（stdinまたは引数として）入力として2つの符号付き整数を取り、最初の数が（1）より大きい、（2）より小さい、または（3）2番目に等しいかどうかに応じて3つの異なる出力を表示する最短プログラムを生成します数。

キャッチ

プログラムでは次のいずれも使用できません。

• 標準の比較演算子：<、>、<=、>=、==、!=。
• 離れてからの任意のライブラリ・ファイルconio、stdioまたはiostream。
• 非ASCIIまたは印刷不能ASCII文字。

勝者

最短文字数のプログラムが勝ちます。

53バイト

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

出力の最初の文字のみが関連します。3つの異なる出力は次のとおりです。

1. b> aの場合は '-'
2. a == bの場合は「0」
3. a> bの場合、その他の文字

sizeof（long）> sizeof（int）であるすべてのプラットフォームで、intの全入力範囲に対して機能します。

編集：ケース3に「+」を一意に印刷させるには、1文字余分にかかります：

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

たぶん私はルールに何かが欠けていますが、...

81バイト

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

Ouputs 00もしa > b、-10あればa == b、と-1-1あればa < b。

90バイト

を使用できる場合stdio、そのフォーマット機能を使用して比較を実行してみませんか？

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

ASCII互換のエンコーディングとリトルエンディアンを想定しています。

72バイト

指数はゼロに丸められますが、右シフトは（実際には）「切り捨てられます」。それは死んだ景品です。

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

65 79バイト

負の数の別の顕著な特性は、負のモジュロを生成することです。これは整数表現にまったく依存しません。それは私の8ビットexcess-127トースターでも動作します！ああ、そしてを使用できるconioので、2バイトを節約してみませんputchか？今、TurboCのコピーしか見つけられなかったら...

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

編集：long longよりも広いと仮定して大きな違いを処理しintます。

64 61文字

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.

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.

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

 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

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.

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()

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

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.

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.

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.

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'.

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) ? ">" : "=");
}