CS:APP datalab

做得有点慢，位运算确实不太容易想到……

写出来的代码如果没有注释也不容易看懂。

/* 
* CS:APP Data Lab 
* 
* <Please put your name and userid here>
* 
* 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.  
*/
#if 0
/*
* 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 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.
    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 you
cannot 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 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;
}
FLOATING POINT CODING RULES
For 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 you
cannot 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 functions
5. 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.
    /*
* 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.
*/
    #endif
    /* 
* bitAnd - x&y using only ~ and | 
*   Example: bitAnd(6, 5) = 4
*   Legal ops: ~ |
*   Max ops: 8
*   Rating: 1
*/
    int bitAnd(int x, int y) {
    //德摩根律
    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) {
    
    x=x>>(n<<3);
    x=0xFF&x;
    return x;
}
/* 
* 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) {
    /* if x is neg, will add 1s on left , change them to 0.
    & a number as 00000...111111
    1<<31->0x80000000 right shift to get 1111...000 then ~
    if n==0 , left shift 1 -> 0, else left shift 0
    the number right shift (n-1) bits(n!=0, right n ,then left 1) 
    or 0 bit(right 0 left 0). */
    /*at first, tried change sign ,then shift, failed at 0*/
    
    return (x>>n)&~(1<<31<<!n>>n<<!!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) {
    //divide conquer
    //due to limiting const(max 0xff), my ops over 40, answer from cnblosg:liqiuhao
    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);//4 bytes:3,2,1,0->3,3+2,2+1,1+0
    x = x + (x>>16);//->3,3+2,3+2+1,3+2+1+0
    return x & 0x3F;//max=0x0010,0000  thus & 0x0011,1111
}
/* 
* bang - Compute !x without using !
*   Examples: bang(3) = 0, bang(0) = 1
*   Legal ops: ~ & ^ | + << >>
*   Max ops: 12
*   Rating: 4 
*/
int bang(int x) {
    return (((~x+1)|x)>>31)+1;
}
/* 
* tmin - return minimum two's complement integer 
*   Legal ops: ! ~ & ^ | + << >>
*   Max ops: 4
*   Rating: 1
*/
int tmin(void) {
    return 1<<31;
}
/* 
* 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) {
    //didn't make out, answer from cnblogs:liqiuhao
    int y;
    n = ~n + 1;
    y = (x << (32 + n)) >> (32 + n);
    //shift to most left,truncate 32-n
    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) {
    /*
    pos:x>>n
    neg:(x+2^n-1)/2^n, (x + (1<<n)-1) >> n
    */
    int mask=((1<<n)+~0)&x>>31; //>> prior of & ,
    return (x + mask) >> 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)&((x>>31)+1);// 0
}
/* 
* 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) {
    /*
    1. x has same sign with y(wont overflow), and y-x>=0
    2. different sign,y is pos.
    */
    int sum=((~x+1+y)>>31)+1;//y-x>=0, (pos>>31)+1=1,(neg>>31)+1=0
    //int xsign=(x>>31)+1;
    int ysign=(y>>31)+1;
    int same=((x^y)>>31)+1;
    return (same&sum)|(!same&ysign);
}
/*
* 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) {
    //binary search
    //1. if in former 16 bits?
    int mask0=0xff<<8;
    int mask1=(mask0+0xff)<<16;
    int res1=!!(mask1&x);//if in?
    int shift1=res1<<4;
    int res2,res3,res4,res5,shift2,shift3,shift4,shift5,ans;
    x=x>>shift1;//mask1>> 0 or 16 bits
    //2. 8 bits
    res2=!!(mask0&x);
    shift2=res2<<3;
    x=x>>shift2;
    //3. 4bits
    res3=!!(0xf0&x);
    shift3=res3<<2;
    x=x>>shift3;
    //4. 2 bits
    res4=!!(0xc&x);
    shift4=res4<<1;
    x=x>>shift4;
    //5. 1 bits
    res5=!!(0x2&x);
    shift5=res5;
    ans=shift1+shift2+shift3+shift4+shift5;
    return ans;
}
/* 
* 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) {
    /*
    if NaN,return uf,else return -uf
    1^A=~A,0^A=A
    judge NaN:==0x7fc00000,0xffc00000
    */
    unsigned mask1=0x80000000;//1<<31;
    unsigned mask2=0x7f800000;//=uf<<1;
    unsigned mask3=0x007fffff;
    unsigned mask4=0x7fc00000;
    //if(!((mask2&uf)^mask2)&&(mask3&uf)) successful
    if(!((mask4&uf)^mask4))//another way
    {return uf;}
    else {
        uf = uf ^ mask1 ;
        return uf;
    }
}
/* 
* 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) {
    
    if(!x){return 0;}
    else{
        int exp=~0;
        int E;
        unsigned mask1=x;
        unsigned mask2,mask3;
        int shift;
        int sign=0x80000000&x;// pos=0, neg=1
        if(sign){
            mask1=-mask1;
        }
        mask3=mask2=mask1;
        while(mask1){
            exp=exp+1;
            mask1=mask1>>1;
        }
        E=exp;
        // exp=(exp+0x7f)<<23;
        shift=31-E-8;
        // mask2=mask2<<(31-exp);
        // mask2=mask2>>8;
        if(shift>=0){//forgot "="
        // left shift
            // mask2<<shift;//mdzz, debug for a long time
            mask3=mask3<<shift; // fotgot "="
        }
        else{
            /* shift < 0, right shift.deal with round.round to even
            if odd,add 0.5 (if deserted frac > 0.5 , +1 ;if = , +1);
            if even ,add 0.499...(if deserted frac > 0.5 , +1 ;if = , +0)*/
            shift=-shift;
            // mask2>>shift;
            mask2=mask2>>shift;
            mask3+=(1<<(shift-1));
            if(mask2&1){
            }
            else{
                mask3+=-1;
            }
            mask3=mask3>>shift;
        }
        exp=((exp+0x7f)<<23)+(mask3>>1);//round may cause overflow
        exp=exp&0xff800000; // get rid of influence of adding frac
        mask3=mask3&0x7fffff;
        return sign|exp|mask3;
    }
}
/* 
* 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) {
    /*
    1. special:inf or NaN
    2. denorm 
    2.1 0
    2.2 close to 0
    3. norm ,exp++*/
    unsigned res;
    int mask1=uf&0x7f800000;
    int sign = uf&0x80000000;
    if(!(mask1^0x7f800000)){//exp all 1
        res= uf;
    }
    else if(!(mask1)){//exp all 0
        if(!(uf&0x7fffffff)){// 0
            res= uf;
        }
        else{
            res=uf<<1;
            res=res|sign;
        }
    }
    else{
        res = uf + 0x800000;
    }
    return res;
}

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