CSAPP:datalab
代码中有具体思路的标注。
/*
* CS:APP Data Lab
*
*
* 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:
1. Integer constants 0 through 255 (0xFF), inclusive. You are
not allowed to use big constants such as 0xffffffff.
2. Function arguments and local variables (no global variables).
3. Unary integer operations ! ~
4. Binary integer operations & ^ | + << >>
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:
1. Use any control constructs such as if, do, while, for, switch, etc.
2. Define or use any macros.
3. Define any additional functions in this file.
4. Call any functions.
5. Use any other operations, such as &&, ||, -, or ?:
6. Use any form of casting.
7. Use any data type other than int. This implies that you
cannot use arrays, structs, or unions.
You may assume that your machine:
1. Uses 2s complement, 32-bit representations of integers.
2. Performs right shifts arithmetically.
3. Has unpredictable behavior when shifting an integer by more
than the word size.
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
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.
You are expressly forbidden to:
1. Define or use any macros.
2. Define any additional functions in this file.
3. Call any functions.
4. Use any form of casting.
5. Use any data type other than int or unsigned. This means that you
cannot use arrays, structs, or unions.
6. Use any floating point data types, operations, or constants.
NOTES:
1. Use the dlc (data lab checker) compiler (described in the handout) to
check the legality of your solutions.
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.
3. Use the btest test harness to check your functions for correctness.
4. Use the BDD checker to formally verify your functions
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.
/*
* 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) {
/*It's equal to find the position that both are 0 and find its reversed version.有点像德摩根律。*/
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) {
/*先算数左移再算数右移，再减去算数右移增加的补位数*/
int t,t1,t2;
t1=x<<((3-n)*2*4);
t2=t1>>7*4>>4<<(2*4);
t=(t1>>(6*4))-t2;
return t;
}
/*
* 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) {
/*找出最高位的符号，然后把算数右移前面补的位数减掉即可*/
int t;
t=x>>31<<(32+(~n+1)+0xffffffff)<<1;
t=(x>>n)&(~t);
return t;
}
/*
* 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) {
/*difficult*/
/*原理：通过同时批量的运算，把操作次数降下来。先理解只有两位的情况，像是将数字两两划分到一个格子里，01比较好理解，就是降次加和，操作完之后每个盒子里都是二进制表示的“1”的个数。再用0011降次加和，把两个盒子合并成一个盒子，以此类推。计数很巧妙！两位比较好想，但是从两位到四位，即把两个盒子合并到一个大盒子，这个方法也很妙！！*/
int mask1=0x55555555;//01(循环至32位)01010101
int mask2=0x33333333;//0011(循环至32位)
int mask3=0x0f0f0f0f;//00001111(循环至32位)
int mask4=0x00ff00ff;//0000000011111111(循环至32位)
int mask5=0x0000ffff;//00000000000000001111111111111111
int t;
t=(x&mask1)+((x>>1)&mask1);
t=(t&mask2)+((t>>2)&mask2);
t=(t&mask3)+((t>>4)&mask3);
t=(t&mask4)+((t>>8)&mask4);
t=(t&mask5)+((t>>16)&mask5);
return t;
}
/*
* bang - Compute !x without using !
* Examples: bang(3) = 0, bang(0) = 1
* Legal ops: ~ & ^ | + << >>
* Max ops: 12
* Rating: 4
*/
int bang(int x) {
/*利用0的特性：相反数等于自身。其他的数，相反数和它本身之间肯定有一个负数。*/
return ((x|(~x+1))>>31)+1 ;
}
/*
* tmin - return minimum two's complement integer
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 4
* Rating: 1
*/
int tmin(void) {
return 1<<31;
}
/*
* 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) {
/*第n位和第n位之前的位数的数字应该都相同，否则就无法表示。*/
return !((x<<(32-n)>>(32-n))^x);
}
/*
* 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) {
/*非整除情况，如果是负数的话，x>>n结果要加一(n非0)。对于对于负数~(x>>31)=0。整除情况如常。*/
int s=!!(x>>31);
int t=!!((x<<(32+~n+1))^0);
return (x>>n)+(s&t&!!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) {
/*x-1算数右移31位，如果x=0则所有位都为0，否则所有位都为1。*/
return !(x>>31)&!!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) {
/*即y-x>=0或y>0,x<0(若溢出)。分类讨论。*/
return (!(y>>31)&!(x>>31)&!((y+~x+1)>>31))|((!(y>>31))&(x>>31))|((y>>31)&(x>>31)&!((y+~x+1)>>31));
}
/*
* 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) {
/*找到x的最高的非0位。二分法处理，两部分有无1，用0、1表示；位置用1位移累加。很妙！！*/
int t;
t=(!!(x>>16))<<4;
t=t+((!!(x>>(8+t)))<<3);
t=t+((!!(x>>(4+t)))<<2);
t=t+((!!(x>>(2+t)))<<1);
t=t+(!!(x>>(1+t)));
return t;
}
/*
* 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) {
/*判断是否为NaN，然后最高位取反。*/
unsigned exp=(uf&0x7ff80000)>>23;//011111111100000000000000000000000
unsigned frac=uf&0x7fffff;//00000000011111111111111111111111
if(!(exp^0xff) && !!(frac^0) )
return uf;
return uf^0x80000000;
}
/*
* 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 i=0,t;
unsigned s=(!!(x>>31))<<31;
if(!(x^0)) return 0;
if(!(x^0x80000000)) return 0xcf000000;
if(s) x=~x+1;//负数转换成正数
while(!((x<<i)&0x80000000)){
i=i+1;
}
t=(x<<i)-0x80000000;
int flag=0;
if((t&0xff)>0x80) flag=1;
else{
if((t&0x180)==0x180) flag=1;
}
t=t>>8;
t=s+((32-i-1+127)<<23)+t+flag;
return t;
}
/* 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 exp=(uf&0x7ff80000)>>23;//011111111100000000000000000000000
unsigned frac=uf&0x7fffff;//00000000011111111111111111111111
unsigned t;
if(!(exp^0xff) ){
return uf;
}
int k=!!(uf&0x7fffffff)&!!exp;
t=(uf&0x80000000)+((exp+k)<<23)+(frac<<!exp);
return t;
}
结果：
./btest