本文主要作为【不周山之读厚 CSAPP】I Data Lab的扩充，小土刀于2016年4月写成，当时总共需要13个函数，而现在需要完成62个函数。
没有阅读过【不周山之读厚 CSAPP】I Data Lab的同学，需要先去阅读。
题目的要求都是一样的，有用的提示小土刀也提示的差不多了。
现在多了两个帮助函数ishow 和 ifshow,另外btest功能增强了,更多功能请阅读datalab-handout目录下的README
- ishow 展示输入的整型数据的详细信息
使用示例：
unix> ./ishow 0x27
Hex = 0x00000027, Signed = 39, Unsigned = 39
unix> ./ishow 27
Hex = 0x0000001b, Signed = 27, Unsigned = 27
- fshow 展示输入的浮点型数据的详细信息
使用示例:
unix> ./fshow 0x15213243
Floating point value 3.255334057e-26
Bit Representation 0x15213243, sign = 0, exponent = 0x2a, fraction = 0x213243
Normalized. +1.2593463659 X 2^(-85)
linux> ./fshow 15213243
Floating point value 2.131829405e-38
Bit Representation 0x00e822bb, sign = 0, exponent = 0x01, fraction = 0x6822bb
Normalized. +1.8135598898 X 2^(-126)
废话不多说，我们进入正题
进入正题
1. absVal
/*
* absVal - absolute value of x
* Example: absVal(-1) = 1.
* You may assume -TMax <= x <= TMax
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 10
* Rating: 4
*/
int absVal(int x) {
int negate = !!(x & (1 << 31));
return (x^(~0 + !negate))+negate;
}
解题思路：
当 x < 0 时
absVal(x) = -x
当 x >= 0 时absVal(x) = x
由小土刀对negate分析，可知
-x = ~x + 1
~x = x ^ (~0)
x = x ^ 0
2.addOK
/*
* addOK - Determine if can compute x+y without overflow
* Example: addOK(0x80000000,0x80000000) = 0,
* addOK(0x80000000,0x70000000) = 1,
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 20
* Rating: 3
*/
int addOK(int x, int y) {
int res = x + y;
int flagx = !(x & (1 << 31));
int flagy = !(y & (1 << 31));
int flagr = !(res & (1 << 31));
return !((flagx & flagy & !flagr) | (!flagx & !flagy & flagr));
}
解题思路：
当 x > 0 且 y > 0 但 x+y < 0时，发生正溢出
当x<0 且 y < 0 但 x+y >= 0 时，发生负溢出
3.allEvenBits
/*
* allEvenBits - return 1 if all even-numbered bits in word set to 1
* where bits are numbered from 0 (least significant) to 31 (most significant)
* Examples allEvenBits(0xFFFFFFFE) = 0, allEvenBits(0x55555555) = 1
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 12
* Rating: 2
*/
int allEvenBits(int x) {
int a = 0x55;
int b = (a << 8) + a; //0x55 55
int c = (b << 8) + a; //0x55 55 55
int d = (c << 8) + a; //0x55 55 55 55
return !((x & d)^d);
}
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) {
int a = 0xAA;
int b = (a << 8) + a; //0xAA AA
int c = (b << 8) + a; //0xAA AA AA
int d = (c << 8) + a; //0xAA AA AA AA
return !((x & d)^d);
}
5.anyEvenBit
/*
* anyEvenBit - return 1 if any even-numbered bit in word set to 1
* where bits are numbered from 0 (least significant) to 31 (most significant)
* Examples anyEvenBit(0xA) = 0, anyEvenBit(0xE) = 1
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 12
* Rating: 2
*/
int anyEvenBit(int x) {
int a = 0x55;
int b = (a << 8) + a; //0x55 55
int c = (b << 8) + a; //0x55 55 55
int d = (c << 8) + a; //0x55 55 55 55
return !!(x & d);
}
6.anyOddBit
/*
* anyOddBit - return 1 if any odd-numbered bit in word set to 1
* where bits are numbered from 0 (least significant) to 31 (most significant)
* Examples anyOddBit(0x5) = 0, anyOddBit(0x7) = 1
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 12
* Rating: 2
*/
int anyOddBit(int x) {
int a = 0xAA;
int b = (a << 8) + a; //0xAA AA
int c = (b << 8) + a; //0xAA AA AA
int d = (c << 8) + a; //0xAA AA AA AA
return !!(x & d);
}
7.bang
/*
* bang - Compute !x without using !
* Examples: bang(3) = 0, bang(0) = 1
* Legal ops: ~ & ^ | + << >>
* Max ops: 12
* Rating: 4
*/
int bang(int x) {
int a = (x | x >> 16);
int b = (a | a >> 8);
int c = (b | b >> 4);
int d = (c | c >> 2);
int e = (d | d >> 1) & 0x01;
return e^0x01;
}
解题思路：
依次折半的将所有位数合并。
step1，将高16位与低16位进行或运算，将高16位中的1藏在低16位中。
step2，将低16位中的高8位与低8位进行或运算，将低16位中的高8位中的1藏于低8位中。
step3，将低8位中的高4位与低4位进行或运算，将低8位中的高4位中的1藏于低4位中。
step4，将低4位中的高2位与低2位进行或运算，将低4位中的高2位中的1藏于低2位中。
step5，将低2位中的高位与低位进行或运算，将低2位中的高位中的1藏于低位中，并与1进行与运算。
如此，step5中e的结果必为0（当且仅当x为0时）或1，将其与1异或便是结果。
改进
参考网上大笨象同学的解法
在补码表示里只有0和它的相反数同时非负，即
0与（~0 + 1）的符号为均为 0；其他任何数|上自己的相反数，符号位都为1；
int bang(int x) {
int a = ~x + 1;
int b = ((a | x) >> 31) + 1 ;
return b;
}
8.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) {
int a = ~x | ~y;
return ~a;
}
9.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) {
//数据结构习题解析第三版 邓俊辉 第一章 12题(b) Page 9
/*
* Warning: 42 operators exceeds max of 40
*
int mask0 = (0x55) | (0x55 << 8) | (0x55 << 16) | (0x55 << 24); //0x55555555
int mask1 = (0x33) | (0x33 << 8) | (0x33 << 16) | (0x33 << 24); //0x33333333
int mask2 = (0x0F) | (0x0F << 8) | (0x0F << 16) | (0x0F << 24); //0x0F0F0F0F
int mask3 = (0xFF) | (0xFF << 16); //0x00FF00FF
int mask4 = (0xFF) | (0xFF << 8); //0x0000FFFF
*/
int a = (0x55) | (0x55 << 8);
int mask0 = a | (a << 16);
int b = (0x33) | (0x33 << 8);
int mask1 = b | (b << 16);
int c = (0x0F) | (0x0F << 8);
int mask2 = c | (c << 16);
int mask3 = (0xFF) | (0xFF << 16);
int mask4 = (0xFF) | (0xFF << 8);
//unsigned int n = (unsigned int)x;
int n = x;
n = (n & mask0) + (n >> 0x01 & mask0);
n = (n & mask1) + (n >> 0x02 & mask1);
n = (n & mask2) + (n >> 0x04 & mask2);
n = (n & mask3) + (n >> 0x08 & mask3);
n = (n & mask4) + (n >> 0x10 & mask4);
return n;
}
BitCount.png
10.
/*
* bitMask - Generate a mask consisting of all 1's
* lowbit and highbit
* Examples: bitMask(5,3) = 0x38
* Assume 0 <= lowbit <= 31, and 0 <= highbit <= 31
* If lowbit > highbit, then mask should be all 0's
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 16
* Rating: 3
*/
int bitMask(int highbit, int lowbit) {
int a = ~0;
//notice: the result of x << 32 is x
// should exec x << 31 then x << 1
int b = a << highbit;
int x = (a << lowbit) & (~(b << 1));
return x;
}
注意事项：
当对int x执行<<32时， 什么事情也没有发生，x并不是0，仍旧是自身。
11.bitMatch
/*
* bitMatch - Create mask indicating which bits in x match those in y
* using only ~ and &
* Example: bitMatch(0x7, 0xE) = 0x6
* Legal ops: ~ &
* Max ops: 14
* Rating: 1
*/
int bitMatch(int x, int y) {
int match0 = ~x & ~y;
int match1 = x & y;
int res = ~(~match0 & ~match1);
return res;
}
12.bitNor
/*
* bitNor - ~(x|y) using only ~ and &
* Example: bitNor(0x6, 0x5) = 0xFFFFFFF8
* Legal ops: ~ &
* Max ops: 8
* Rating: 1
*/
int bitNor(int x, int y) {
//int res = ~(~(~x & ~y));
int res = ~x & ~y;
return res;
}
13.bitOr
/*
* bitOr - x|y using only ~ and &
* Example: bitOr(6, 5) = 7
* Legal ops: ~ &
* Max ops: 8
* Rating: 1
*/
int bitOr(int x, int y) {
int res = ~(~x & ~y);
return res;
}
14.bitParity
/*
* bitParity - returns 1 if x contains an odd number of 0's
* Examples: bitParity(5) = 0, bitParity(7) = 1
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 20
* Rating: 4
*/
int bitParity(int x) {
int a = x ^ (x >> 16);
int b = a ^ (a >> 8);
int c = b ^ (b >> 4) ;
int d = c ^ (c >> 2);
int e = (d ^ (d >> 1)) & 0x01;
return e;
}
解题思路类似第7题
15.bitReverse
/*
* bitReverse - Reverse bits in a 32-bit word
* Examples: bitReverse(0x80000002) = 0x40000001
* bitReverse(0x89ABCDEF) = 0xF7D3D591
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 45
* Rating: 4
*/
int bitReverse(int x) {
/*
* Warning: 50 operators exceeds max of 45
*
int mask0 = 0xFF | 0xFF << 8; //0x0000FFFF
int mask1 = 0xFF | 0xFF << 16; //0x00FF00FF
int mask2 = 0x0F | 0x0F << 8 | 0x0F << 16 | 0x0F << 24; //0x0F0F0F0F
int mask3 = 0x33 | 0x33 << 8 | 0x33 << 16 | 0x33 << 24; //0x33333333
int mask4 = 0x55 | 0x55 << 8 | 0x55 << 16 | 0x55 << 24; //0x55555555
*/
int mask0 = (0xFF) | (0xFF << 8);
int mask1 = (0xFF) | (0xFF << 16);
int a = (0x0F) | (0x0F << 8);
int mask2 = a | (a << 16);
int b = (0x33) | (0x33 << 8);
int mask3 = b | (b << 16);
int c = (0x55) | (0x55 << 8);
int mask4 = c | (c << 16);
int n = (x >> 16 & mask0) | (x << 16);
n = (n >> 8 & mask1) | (n << 8 & ~mask1);
n = (n >> 4 & mask2) | (n << 4 & ~mask2);
n = (n >> 2 & mask3) | (n << 2 & ~mask3);
n = (n >> 1 & mask4) | (n << 1 & ~mask4);
return n;
}
解题思路：
可借鉴第7题、第9题、 第14题
折半的交换左右两部分
例如 12345678
第一次交换后： 5678 | 1234
第二次交换后： 78 | 56 || 34 | 12
第三次交换后： 8|7|| 6|5 ||| 4|3 || 2|1
16.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) {
int a = x & ~y;
int b = ~x & y;
int c = ~(~a & ~b);
return c;
}
17.byteSwap
/*
* byteSwap - swaps the nth byte and the mth byte
* Examples: byteSwap(0x12345678, 1, 3) = 0x56341278
* byteSwap(0xDEADBEEF, 0, 2) = 0xDEEFBEAD
* You may assume that 0 <= n <= 3, 0 <= m <= 3
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 25
* Rating: 2
*/
int byteSwap(int x, int n, int m) {
int n8 = n << 3;
int m8 = m << 3;
int a = ((x >> n8) & 0xFF) << m8;
int b = ((x >> m8) & 0xFF) << n8;
int c = (x & ~(0xFF << m8) & ~(0xFF << n8));
return (a | b | c);
}
解题思路：
将需要交换的2个byte分别扣（与运算）出来，再错位放入（或运算)
18.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) {
int flag = !!x;
int a = y ^ z;
int b = a ^ ( y & (~0 + flag)) ^ ( z & (~0 + !flag));
return b;
}
19.copyLSB
/*
* copyLSB - set all bits of result to least significant bit of x
* Example: copyLSB(5) = 0xFFFFFFFF, copyLSB(6) = 0x00000000
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 5
* Rating: 2
*/
int copyLSB(int x) {
int n = x & 0x01;
n = ~n + 1;
return n;
}
20.distinctNegation
/*
* distinctNegation - returns 1 if x != -x.
* and 0 otherwise
* Legal ops: ! ~ & ^ | +
* Max ops: 5
* Rating: 2
*/
int distinctNegation(int x) {
return !!(x+x);
}
解题思路：
只有 0 与 Tmin 的相反数等于自身,另外有如下特性
0 + 0 = 0
Tmin + Tmin = 0
21.dividePower2
/*
* dividePower2 - Compute x/(2^n), for 0 <= n <= 30
* Round toward zero
* Examples: dividePower2(15,1) = 7, dividePower2(-33,4) = -2
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 15
* Rating: 2
*/
int dividePower2(int x, int n) {
int flag = !(x & (1 << 31));
x = x + (1 << (n & (~0 + flag))) + ~0;
return x >> n;
}
解题思路：
明白整数乘法总是舍入到零（CSAPP 中文第二版 P64)
因此有如下除法公式：
x/pwr2k = (x < 0 ? ( x + (1 << k) -1) : x) >> k; （CSAPP 中文第二版 P66)
22.evenBits
/*
* evenBits - return word with all even-numbered bits set to 1
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 8
* Rating: 1
*/
int evenBits(void) {
int mask = 0x55 | 0x55 << 8 | 0x55 << 16 | 0x55 << 24;
return mask;
}
23.ezThreeFourths
/*
* ezThreeFourths - multiplies by 3/4 rounding toward 0,
* Should exactly duplicate effect of C expression (x*3/4),
* including overflow behavior.
* Examples: ezThreeFourths(11) = 8
* ezThreeFourths(-9) = -6
* ezThreeFourths(1073741824) = -268435456 (overflow)
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 12
* Rating: 3
*/
int ezThreeFourths(int x) {
int a = (x << 1) + x; //a = x * 3;
int flag = !!(a & (0x1 << 31));
int b = a + (0x3 & (flag | flag << 1));
return b >> 2;
}
解题思路：
先计算a = x * 3,再对a/4 向零取整，当x*3 溢出时，也符号期望值。
如果先计算x/4 会损失精度
24.fitsBits
/*
* 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) {
int positive = !(x & (1 << 31));
int mask = ~0 << (n + ~0);
int a = !((x^(~0 + positive)) & mask);
return a;
}
解题思路：
这道题目对我很难。我思考了并修改了很久，才得出正确的结果
当有1个bit时， 能表示的数 只有 {-1 ， 0}；
当有2个bit时，能表示的数有 {-2， -1， 0， 1}；
当有3个bit时，能表示的数有{-4，-3，-2，-1，0，1，2，3}；
刚开始我的思路是：既然Tmin = -Tmax - 1;那求出Tmin，然后加上x的绝对值小于0，就说明可以放下x。但是 ，当x = Tmin32时，无法求得x的绝对值。
最后通过观察，发现能表示的数从n-1位左边所有的数必为全1或全0（正负两种情况）。
优化
参考网上大笨象同学的解法
想法：有两个规律假设n=4只有当
0x11111...[1xxx]或0x00000...[0xxx]我们才能用n个二进制位来表式x，想法出来了我就将x往右移n-1位~((~n)+1)，判断剩下的数字是不是全1或者全0不就好了啊？是的就是这样辣。
int fitsBits(int x, int n)
{
int tmp = ~((~n)+1);
int tmpx = x >> tmp;
int ans = ( !tmpx | !(tmpx+1) );
return ans;
}
25.fitsShort
/*
* fitsShort - return 1 if x can be represented as a
* 16-bit, two's complement integer.
* Examples: fitsShort(33000) = 0, fitsShort(-32768) = 1
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 8
* Rating: 1
*/
int fitsShort(int x) {
int a = x >> 15;
int res = !a | !(a + 1);
return res;
}
解题思路：
我采用的是和24题一样的解法。将16替换为n。使用了10个操作符，超出了限制。
以上为大笨象同学的解法。
浮点数
浮点数采用和整数完全不同的表示法。
可以参考https://wdxtub.com/2016/04/16/thin-csapp-1/
image
浮点数的实际表示.png
26.floatAbsVal
/*
* floatAbsVal - Return bit-level equivalent of absolute value of 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 floatAbsVal(unsigned uf) {
unsigned a = uf & ~(1 << 31);
unsigned B = (0x7F << 24) | (0x80 << 16);
if (((a & B) == B) && (a ^ B))
return uf;
return a;
}
解题思路：
主要难点在于理解
NaN即可
（未完待续...）