Datalab實驗

Datalab實驗報告

Mizersy 3017216*** 計科*班

實驗內容

1. bitAnd

1.1實驗要求

• bitAnd - x&y using only ~ and |
• Example: bitAnd(6, 5) = 4
• Legal ops: ~ |
• Max ops: 8
• Rating: 1

1.2代碼

int bitAnd(int x, int y) {
  return ~((~x) | (~y));
}

1.3解題思路

根據德摩根定律，x&y <==> ~((~x) | (~y))

2. getByte

2.1 實驗要求

• 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

2.2 代碼

int getByte(int x, int n) {
	n = n << 3;
	x = x >> n;
  	return x&(0xff);
}

2.3 解題思路

令x右移8*n位，使得目標為變為二進制下最低的8位，在與0xff相與，將高位清零。

3. logicalShift

3.1 實驗要求

• 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

3.2 代碼

int logicalShift(int x, int n) {
	int y = (1 << 31) >> n;
	x = x >> n;
	y = y << 1;
	y = ~y; 
  	return x&y;
}

3.3 解題思路

先將x算術右移n位，再構造低n位為1，其余位均為0的y，返回x&y的值即為所求

4. bitCount

4.1 實驗要求

• bitCount - returns count of number of 1's in word
• Examples: bitCount(5) = 2, bitCount(7) = 3
• Legal ops: ! ~ & ^ | + << >>
• Max ops: 40
• Rating: 4

4.2 代碼

int bitCount(int x) {
	int count,tmp1,tmp2,tmp3,tmp4,tmp5;
	count = 0;
	tmp1 = (0x55) | (0x55 << 8);
	tmp1 = tmp1 | (tmp1 << 16);
	tmp2 = (0x33) | (0x33 << 8);
	tmp2 = tmp2 | (tmp2 << 16);
	tmp3 = (0x0f) | (0x0f << 8);
	tmp3 = tmp3 | (tmp3 << 16);
	tmp4 = (0xff) | (0xff << 16);
	tmp5 = (0xff) | (0xff << 8);
	count = (x & tmp1) + ((x >> 1) & tmp1);
	count = (count & tmp2) + ((count >> 2) & tmp2);
	count = (count & tmp3) + ((count >> 4) & tmp3);
	count = (count & tmp4) + ((count >> 8) & tmp4);
	count = (count & tmp5) + ((count >> 16) & tmp5); 
 	return count;
}

4.3 解題思路

本題難點在於只允許使用40個操作符，考慮采用二分法。先將相鄰兩位相加，再對上一步所得結果的每兩位看作一個整體，將相鄰位相加......以此類推。最終使用36個操作符。

5. bang

5.1 實驗要求

• bang - Compute !x without using !
• Examples: bang(3) = 0, bang(0) = 1
• Legal ops: ~ & ^ | + << >>
• Max ops: 12
• Rating: 4

5.2 代碼

int bang(int x) {
	int tmp = (~x) + 1;
	tmp = tmp | x;
	tmp = tmp >> 31;
  return tmp+1;
}

5.3 解題思路

如果x非0，則x或-x必有一個最高為為1，將x|(-x)算術右移31位可得到所有位全為1的數，再+1即可得到0.

如果x為0，則x和-x最高為均為0，將x|(-x)算術右移31位得到的還是0，再+1即可得到1.

6. tmin

6.1 實驗要求

• tmin - return minimum two's complement integer
• Legal ops: ! ~ & ^ | + << >>
• Max ops: 4
• Rating: 1

6.2 代碼

int tmin(void) {
  return (1 << 31);
}

6.3 解題思路

最小的二進制補碼整數最高位為1，其余全為0.

7. fitsBits

7.1 實驗要求

• 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

7.2 代碼

int fitsBits(int x, int n) { 
	int shift = 33 + (~n);
	return !(((x << shift) >> shift) ^ x);
}

7.3 解題思路

假設n=3只有當0x11111...[1xx]或0x00000...[0xx]我們才能用n個二進制位來表式x.

先將x左移32-n位，再算術右移32-n位，然后與x異或，接着取“！”即可

8. divpwr2

8.1 實驗要求

• 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

8.2 代碼

int divpwr2(int x, int n) {
	int sign,tmp;
	sign = x >> 31;//判斷符號 
	tmp = sign & ((1 << n) + (~0)); 
    return (x+tmp) >> n;
}

8.3 解題思路

如果x是正數，直接算術右移n位即可。

如果x是負數，算術右移n位后需要向上取整。

• 向上取整方法：

  在算術右移之前加上一個低n位全為1，其余為全為0的數。

9. negate

9.1 實驗要求

• negate - return -x
• Example: negate(1) = -1.
• Legal ops: ! ~ & ^ | + << >>
• Max ops: 5
• Rating: 2

9.2 代碼

int negate(int x) {
  return (~x)+1;
}

9.3 解題思路

二進制補碼求負數，直接取反加1即可

10. isPositive

10.1 實驗要求

• isPositive - return 1 if x > 0, return 0 otherwise
• Example: isPositive(-1) = 0.
• Legal ops: ! ~ & ^ | + << >>
• Max ops: 8
• Rating: 3

10.2 代碼

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

10.3 解題思路

如果符號位為1，~(x >> 31)可得到0x00000000。將0x00000000&x后可得到0，在取兩次！！后依舊為0.

如果符號位為0，~(x >> 31)可得到0xffffffff。如果x為0，將0xffffffff&x后可得到0在取兩次！！后依舊為0. 如果x>0，將0xffffffff&x后可得到x,在取兩次！！后即可得到1.

11. isLessOrEqual

11.1 實驗要求

• isLessOrEqual - if x <= y then return 1, else return 0
• Example: isLessOrEqual(4,5) = 1.
• Legal ops: ! ~ & ^ | + << >>
• Max ops: 24
• Rating: 3

11.2 代碼

int isLessOrEqual(int x, int y) {
	int sign,ret1,ret2;
	sign = (x^y) >> 31;
	ret1 = ((~x+1+y)>>31)&(~sign);//如果x和y同號 
	ret2 = (y>>31) & sign;
  	return !(ret1|ret2);
}

11.3 解題思路

本題難點在於如果直接用y-x>=0來判斷，存在y-x溢出的情況。

但我們很容易發現，只有在y和x異號時才會出現溢出的情況，所以只需先判斷y和x的符號，如果符號相同，則通過y-x是否非負來判斷結果；如果符號不同，則直接判斷符號位即可比較大小。

12. ilog2

12.1 實驗要求

• ilog2 - return floor(log base 2 of x), where x > 0
• Example: ilog2(16) = 4
• Legal ops: ! ~ & ^ | + << >>
• Max ops: 90
• Rating: 4

12.2 代碼

int ilog2(int x) {
	int ret = 0;
	ret = (!!(x >> 16)) << 4;
	ret = ret + ((!!(x >> (ret + 8))) << 3);
	ret = ret + ((!!(x >> (ret + 4))) << 2);
	ret = ret + ((!!(x >> (ret + 2))) << 1);
	ret = ret + (!!(x >> (ret + 1)));
  	return ret;
}

12.3 解題思路

本題可以轉化為尋找最高位的1在哪個位置的問題，很容易可以想到采用二分查找難點在於如何在不使用循環體的情況下實現二分。維護一個ret，先將32位二進制數x右移16位，判斷移位后的結果是否為0來判斷高16位是否含1.如果含1，最終結果必然在16~31之間則令ret+=16，然后再用同樣方法以此類推。最終找到最高為1的位置。

13. float_neg

13.1 實驗要求

• 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

13.2 代碼

unsigned float_neg(unsigned uf) {
	unsigned tmp;
	tmp = uf & (0x7fffffff);
	if (tmp > 0x7f800000)
		return uf;
 	return uf ^ (0x80000000);
}

13.3 解題思路

如果uf 為nan，直接返回uf。否則直接將符號位取反即可。

14. float_i2f

14.1 實驗要求

• 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

14.2 代碼

unsigned float_i2f(int x) {
    unsigned shiftLeft=0;
    unsigned afterShift, tmp, flag;
    unsigned absX=x;
    unsigned sign=0;
    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;
}

14.3 解題思路

整數轉浮點數經典題。

如果x為0，直接return 0.

如果x為負數，記錄符號，然后把x變為正數。

一直左移，直到溢出，並記錄移位次數。

afterShift記錄尾數部分，159-shiftleft表示階碼

如果大於一半進一，少於一半則舍棄，等於一半則向偶數舍入

15. float_twice

15.1 實驗要求

• 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

15.2 代碼

unsigned float_twice(unsigned uf) {
    unsigned f = uf;
    if ((f & 0x7F800000) == 0) 
        f = ((f & 0x007FFFFF) << 1) | (0x80000000 & f);
    else if ((f & 0x7F800000) != 0x7F800000)
        f =f + 0x00800000;
    return f;
}

15.3 解題思路

如果階碼為0，則直接將尾數左移一位，如果尾數溢出則恰好移入階碼，再加上符號位即可。

如果階碼全為1，則直接返回f。

如果階碼不為0且不全為1，則直接給階碼+1。

實驗結果