由于http://csapp.cs.cmu.edu/并未完全開放實驗，很多附加實驗做不了，一些環境也沒辦法搭建，更沒有標準答案。做了這個實驗的朋友可以和我對對答案；）

實驗內容和要求可在http://csapp.cs.cmu.edu/3e/labs.html獲得。

Data Lab [Updated 5/4/16]

bits.c

/* * CS:APP Data Lab * * <李秋豪 Richard Li>* * 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. *//** Instructions to Students:** STEP 1: Read the following instructions carefully.*//* You will provide your solution to the Data Lab by editing the collection of functions in this source file.INTEGER CODING RULES:Replace the "return" statement in each function with oneor 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 arenot 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 toone 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.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.7. Use any data type other than int. This implies that youcannot use arrays, structs, or unions.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 morethan the word size.EXAMPLES OF ACCEPTABLE CODING STYLE:pow2plus1 - returns 2^x + 1, where 0 <= x <= 31int pow2plus1(int x) {exploit ability of shifts to compute powers of 2return (1 << x) + 1;}pow2plus4 - returns 2^x + 4, where 0 <= x <= 31int pow2plus4(int x) {exploit ability of shifts to compute powers of 2 int result = (1 << x);result += 4;return result;}FLOATING POINT CODING RULESFor the problems that require you to implent floating-point operations, the coding rules are less strict. You are allowed to use looping and conditional control. You are allowed to use both ints and unsigneds. You can use arbitrary integer and unsigned constants.You are expressly forbidden to:1. Define or use any macros.2. Define any additional functions in this file.3. Call any functions.4. Use any form of casting.5. Use any data type other than int or unsigned. This means that youcannot use arrays, structs, or unions.6. Use any floating point data types, operations, or constants.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. Use the BDD checker to formally verify your functions5. The maximum number of ops for each function is given in theheader comment for each function. If there are any inconsistencies between the maximum ops in the writeup and in this file, considerthis file the authoritative source.* STEP 2: 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 BDD checker to formally verify that your solutions produce * the correct answers. */ /*----------------------------------------------------*//* * 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); } /* * 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) {n = n << 3; /* n = n*8 */return (x & (0xFF<<n)) >> n; } /* * 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) {int mask = ~0 << n;return (mask & x) >> n; } /** 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) {/*這個題我一開始的思路是逐個位的將x向右移動，每次移動后將和0x01與的結果和sum相加。但是這樣操作符的數量會超過40個。后來想到，其實沒有必要非要每次把加法放在起始位，其他位置也是符合加法規律的，所以可以在多個位置同時進行加法，最后“匯總”到首位的地方*//*同時要注意兩個問題：1.第一次一位相加的時候可能會溢出到兩位”10“，第二次兩位相加的時候可能會溢出到三位”100“，第三次相加的時候是不會溢出超過四位的”1000“，所以會產生0x0y0z0a這種格式。將這四個四位相加即可。2.四個四位相加時不能使用x+(x>>24)因為之前x = x + (x>>8)已經將第三個和第四個4位相加了，最后只需要將新的第三位和第四位相加。*/int mask1 = (((((0x55 << 8) + 0x55) << 8) + 0x55) << 8) + 0x55;int mask2 = (((((0x33 << 8) + 0x33) << 8) + 0x33) << 8) + 0x33;int mask3 = (((((0x0f << 8) + 0x0f) << 8) + 0x0f) << 8) + 0x0f;x = (x & mask1) + ((x >>1) & mask1);x = (x & mask2) + ((x >>2) & mask2);x = (x & mask3) + ((x >>4) & mask3);x = x + (x>>8);x = x + (x>>16);return x & 0x3F; } /* * bang - Compute !x without using !* Examples: bang(3) = 0, bang(0) = 1* Legal ops: ~ & ^ | + << >>* Max ops: 12* Rating: 4 */ int bang(int x) {x = x | (x >> 16);x = x | (x >> 8);x = x | (x >> 4);x = x | (x >> 2);x = x | (x >> 1);return x & 1; } /* * tmin - return minimum two's complement integer * Legal ops: ! ~ & ^ | + << >>* Max ops: 4* Rating: 1*/ int tmin(void) {int x = 0x80 << 24;return x; } /* * 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*/ int fitsBits(int x, int n) {n = ~n + 1; /* 取得-n */int y = x << (32 + n) >> (32 + n); /* 將縮短后的”符號位“擴展”*/return !(y ^ x);} /* * 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*/ int divpwr2(int x, int n) {int bias = (x >> 31) & ((1 << n) + ~0); /* 如果x為正數，與運算后bias為0 */x = x + bias;return x >> n; } /* * negate - return -x * Example: negate(1) = -1.* Legal ops: ! ~ & ^ | + << >>* Max ops: 5* Rating: 2*/ int negate(int x) {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) {return (!((x >> 31) & 1)) & (!!x); } /* * 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 a = x >> 31;int b = y >> 31; /*正數的話為0，負數的話為-1*/int c = a ^ b; /*分同異號來處理，異號的話c為-1，同號的話c為0*//*接下來對同號和異號兩種情況分別處理，處理的結果不為零（準確講是全f）代表x<=y，最后將兩種情況或（實際只有一個不為零）*/int case1 = c & a; /*異號，只有當x為負數的時候返回1*/int case2 = ~c & ( ~((y + (~x + 1)) >> 31));/*同號，x-y的結果非負即返回1*/int result = case1 | case2;return !!result; /*由于result可能為全f，需要用!處理一下*/ } /** 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) {/*開始判斷是在32位里的高16位還是低16位*/int a16 = !!(x >> 16); /*a16為1代表1在高16位，為0代表在低十六位*/int b16 = a16 << 4; /*1在高16位時b16 = 16*//*開始判斷是在16里的高8位還是低8位*/int a8 = !!(x >> (8 + b16));/*a8為1代表1在高8位，為0代表在低8位*/int b8 = a8 << 3;/*開始判斷是在8里的高4位還是低4位*/int a4 = !!(x >> (4 + b8 + b16));/*a8為1代表1在高8位，為0代表在低8位*/int b4 = a4 << 2;/*開始判斷是在4里的高2位還是低2位*/int a2 = !!(x >> (2 + b4 + b8 + b16));/*a8為1代表1在高8位，為0代表在低8位*/int b2 = a2 << 1;/*開始判斷是在16里的高1位還是低1位*/int a1 = !!(x >> (1 + b2 + b4 + b8 + b16));/*a8為1代表1在高8位，為0代表在低8位*/int b1 = a1 << 0;return b1 + b2 + b4 + b8 + b16;} /* * 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*/ unsigned float_neg(unsigned uf) {unsigned mask = ~0 << 31; return mask ^ x; } /* * 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*/ unsigned float_i2f(int x) {int position = 0; /*計算最高有效位距離第0位的位置*/unsigned result = 0;unsigned bias = 127;int tmp = 0;if (x == 0){return result;}if (x == 0x80000000) /*由于-x = x，此種情況單獨處理*/{result = 0xcf000000;return result;}for(int i = 16; i >= 1; i /= 2) /*計算距離*/{tmp = i + positon;if (x << (tmp) >> (tmp) == x){position += i;}}result += (((30 - position) + bias) << 23);/*放置exp位*/if (x < 0){x = -x;result | 0x80000000; /*放置符號位*/}tmp = 7 - position;if (tmp > 0){x >>= tmp;}else{x = (x <<-tmp) & 0x007fffff; /*注意移到第24位的數字要變為0*/}return result | x; /*放置frac位*//*剛好30個，O(∩_∩)O哈哈~*/ } /* * 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*/ unsigned float_twice(unsigned uf) {int tmp = 0x7f800000;int tmpp = uf & tmp; /*切割出exp位*/int tmppp = uf & 0x80000000; /*切割出符號位*/if (tmpp == tmp) /* infinity or NaN */{return uf;}if (tmpp == 0) /* Denormalized */{return (uf << 1) | tmppp; /*恢復符號位*/}return (uf & 0x807fffff) | (((tmpp >> 23) + 1) << 23); /* Noemalized */ }

轉載于:https://www.cnblogs.com/liqiuhao/p/7325133.html

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