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）-拓冰建站

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，Emmmm......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: ! ~ & ^ | + << >>*   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: ! ~ & ^ | + << >>*   Max ops: 10*   Rating: 4*/
int absVal(int x) {int s = x>>31;return (x+s)^s;
}
/* * bitAnd - x&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: ! ~ & ^ | + << >>*   Max ops: 10*   Rating: 3*/
int replaceByte(int x, int n, int c) {int s = n<<3;return (c<<s)|(x&~(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: ! ~ & ^ | + << >>*   Max ops: 12*   Rating: 2*/
int mult3div2(int x) {x += (x<<1);return (x+((x>>31)&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。

计算机的原理是基于二进制数制和逻辑电路。计算机中的所有数据都是以二进制形式存储和处理的，逻辑电路则负责对这些数据进行运算和控制。计算机的核心是中央处理器(CPU)，它包含算术逻辑单元(ALU)、控制单元(CU)和寄存器等组件，负责执行指令和处理数据。

除了硬件，计算机还需要软件来实现各种功能。软件可以分为系统软件和应用软件两类。系统软件包括操作系统、编译器和驱动程序等，它们负责管理计算机的资源和提供基本的功能。应用软件则是为了满足用户的需求而开发的各种程序，例如办公软件、游戏和浏览器等。

总之，计算机是一种非常重要的电子设备，它的发展历程经历了多个阶段，从机械计算机到集成电路计算机，每一代计算机都有着自己的特点和优势。计算机的原理基于二进制数制和逻辑电路，它需要硬件和软件的配合才能实现各种功能。