Compare two numbers without if

I. Problems

There are two variables, a and B, which do not need to be "if", "? : ", Switch or other judgment statements, find the two numbers that are relatively large

Ii. Solutions

/* Method 1: calculate the average value (a + B) + abs (a-B )) /2 is (a + B-abs (a-B)/2 */int fMax1 (int a, int B) {return (a + B) + abs (a-B)/2 ;}

/* Method 2: When abs () a is not used <B, a/B = 0, so the front is B * (B/a), followed by B/, if the result is ba = B, a/B = 1, so the first is a + B = 2a, followed by 2, then the result is aa> B, b/a = 0, so the first is a * (a/B), followed by a/B, then the result is a */int fMax2 (int a, int B) {int larger = (a * (a/B) + B * (B/a)/(a/B + B/); // long smaller = (B * (a/B) + a * (B/a)/(a/B + B/a); return larger ;}

/* Method 3: if the remainder of a/B is not 0, it means that a> B is a good method, but the question says, "? : "*/Int fMax3 (int a, int B) {return (a/B )? A: B ;}

/* Method 4: shift method. If B is less than 0, it is saved as a complement, so the highest bit is 1, so the right shift 31 bits B> 31 is actually the highest bit value B> = 0, the highest bit is 0, so B and 1 are B, a = a-(a-B) = bb and 1 are 0, equivalent to a = a-0 = a */int fMax4 (int a, int B) {B = a-B; a-= B & (B> 31); return a ;}// shift method int fMax5 (int a, int B) {int c [2] = {a, B}; int z = a-B; z = (z> 31) & 1; return c [z];} // shift method int fMax6 (int a, int B) {int flag = (a-B)> 31) & 1; return a-(a-B) * flag ;}

// I think this should be the best of B's int fMax7 (int a, int B) {int pair [2] = {a, B }; return pair [a <B];}

Iii. Feelings

The algorithm is so amazing that you need to think more in the future ···