isGreater bitwise C manipulation - greatest of two numbers

问题

Bitwise manipulation and I have to find the greatest of two numbers. These are the rules for the same:

 /* 
  * isGreater - if x > y  then return 1, else return 0 
  * Example: isGreater(4,5) = 0, isGreater(5,4) = 1
  *   Legal ops: ! ~ & ^ | + << >>
  *   Max ops: 24
 */

This is the code I've written for this:

int isGreater(int x, int y) {
/* to find greater, subtract y from x using 2's complement method.
 * then find 2's complement of the answer and shift right by 31 to give MSB
 * which is 1 if x>y and 0 if x<y */
 int ydash=(~y)+0x01;
 int diff=x+ydash;
 int a=(~diff)+0x01;
 int b=a>>31;
 int c=b&0x01;  
 return c;
}

For which I get this error:

ERROR: Test isGreater(-2147483648[0x80000000],2147483647[0x7fffffff]) failed... ...Gives 1[0x1]. Should be 0[0x0]

I'm allowed to use unsigned int, but no other data types. I'm not sure how that'd help though. How do I get around this?

回答1:

Here you may not be allowed use if-else or any of the control statements. But you can "construct" some if-else-like statement.

int isGreater(int x, int y) {        
    int sgnext_x = x >> 31;  //sgnext_x = x >= 0 ? 00000000 : 11111111
    int sgnext_y = y >> 31;
    int sgn_y = sgnext_y & 1;  //sgn_y = y >= 0 ? 0 : 1
    int minus = x + (~y + 1);  // a - b = a + (~b+1)
    int sgn_minus =(minus >> 31) & 1;

    //A control-like statement. ((statement & a) | (!(statement | (b))))
    //Returns a if statment = 11111111
    //Returns (b&1) if statment = 00000000

    int which = sgnext_x ^sgnext_y;
    int result = (!!(x^y)) & ((which & sgn_y) | (!(which | (sgn_minus))));
  return result;
}

回答2:

With x = -2147483648 and y = 2147483647 then x - y = -4,294,967,295 which is outside the range of int, hence the result cannot be represented in the variable and you got undefined behavior.

To get over this you need to use a type wider than int. As you are only allowed to use unsigned int, you'll have to implement big int operations yourself if you want to use a bigger type. You can also use another way like checking the overflow condition separately

if ((x ^ y) & 0x80000000) // x and y have different sign
{
    return (y >> 31) & 1; // return 1 if y is negative
}
else     // x and y have the same sign, overflow never occurs
{
    unsigned int xu = x, yu = y;
    unsigned int xmu = ~xu + 1U;
    unsigned int diffu = yu + xmu;
    return diffu >> 31;
}

If you aren't allowed to use conditionals you can use a muxer to mux the values

unsigned int r1 = (y >> 31) & 1U; // 1st case

unsigned int xu = x, yu = y;
unsigned int xmu = ~xu + 1U;
unsigned int diffu = yu + xmu;
unsigned int r2 = diffu >> 31;    // 2nd case

unsigned int s = ((x ^ y) >> 31) & 1U; // s = 1 if x and y have different sign, 0 otherwise
unsigned int mask = 0xFFFFFFFFU + s;
return (r1 & ~mask) | (r2 & mask);

回答3:

The algorithm you came up with fundamentally doesn't work, since if you subtract, almost all possible answers can be both the result of a subtraction where a < b and of a subtraction where a >= b. So essentially, the result doesn't tell you anything, except when it's zero then you know that a == b.

Hacker's Delight has this answer in chapter 2.11, Comparison Predicates:

x < y: (x & ~y) | ((x ^ ~y) & (x - y))

(verification)

You know how to implement subtraction in terms of addition, and swapping x and y shouldn't be a problem either. The result appears in the sign bit.

来源：https://stackoverflow.com/questions/28141981/isgreater-bitwise-c-manipulation-greatest-of-two-numbers

标签

bit-manipulation