Compare two integers in C or C++ without comparison operators

Question

Produce the shortest program which takes two signed integers as input (through stdin or as arguments) and displays 3 different outputs depending upon whether the first number is (1) greater than, (2) smaller than, or (3) equal to the second number.

The Catch

You cannot use any of the following in your program:

• The standard comparison operators: <, >, <=, >=, ==, !=.
• Any library file apart from conio, stdio, or iostream.
• Any non-ASCII or non-printable ASCII character.

The Winner

The program with the shortest number of characters wins.

Answer 1

53 bytes

main(a,b){scanf("%d%d",&a,&b);printf("%ld",0l+a-b);}

Only the first character of the output is relevant. The three different outputs are:

1. '-' if b > a
2. '0' if a == b
3. any other character if a > b

It works for the full input range of int on all platforms where sizeof(long) > sizeof(int).

Edit: it costs one extra character to make case 3 print a '+' uniquely instead:

main(a,b){scanf("%d%d",&a,&b);printf("%+ld",0l+a-b);}

Answer 2

90 bytes

If we can use stdio, why not use its formatting capabilities to perform comparison?

main(a,b){scanf("%d%d",&a,&b);snprintf(&a,2,"%d",b-a);a&=63;putchar(51-!(a-45)-!!(a-48));}

Assumes ASCII-compatible encoding and little-endianness.

72 bytes

Quotients are rounded toward zero but right-shifts are (in practice) "rounded down". That's a dead giveaway.

main(a,b){scanf("%d%d",&a,&b);a-=b;putchar(a?a|=1,a/2-(a>>1)?60:62:61);}

65 79 bytes

Another distinguishing property of negative numbers is that they produce negative modulo. This one doesn't depend on integer representation at all; it even works on my 8-bit excess-127 toaster! Oh, and since we can use conio, why not save two bytes with putch? Now, if I could only find my copy of TurboC...

main(a,b){scanf("%d%d",&a,&b);long long d=a;d-=b;putch(d?d|=1,d%2-1?60:62:61);}

EDIT: Handle large differences assuming long long is wider than int.

Answer 3

Maybe I'm missing something in the rules, but...

81 bytes

main(a,b){scanf("%d%d",&a,&b);long long l=a;l-=b;printf("%lld%d",--l>>63,l>>63);}

Ouputs 00 if a > b, -10 if a == b, and -1-1 if a < b.

Answer 4

64 61 characters

main(a,b){scanf("%d%d",&a,&b);for(a-=b;a/2;a/=2);putchar(a);}

Prints the character values of -1, 0, and 1 for less than, equal to, or greater than, respectively.

This implementation relies on undefined behavior for b being of type int and for inputs outside the range INT_MIN / 2 to INT_MAX / 2. On platforms where signed overflow wraps around, whether 2s-complement (basically all of them) or sign-magnitude, it will fail for 25% of possible pairs of valid int. Interestingly (to me anyway), it will work correctly on platforms where signed overflow saturates.

Answer 5

66 102 bytes

main(a,b,c,d,e){scanf("%d %d",&a,&b);e=1<<31;c=a&e;d=b&e;putchar(a-b?c&~d?48:d&~c?49:a-b&e?48:49:50);}

Reads the integers from STDIN and prints 0 (a < b), 1 (a > b) or 2 (a == b).

Edit: Now it should also work for differences which are too large to fit into a 32-bit integer. I'm sure That nested ternary can be shortened with some more bit-magic.

Answer 6

52 bytes

Sadly this one only works for positive integers, but I thought the concept of using purely arithmetic operators was interesting:

main(a,b){scanf("%d%d",&a,&b);putchar(b%a/b-a%b/a);}

Outputs:

• ASCII code 0xFF: a less than b
• ASCII code 0x00: a equal to b
• ASCII code 0x01: a greater than b

Answer 7

 59   54 characters

54 charcters with a compiler like gcc which doesn't balk at main(x,y):

main(x,y){scanf("%d%d",&x,&y);y-=x;putchar(y>>31|!y);}

59 characters otherwise:

main(){int x,y;scanf("%d%d",&x,&y);y-=x;putchar(y>>31|!y);}

Output:

• ASCII code 0x00 if x < y
• ASCII code 0xFF if x > y
• ASCII code 0x01 if x == y

Answer 8

66 bytes

main(a,b){scanf("%d%d",&a,&b);putchar((0l+b-a>>63)-(0l+a-b>>63));}

Prints the byte 0x00 if a == b, 0x01 if a < b and 0xff if a > b.

Since non-ASCII or non-printable ASCII character in [my] program and if anything is not explicitly forbidden in the question, then it is permitted, unprintable character in the output should be completely fine.

Answer 9

87 characters

main(a,b,c){scanf("%d%d",&a,&b);c=1<<31;a+=c;b+=c;puts(a^b?(unsigned)a/b?">":"<":"=");}

Using the 2^31 trick to convert to unsigned int's

Casting the division to unsigned to handle the upper bit as data, not sign

Using ^ to XOR a and b, when they are equal this returns 0

Using nested conditionals (?) to get "<",">", or "=" to feed to puts()

Answer 10

In 64 chars without stdio.h

a,b;main(){scanf("%d%d",&a,&b);puts((a-b)>>31?"<":a^b?">":"=");}

prints '>' if a > b, '<' if a < b, '=' if a == b int overflow is UB. Just don't overflow.

// declare a and b as ints
a,b;

// defaults to int
main()
{
  scanf("%d%d",&a,&b);
   /*
    * (a-b)>>31 signbit of a-b
    * a^b a xor b -> 0 if they are equal
    */
  puts(((a-b)>>31) ? "<" : (a^b) ? ">" : "=");
}

Answer 11

88 89 bytes

main(a,b){scanf("%d%d",&a,&b);a+=1<<31;b+=1<<31;for(;a&&b;a--)b--;putchar(a?b?48:49:50);}

This starts by adding 1<<31 (INT_MIN) to a and b, so that 0 now corresponds INT_MIN. Then loops and decrements a and b every loop until either is 0, then prints 0, 1 or 2 depending on whether a, b or both are 0.

120 119 bytes

main(a,b,c){scanf("%d%d",&a,&b);c=1<<31;a^=c;b^=c;for(c~=c;!c;c/=2)if(a&c^b&c){putchar(a?48:49);return;}putchar(50);}

It's not the shortest solution as it is but might be golfed down a bit by better golfer than me. (Or just people with more knowledge of C than me)

The idea is to mask each bit, starting from the left one and checking for inequality. The rest should explain itself. Since negative numbers start with a 1 bit, I first invert the first bit with a^=1<<31.

Answer 12

71 bytes

main(x,y,z){scanf("%d%d",&x,&y);putchar((z=x-y)?(z&(z>>31))?50:49:51);}

http://ideone.com/uvXm6c

Answer 13

68 characters

int main(a,b){scanf("%d%d",&a,&b);putchar(a-b?((unsigned)a-b)>>31:2);}

Puts ASCII character 1, 2 or 3 for less than, greater than or equal, respectively.

Answer 14

I think I'm not even going to try at writing short code. What I will attempt is to perform this comparison in a way that is portable per the C99 spec.

int a, b;   // Let's assume these are initialized
int sign_a = a ? ((a|7)^2)%2 + ((a|7)^3)%2 : 0;

The modulo operator preserves sign, but it may well produce a zero (including negative zero), so we ensure we have both an odd and even value for it to check (even without knowing if we are using ones complement). Arithmetic operations may overflow, but bitwise will not, and by ensuring there are bits both set and cleared we avoid inadvertently converting our number into negative zero or a trap value. The fact that two operations are required to do so oddly should not matter, because the possible trap representation doesn't cause undefined behaviour until put in an lvalue. Doing the operation with bit 0 toggled ensures we get exactly one non-zero remainder. Armed with the knowledge of both signs, we can decide how to proceed with the comparison.

char result="\0<<>=<>>\0"[4+3*sign_a+sign_b]
if (!result) {   // signs matching means subtraction won't overflow
  int diff=a-b;
  int sign_diff=diff ? (diff|7^2)%2 + (diff|7^3)%2 : 0;
  result = ">=<"[1-sign_diff];
}

This method may be one of a few that permit extracting the sign of an integer negative zero. We solve that by explicitly checking for zero. If we were golfing for real, we could of course allow the comparison of two zeros to also perform the subtraction.

Answer 15

C 80 chars

a,b,c=1<<31;main(){scanf("%d%d",&a,&b);putchar(a^b?(a&c^b&c?a:a-b)&c?60:62:61);}

It prints '<', '>' or '=', as it should.

C 63 chars

A new approach:

a;main(b){scanf("%d%d",&a,&b);putchar(50+(0L+a-b>>42)+!(a-b));}

Prints '1', '2' or '3'.