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CSAPP:位操作实现基本运算

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实验要求

给出15个函数,规定了实现每个函数需要的逻辑和算术操作符(规定数量)。
只能使用规定的操作符! ˜ & ˆ | + << >>
1-12题不能使用循环或者条件语句
不能使用超过8位的常数(ff)

转载请注明出处:https://www.cnblogs.com/ustca/p/11740382.html

实现代码

1、pow2plus1

/*
 * 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;
}

2、pow2plus4

/*
* 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;
}

3、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) {
	return (~(x&y))&(~(~x&~y));//列出真值表
}

4、tmin

/*   tmin - return minimum two's complement integer 
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 4
 *   Rating: 1
 */
int tmin(void) {
	return 1<<31;//0x80000000
}

5、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) {
	return !(x+x+2) & !!(~x);//x+1+x+1溢出并且非全一
	//x:		0111 1111 1111 1111 1111 1111 1111 1111
	//x+1:	1000 0000 0000 0000 0000 0000 0000 0000
	//x+1+x:	1111 1111 1111 1111 1111 1111 1111 1111
	//x+1+x+1:0000 0000 0000 0000 0000 0000 0000 0000
}

6、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) {	
	x = (x>>16) & x;	
	x = (x>>8) & x;		
	x = (x>>4) & x;		
	x = (x>>2) & x;		
	return (x>>1)&1;	
	// &运算符的“归一性”
	//1010 1010 1010 1010 1010 1010 1010 1010
	//0000 0000 0000 0000 1010 1010 1010 1010
	//0000 0000 0000 0000 0000 0000 1010 1010
	//0000 0000 0000 0000 0000 0000 0000 1010
	//0000 0000 0000 0000 0000 0000 0000 0010 
	// 可以反推理解:后四位四次翻转得第一行
	// 只要倒数第二位为1成立,反推后所有的奇数位都为1
}

7、negate

/*   negate - return -x 
 *   Example: negate(1) = -1.
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 5
 *   Rating: 2
 */
int negate(int x) {
	return ~x+1;	//带符号位取反加一即为相反数
}

8、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) {
	// 0x30 = 0011 0000b   0x39 = 0011 1001b
	int a = (x>>4) ^ 0x3;	// 判断5、6位是否全1
	int b0 = (x>>3) & 1;	// 判断第4位是否为1
	int b1 = (x>>2) ^ 1;	// 判断第3位是否为1
	int b2 = (x>>1) ^ 1;	// 判断第2位是否为1
	return (!a) & ((!b0) | (b0&b1&b2)); 
	// 如果5、6位全1 且 (4位为0或4位为1,2、3位为0)
}

9、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) {
	x = !x;	// 将x设置为0或1
	x = (x<<31)>>31; // 将x的0或1拓展到32位全0或全1
	return (~x&y) | (x&z); // x为真则~x全1返回y,为假则x全1返回z
}

10、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) {
	int z,s,sx,sy;
	sx = (x>>31)&1;	// 	取x的符号位
	sy = (y>>31)&1; //	取y的符号位
	z = x + ~y + 1; //	z = x-y
	s =  ((z>>31) & 1) | (!(z^0));
	// 取z的符号位,s为真时x<y或者z全0(x==y)
	return  ((!(sx^sy))&s) | (sx&(!sy));
	// xy同号且z<=0 或 x<=0 y>=0
}

11、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) {
	// |运算符的“分裂性”
	x |= x>>16; // 若高16位有1,则传递给低16位的对应位
	x |= x>>8;	// 若低16位的高8位有1,则传递给低8位的对应位
	x |= x>>4;	// 若低8位的高4位有1,则传递给低4位的对应位
	x |= x>>2;	// 若低4位的高2位有1,则传递给低2位的对应位
	x |= x>>1;	// 若低2位的高1位有1,则传递给最低1位
	x ^= 1; 	// 只要x包含1,则必定会导致此时的x为1,x^=1即取反
	return x&1; 
}

12、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 s,c1,c2,c3,c4,c5,c6;
	int cnt = 0;	// 	计数
	s = (x>>31)&1;	//	符号位
	x = ((s<<31)>>31) ^ x; // 取反x
	s = !!(x>>16);	// 判断高16位是否有1,有则s为1
	c1 = s<<4;		// 若高16位有1,则低16位可以计数16
	x >>= c1;		// 右移将已经计数的位移除,c1若为0,则用折半的长度判断
	s = !!(x>>8);	// 用8位的长度去判断,有效位的个数计入c2
	c2 = s<<3;
	x >>= c2;
	s = !!(x>>4);	// 用4位的长度去判断,有效位的个数计入c3
	c3 = s<<2;
	x >>= c3;
	s = !!(x>>2);	// 用2位的长度去判断,有效位的个数计入c4
	c4 = s<<1;
	x >>= c4;
	s = !!(x>>1);	// 用1位的长度去判断,有效位的个数计入c5
	c5 = s;
	x >>= c5;
	c6 = !!x;		// 判断最低位是否为1
	cnt = c1+c2+c3+c4+c5+c6+1;	// 将每次获得的低位有效位相加,再加1位符号位
	return cnt;
}

13、floatScale2

/*   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) {
	unsigned f = uf;
	if ((f & 0x7F800000) == 0)// 如果阶码为0
		f = ((f & 0x007FFFFF) << 1) | (0x80000000 & f); 
		// 尾数不为0则尾数左移1位,尾数第1位为1则阶码加1,尾数为0则uf为0返回0
	else if ((f & 0x7F800000) != 0x7F800000)// 如果阶码不为0,且非全1
		f = f + 0x00800000;// 阶码加1
	return f;
}

14、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 INF = 1<<31;	// INF = MaxInt+1
	int e = (uf>>23) & 0xff;// 阶码
	int s = (uf>>31) & 1;	// 符号位
	if (uf == 0) return 0;
	uf <<= 8;		// 左移保留至阶码最后1位
	uf |= 1<<31;	// 阶码最后一位设为1
	uf >>= 8;		// 高八位全0
	e -= 127;		// 阶数
	if ((uf & 0x7f80000) == 0x7f80000 || e >= 32)
		return INF; // 超过int范围返回INF
	if (e < 0) // 小数返回0
		return 0;
	if (e <= 22) // 位数小于等于22位,尾数位右移
		uf >>= 23-e;
	else 
		uf <<= e-23; // 尾数大于22位,尾数为左移
	if (s) 
		uf = ~uf + 1;// 若原uf为负数,则对此处的正数uf取反加1得其相反数
	return uf;
}

15、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) {
	unsigned INF = 0xff << 23; // 阶码全1
	int e = 127 + x;	// 得到阶码
	if (x < 0) // 阶数小于0直接返回0
		return 0;
	if (e >= 255) // 阶码>=255直接返回INF
		return INF;
	return e << 23;
	// 直接将阶码左移23位,尾数全0,规格化时尾数隐藏有1个1作为底数
}

posted on 2019-10-25 19:49  东寻  阅读(...)  评论(...编辑  收藏