CSAPP:數據實驗

   書是要看的，但實驗也是要做的！接下來將《深入理解計算機系統》裏面的幾個實驗做一下，鞏固一下理論知識。

    數據實驗。這個實驗要求學生們實現簡單的邏輯和算術函數，但是隻能使用一個高度受限的c的子集。比如，他們必須只能用位級操作來計算一個數字的絕對值。這個實驗幫助學生們瞭解c數據類型的位級表示，和數據操作的位級行爲。
    說實話，這個實驗對於有些人可能會很有興趣，因爲它屬於智力型實驗。有點像奧林匹克數學競賽上面的題目。如果以後不是從事太過於底層的工作，實驗的過程瞭解一下就行了。
    bits.c的代碼如下：

/* * CS:APP Data Lab * * bits.c - Source file with your solutions to the Lab. * This is the file you will hand in to your instructor. * * WARNING: Do not include the <stdio.h> header; it confuses the dlc * compiler. You can still use printf for debugging without including * <stdio.h>, although you might get a compiler warning. In general, * it's not good practice to ignore compiler warnings, but in this * case it's OK. */ #include "btest.h" #include <limits.h> /* * Instructions to Students: * * STEP 1: Fill in the following struct with your identifying info. */ team_struct team = { /* Replace this with your full name */ "lycos", /* Replace this with your Andrew login ID */ "ac00" }; #if 0 /* * STEP 2: Read the following instructions carefully. */ You will provide your solution to the Data Lab by editing the collection of functions in this source file. CODING RULES: Replace the "return" statement in each function with one or more lines of C code that implements the function. Your code must conform to the following style: int Funct(arg1, arg2, ...) { /* brief description of how your implementation works */ int var1 = Expr1; ... int varM = ExprM; varJ = ExprJ; ... varN = ExprN; return ExprR; } Each "Expr" is an expression using ONLY the following: 1. Integer constants 0 through 255 (0xFF), inclusive. You are not allowed to use big constants such as 0xffffffff. 2. Function arguments and local variables (no global variables). 3. Unary integer operations ! ~ 4. Binary integer operations & ^ | + << >> Some of the problems restrict the set of allowed operators even further. Each "Expr" may consist of multiple operators. You are not restricted to one operator per line. You are expressly forbidden to: 1. Use any control constructs such as if, do, while, for, switch, etc. 2. Define or use any macros（宏）. 3. Define any additional functions in this file. 4. Call any functions. 5. Use any other operations, such as &&, ||, -, or ?: 6. Use any form of casting. You may assume that your machine: 1. Uses 2s complement, 32-bit representations of integers. 2. Performs right shifts arithmetically. 3. Has unpredictable behavior when shifting an integer by more than the word size. EXAMPLES OF ACCEPTABLE CODING STYLE: /* * 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; } /* * 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; } NOTES: 1. Use the dlc (data lab checker) compiler (described in the handout) to check the legality of your solutions. 2. Each function has a maximum number of operators (! ~ & ^ | + << >>) that you are allowed to use for your implementation of the function. The max operator count is checked by dlc. Note that '=' is not counted; you may use as many of these as you want without penalty. 3. Use the btest test harness to check your functions for correctness. 4. The maximum number of ops for each function is given in the header comment for each function. If there are any inconsistencies between the maximum ops in the writeup and in this file, consider this file the authoritative source. #endif /* * STEP 3: Modify the following functions according the coding rules. * * IMPORTANT. TO AVOID GRADING SURPRISES: * 1. Use the dlc compiler to check that your solutions conform * to the coding rules. * 2. Use the btest test harness to check that your solutions produce * the correct answers. Watch out for corner cases around Tmin and Tmax. */ /* * bitAnd - x&y using only ~ and | * Example: bitAnd(6, 5) = 4 * Legal ops: ~ | * Max ops: 8 * Rating: 1 */ int bitAnd(int x, int y) { /* 狄摩根定律 （A∪B）’=A’∩B’ （A∩B）’=A’∪B’ 其中A’表示集合A對全集的補集*/ return ~((~x)|(~y)); } /* * bitXor - x^y using only ~ and & * Example: bitXor(4, 5) = 1 * Legal ops: ~ & * Max ops: 14 * Rating: 2 */ int bitXor(int x, int y) { /* 狄摩根定律 （A∪B）’=A’∩B’ （A∩B）’=A’∪B’ 其中A’表示集合A對全集的補集*/ return (~(~x&~y))&(~(x&y)); } /* * evenBits - return word with all even-numbered bits set to 1 * Legal ops: ! ~ & ^ | + << >> * Max ops: 8 * Rating: 2 */ int evenBits(void) { /*置所有偶數爲爲1，則得到0x55555555即可*/ int i = 0x55; return (i << 24) | (i << 16) | (i << 8) | i; } /* * 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 */ int getByte(int x, int n) { return 0xff & (x >> (n << 3)); } /* * bitMask - Generate a bitmask consisting of all 1's * from lowbit to highbit and 0's everywhere else. * 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) { /* 生成一個int值bitmask，它的lowbit到highbit位之間爲1 其餘位爲0，返回bitmask 注意，最後還要&上(i << lowbit)以防止lowbit大於highbit的情況*/ int i = ~0; //return ((i << ( highbit + 1)) ^ (i << lowbit)) & (i << lowbit); return (((i << highbit) << 1) ^ (i << lowbit)) & (i << lowbit); } /* * reverseBytes - reverse the bytes of x * Example: reverseBytes(0x01020304) = 0x04030201 * Legal ops: ! ~ & ^ | + << >> * Max ops: 25 * Rating: 3 */ int reverseBytes(int x) { return ((x>>24)&0xff) | (((x>>16)&0xff)<<8) | (((x>>8)&0xff)<<16) | ((x&0xff)<<24); /* 我覺得下面的做法也行！因爲左移時無論啥種情況，都補0. ((x>>24)&0xff) | ((x>>16)&(0xff<<8)) | ((x<<8)&(0xff<<16)) | (x<<24) */ } /* * leastBitPos - return a mask that marks the position of the * least significant 1 bit. If x == 0, return 0 * Example: leastBitPos(96) = 0x20 * Legal ops: ! ~ & ^ | + << >> * Max ops: 6 * Rating: 4 */ int leastBitPos(int x) { /* 技巧性相當高！利用了補碼的特性， x的補碼中有效位爲1的最低的位置是一個特殊的位置， 它之前的位與x相反，之後的爲全爲0*/ return (~x+1)&x; //這個x&(x^(x+(~0)))也不錯！ } /* * 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) { /* 它使用了0的特性~x+1 == ~x*/ return ((~(~x+1)&(~x))>>31)&1; } /* * 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 */ int logicalShift(int x, int n) { /*First shift the x by n and clean the first k bits of "1" caused by the sign bit. B/c we know how many bits we * shifted, then we know how many bits we need to clean. */ int i = 1 << 31; return ((~(i>>(n+(~0))))&(x>>n)&(!(!n)<<31>>31))|(((!n)<<31>>31)&x); } /* * minusOne - return a value of -1 * Legal ops: ! ~ & ^ | + << >> * Max ops: 2 * Rating: 1 */ int minusOne(void) { /* 0xffffffff的值就是-1*/ return ~0; } /* * tmin - return minimum two's complement integer * Legal ops: ! ~ & ^ | + << >> * Max ops: 4 * Rating: 1 */ int tmin(void) { /* TMIN is 1 followed by 31 zeros */ return (1 << 31); } /* * TMax - return maximum two's complement integer * Legal ops: ! ~ & ^ | + << >> * Max ops: 4 * Rating: 1 */ int tmax(void) { return ~(1 << 31); } /* * negate - return -x * Example: negate(1) = -1. * Legal ops: ! ~ & ^ | + << >> * Max ops: 5 * Rating: 2 */ int negate(int x) { /* 求補碼 */ return ~x + 1; } /* * isPositive - return 1 if x > 0, return 0 otherwise * Example: isPositive(-1) = 0. * Legal ops: ! ~ & ^ | + << >> * Max ops: 8 * Rating: 3 */ int isPositive(int x) { //0以及0x80000000的特性：~x+1 //錯誤答案：return !((x>>31)&0x1); return (~x>>31)&((~x+1)>>31)&0x1 ; } /* * 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 xsign = (x >> 31) & 0x1; int ysign = (y >> 31) & 0x1; /* 分爲三種情況： 1，x > 0 ,y < 0; 2, x > 0 ,y > 0;或x < 0,y < 0;(利用y-x是否爲0求） 3, x < 0 ,y > 0 */ //return (xsign & ~ysign) | !(((~x+1+y)>>31)&0x1) & !(~xsign & ysign); /* *上述答案有一個小bug，那就是x == y時它返回1，與要求不符。正解如下 *((xsign & ~ysign) | !((~x+y)>>31)) & !(~xsign & ysign)。 */ return (xsign & ~ysign) | !(((~x+y)>>31)&0x1) & !(~xsign & ysign); } /* * sm2tc - Convert from sign-magnitude to two's complement * where the MSB is the sign bit * Example: sm2tc(0x80000005) = -5. * Legal ops: ! ~ & ^ | + << >> * Max ops: 15 * Rating: 4 */ int sm2tc(int x) { /*首先理解題意：sign-magnitude爲有符號數，two's complement爲二進制補碼錶示 * 0x80000005爲一個有符號數，-5的二進制補碼爲0xfffffff3 ，我們要做的就是把 * 0x80000005 ——> 0xfffffff3 */ int sign = x >> 31; return (x ^ sign) + (((1 << 31) + 1) & sign); //其他答案：return (sign & (~(x ^ (x & (1 << 31))) + 1)) | (~sign & x); } /* * tc2sm - Convert from two's complement to sign-magnitude * where the MSB is the sign bit * You can assume that x > TMin * Example: tc2sm(-5) = 0x80000005. * Legal ops: ! ~ & ^ | + << >> * Max ops: 15 * Rating: 4 */ int tc2sm(int x) { /*首先理解題意：sign-magnitude爲有符號數，two's complement爲二進制補碼錶示 * 0x80000005爲一個有符號數，-5的二進制補碼爲0xfffffff3 ，我們要做的就是把 * 0xfffffff3 ——> 0x80000005 */ int sign = (x>>31) & 0x1; return 2; } /* * isNonZero - Check whether x is nonzero using * the legal operators except ! * Examples: isNonZero(3) = 1, isNonZero(0) = 0 * Legal ops: ~ & ^ | + << >> * Max ops: 10 * Rating: 4 */ int isNonZero(int x) { /*If x isn't zero, either x or -x is negative*/ int ret = ~x + 1; ret = ret | x; return (ret >> 31) & 1; } /* * abs - absolute value of x (except returns TMin for TMin) * Example: abs(-1) = 1. * Legal ops: ! ~ & ^ | + << >> * Max ops: 10 * Rating: 4 */ int abs(int x) { /* exploit arithmetic right shift */ int a = x >> 31; int b = x ^ a; return b + (~a + 1); } /* * 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 mask = !x; mask = (mask << 31) >> 31; return (mask & z) | (~mask & y); } /* * isEqual - return 1 if x == y, and 0 otherwise * Examples: isEqual(5,5) = 1, isEqual(4,5) = 0 * Legal ops: ! ~ & ^ | + << >> * Max ops: 5 * Rating: 2 */ int isEqual(int x, int y) { /* * My solution 1: * * int ret = x + (~y + 1); * * return !ret; * * which check if x-y==0 (overflow does not matter),but it used * too much operators.Later I found that x^y==0 if and only if x==y. * So here is My Solution 2: */ return !(x ^ y); }

每個函數看起來都很簡單，如果用心做的話也是要花不少時間的。。。