C Puzzles

Unit-test
----------
spyglass: $cat my_version2.c


int max(int a,int b)
{
int g=0 ;

!a && b && ( ((b+1)/b ) && (g = b) || (g=a));
!b && a && ( ((a+1)/a ) && (g = a) || (g=b));

(!(!a || !b)) &&
( (( ( !((b+1)/b) && (!( a/b) ) )
|| ( ((a+1)/a) && (a/b) )
) && (g=a))
|| (g=b)
) ;
return g;
}

int main()
{
int a,b;
int g;
int i=15; // max no of expected data

printf("Enter two numbers:");
scanf("%d %d", &a , &b );
while( --i ){
g = max(a,b);
printf("greatest = %d ,small =%d \n", g, a+b-g );

printf("Enter two numbers:");
scanf("%d %d", &a , &b );
}
}

spyglass: $gcc my_version2.c
spyglass: $cat test_data.txt
10 2
2 10
-2 10
10 -2
-2 -10
-10 -2
-10 2
2 -10
0 0
0 10
10 0
-10 0
0 -10


spyglass: $./a.out < test_data.txt
Enter two numbers:greatest = 10 ,small =2
Enter two numbers:greatest = 10 ,small =2
Enter two numbers:greatest = 10 ,small =-2
Enter two numbers:greatest = 10 ,small =-2
Enter two numbers:greatest = -2 ,small =-10
Enter two numbers:greatest = -2 ,small =-10
Enter two numbers:greatest = 2 ,small =-10
Enter two numbers:greatest = 2 ,small =-10
Enter two numbers:greatest = 0 ,small =0
Enter two numbers:greatest = 10 ,small =0
Enter two numbers:greatest = 10 ,small =0
Enter two numbers:greatest = 0 ,small =-10
Enter two numbers:greatest = 0 ,small =-10
Enter two numbers:greatest = 0 ,small =-10 <<<< ignore the tail
Enter two numbers:spyglass: $
spyglass: $
 

Other-ideas :
____________________________

void compare (int a, int b)
{
int diff, msb;

msb = 1 << (sizeof(int)*8 - 1);

if ((a & msb) && !(b &msb)) /* a is negative and not b */
return b;
else if ((b & msb) && !(a &msb)) /* b is negative and not a */
return a;

/* a and b are of same sign */
diff = a-b;
if (!diff || !(diff & msb)) /* a==b and a > b */
return a;
return b;
}


Regards,
Arul Kumar C
_________________________________

void compare (int a, int b)
{
int i, aset, bset;

if (!(a ^ b))
printf("a=b\n");
else { /* either a < b or a > b */
i = 31;
while (i ^ -1) { /* while i != -1 */
aset = bset = 0;
if (a & (1 << i)) /* ith bit from left is set in a */
aset = 1;
if (b & (1 << i)) /* ith bit from left is set in b */
bset = 1;
if (!(aset ^ bset)) { /* bits in both a & b set or both are not set */
i--; continue;
}
else { /* there is a mismatch */
if (!(i ^ 31)) { /* i is 31, take care of negative nos */
if (aset) printf("a < b\n");
else printf("a > b\n");
}
else {
if (aset) printf("a > b\n");
else printf("a < b\n");
}
return;
}
}
}
}


int main()
{
int a,b;
printf("Enter a:");
scanf("%d", &a);
printf("Enter b:");
scanf("%d", &b);
compare(a,b);
}

Thanks,
Anand
 

Hi Vinod,

I forgot to add comments.

Is it a implementation independent code?Especially, checking the MSB for -ve values ?Page 36 in K&R (2.2 Data Types and Sizes) says about 'char'.he is saying that it is machine-dependent.Am not sure about 'int'.

if x is a large +ve and y is a large -ve, it might result in overflow.
 

yes, checking a MSB depends on compilers also. ex:- notation of compiler whether it is a little indian and big indian etc.,

so i am thinking of this solution

int max_two(int i, int j)
{
int temp = i - j;
if(!temp) return i; // if both are equal
return (((temp+1)/ temp) ? i : j ); // if not equal
}