正数和负数都变成非负数,即0..1..形式,方便计算最高有效位位置
//利用二分法求int x的最高位1所在的位置
b16=!!(x>>16) <<4; //如果高16位不为0,则我们让b16=16
x>>=b16; //如果高16位不为0 则我们右移动16位 来看高16位的情况
//下面过程基本类似
b8=!!(x>>8)<<3;
x >>= b8;
b4 = !!(x >> 4) << 2;
x >>= b4;
b2 = !!(x >> 2) << 1;
x >>= b2;
b1 = !!(x >> 1);
x >>= b1;
b0 = x;
return b0+b1+b2+b4+b8+b16+1;
}
11.floatScale2
这里提供了两个解决方案,第一个实现方式不如第二个漂亮,大家参考吧
/*
* 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) {
// //float指数部分为8位,小数部分23位,还有最开始的符号位,构造掩码获取其指数部分和符号位
// //0111 1111 1000 ...(...表示20个0)=0x7f800000,指数8位全为1的掩码
// int exp =(uf&0x7f800000)>>23;
// int sign = uf&(1<<31);
// if(exp==0) return uf<<1|sign;//0和非规格化最小数乘2后,也不会越界
// if(exp==255) return uf;//按题目需求返回参数
// exp++;//指数部分+1,表示乘2
// if(exp==255) return 0x7f800000|sign;//乘2后,变为了无穷大,返回无穷大
// //0x807fffff,构造掩码,保留符号位和小数部分,然后和指数部分合并
// return (exp<<23)|(uf&0x807fffff);
// }
unsigned floatScale2(unsigned uf) {
unsigned exp = (uf&0x7f800000)>>23;
unsigned sign=uf>>31&0x1;
unsigned frac=uf&0x7FFFFF;
unsigned res;
if(exp==0xFF)return uf;
else if(exp==0){
frac <<= 1;//这里小数部分乘2后有可能"溢出"到阶码部分,frac=0x7fffff&uf,
//frac十六进制第一位是7(0111)溢出后第一位变为1,和阶码部分再合并,也是可行的,这里处理比较巧妙,也自然的从非规格化数变成了规格化数
res = (sign << 31) | (exp << 23) | frac;
}
else{
exp++;
res = (sign << 31) | (exp << 23) | frac;
}
return res;
}
12.floatFloat2Int
/*
* 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) {
unsigned exp = (uf&0x7f800000)>>23;
int sign=uf>>31&0x1;
unsigned frac=uf&0x7FFFFF;
int E=exp-127;
if(E<0)return 0;
else if(E >= 31){
return 0x80000000u;
}
else{
frac=frac|1<<23;//规格化小数部分隐含的1合并进来
if(E<23) {//需要舍入
frac>>=(23-E);
}else{
frac <<= (E - 23);
}
}
if (sign)
return -frac;
else
return frac;
}
13. floatPower2
/*
* 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) {
if(x<-149){//2^-126 * 2^-23 如果小于最小的非规格化数
return 0;
}else if(x<-126){//小于最大非规格化数
//E=1-127=-126
int shift = 23+(x+126);
return 1 <<shift;
}else if(x<=127){//小于最大规格化数
//x=expr-bias
int expr=x+127;
return expr << 23;
}else{
return (0xFF)<<23;
}
}
标签:Rating,DataLab,return,ops,int,31,CSAPP,unsigned,Lab1
From: https://blog.csdn.net/qq_33811080/article/details/141366776