西工大网络空间安全学院计算机系统基础实验一（61分答案）

本文介绍: 只能说没有一道题是自己写的，都是在网上查资料抄别人的，也不知道这有什么意思。

只能说没有一道题是自己写的，都是在网上查资料抄别人的，也不知道这有什么意思。也不知道自己学到了什么，怎么说呢，Emm mm……对了，文末最后的几段话是为了凑字数，大家简单忽略掉就好。

/* 
 * tmin - return minimum two's complement integer 
 *   Legal ops: ! ~ &amp; ^ | + &lt;&lt; &gt;&gt;
 *   Max ops: 4
 *   Rating: 1
 */
int tmin(void) {
  return 1<<31;
}
/* 
 * absVal - absolute value of x
 *   Example: absVal(-1) = 1.
 *   You may assume -TMax <= x <= TMax
 *   Legal ops: ! ~ &amp; ^ | + << &gt;&gt;
 *   Max ops: 10
 *   Rating: 4
 */
int absVal(int x) {
  int s = x>>31;
  return (x+s)^s;
}
/* 
 * bitAnd - x&amp;y using only ~ and | 
 *   Example: bitAnd(6, 5) = 4
 *   Legal ops: ~ |
 *   Max ops: 8
 *   Rating: 1
 */
int bitAnd(int x, int y) {
  return ~(~x|~y);
}
/* 
 * replaceByte(x,n,c) - Replace byte n in x with c
 *   Bytes numbered from 0 (LSB) to 3 (MSB)
 *   Examples: replaceByte(0x12345678,1,0xab) = 0x1234ab78
 *   You can assume 0 <= n <= 3 and 0 <= c <= 255
 *   Legal ops: ! ~ &amp; ^ | + << >>
 *   Max ops: 10
 *   Rating: 3
 */
int replaceByte(int x, int n, int c) {
  int s = n<<3;
  return (c<<s)|(x&amp;~(0xff<<s));
}
/*
 * mult3div2 - multiplies by 3/2 rounding toward 0,
 *   Should exactly duplicate effect of C expression (x*3/2),
 *   including overflow behavior.
 *   Examples: mult3div2(11) = 16
 *             mult3div2(-9) = -13
 *             mult3div2(1073741824) = -536870912(overflow)
 *   Legal ops: ! ~ &amp; ^ | + << >>
 *   Max ops: 12
 *   Rating: 2
 */
int mult3div2(int x) {
  x += (x<<1);
  return (x+((x>>31)&amp;1))>>1;
}
/*
 * multFiveEighths - multiplies by 5/8 rounding toward 0.
 *   Should exactly duplicate effect of C expression (x*5/8),
 *   including overflow behavior.
 *   Examples: multFiveEighths(77) = 48
 *             multFiveEighths(-22) = -13
 *             multFiveEighths(1073741824) = 13421728 (overflow)
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 12
 *   Rating: 3
 */
int multFiveEighths(int x) {
  x += (x<<2);
  return (x+((x>>31)&7))>>3;
}
/* 
 * 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) {
  return (((x^y)>>31)|~(((x+y)^x)>>31))&1;
}
/*
 * 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) {
  int m2 = (0x55<<8)|0x55;
  int m4 = (0x33<<8)|0x33;
  int m8 = (0x0f<<8)|0x0f;
  m2 |= (m2<<16);
  m4 |= (m4<<16);
  m8 |= (m8<<16);
  x += ~((x>>1)&m2)+1;
  x = ((x>>2)&m4)+(x&m4);
  x = (x+(x>>4))&m8;
  x += (x>>8);
  x += (x>>16);
  return x&0x3f;
}
/* 
 * isLess - if x < y  then return 1, else return 0 
 *   Example: isLess(4,5) = 1.
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 24
 *   Rating: 3
 */
int isLess(int x, int y) {
  int ny = ~y;
  return ((((x+ny+1)&(x^ny))|(x&ny))>>0x1f)&1;
}
/* 
 * 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 ny = ~y;
  return ((((((x+ny+1)&(x^ny))|(x&ny))>>0x1f))&1)|!(x^y);
}
/*
 * trueFiveEighths - multiplies by 5/8 rounding toward 0,
 *  avoiding errors due to overflow
 *  Examples: trueFiveEighths(11) = 6
 *            trueFiveEighths(-9) = -5
 *            trueFiveEighths(0x30000000) = 0x1E000000 (no overflow)
 *  Legal ops: ! ~ & ^ | + << >>
 *  Max ops: 25
 *  Rating: 4
 */
int trueFiveEighths(int x) {
int const eights = x >> 3;
int const rem = x & 7;

return eights + (eights << 2) + (rem + (rem << 2) + (x >> 31 & 7) >> 3);
}
  /*
  int s = (x>>31)&1;
  int c = (s<<3)+~s+1;
  int h = ((x&(0xFF<<24))+c)>>3;
  int l = (x&~(0xFF<<24));
  return (h<<2)+h+((((l<<2)+l)+c)>>3);
  */
/*
 * parityCheck - returns 1 if x contains an odd number of 1's
 *   Examples: parityCheck(5) = 0, parityCheck(7) = 1
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 20
 *   Rating: 4
 */
int parityCheck(int x) {
  x ^= (x>>16);
  x ^= (x>>8);
  x ^= (x>>4);
  x ^= (x>>2);
  x ^= (x>>1);
  return x&1;
}
/* 
 * rempwr2 - Compute x%(2^n), for 0 <= n <= 30
 *   Negative arguments should yield negative remainders
 *   Examples: rempwr2(15,2) = 3, rempwr2(-35,3) = -3
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 20
 *   Rating: 3
 */
int rempwr2(int x, int n) {
  int s = x>>31;
  x = (x+s)^s;
  x &= ((~0)+(1<<n));
  return (x^s)+~s+1;
}
/* 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 temp = x ^ (x << 1);
	int bit_16,bit_8,bit_4,bit_2,bit_1;
	bit_16 = !!(temp >> 16) << 4;
	temp = temp >> bit_16;
	bit_8 = !!(temp >> 8) << 3;
	temp = temp >> bit_8;
	bit_4 = !!(temp >> 4) << 2;
	temp = temp >> bit_4;
	bit_2 = !!(temp >> 2) << 1;
	temp = temp >> bit_2;
	bit_1 = !!(temp >> 1);
	return 1 + bit_1 + bit_2 + bit_4 + bit_8 + bit_16;
}
/*
 * 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) {
 int count = 0;
 count = (!!(x>>16)) << 4;
 count =count+((!!(x>>(8 + count)))<<3);
 count =count+((!!(x>>(4 + count)))<<2);
 count =count+((!!(x>>(2 + count)))<<1);
 count =count+((!!(x>>(1 + count)))<<0);
 return count;
}

计算机是一种能够按照程序自动进行数据处理和运算的电子设备。计算机的发展可以分为四个阶段：机械计算机、电子管计算机、晶体管计算机和集成电路计算机。

机械计算机是最早的计算机，它们使用齿轮和滑动杆等机械装置进行计算。电子管计算机是第一代计算机，它们使用电子管代替了机械装置，大大提高了计算速度。晶体管计算机是第二代计算机，它们使用晶体管代替了电子管，使得计算机更加小型化和可靠化。集成电路计算机是第三代计算机，它们使用集成电路代替了晶体管，使得计算机更加高效和便携。