modulus

问题 The part I keep getting stuck on is boolean(0 % 2 !=0) == false. I mean if 2 goes into 0, 0 times then the remainder would be 2, and 2 does not equal 0. So it should be true. Yet but when I put the boolean in my java program it is treating it as false. Anyone know why? The only logical answer I can wrap my head around is that maybe integers go into 0 and infinite number of times and so are recognized as false, anyone? 回答1: There are two steps: 0 % 2 evaluates to 0 . 0 != 0 evaluates to false

问题 For example: int x = 35; if( x%2==1) { //do something } I just want to check the modulus value without assigning the result to x . Assume that value is in eax , so I can use DIV instruction, then put back the original value to eax etc.. but that seems inefficient. Can you suggest a better way? 回答1: To branch according to the modulo 2 of a value in al / ax / eax / rax : test al,1 jnz is_odd is_even: ; do something for even numbers. is_odd: ; do something for odd numbers. However, if all you

问题 Is there a trick for creating a faster integer modulus than the standard % operator for particular bases? For my program, I'd be looking for around 1000-4000 (e.g. n%2048). Is there a quicker way to perform n modulus 2048 than simply: n%2048 ? 回答1: If the denominator is known at compile time to be a power of 2, like your example of 2048, you could subtract 1 and do a bitwise-and. That is: n % m == n & (m - 1) ...where m is a power of 2. For example: 22 % 8 == 22 - 16 == 6 Dec Bin ----- -----

问题 Could anyone tell me why these two modulus calculations yield two different outcomes? I just need to blame someone or something but me for all those hours I lost finding this bug. public void test1() { int stepAmount = 100; float t = 0.02f; float remainder = t % (1f / stepAmount); Debug.Log("Remainder: " + remainder); // Remainder: 0.01 float fractions = 1f / stepAmount; remainder = t % fractions; Debug.Log("Remainder: " + remainder); // Remainder: 0 } Using VS-2017 V15.3.5 回答1: My best bet

问题 I remember from assembly that integer division instructions yield both the quotient and remainder. So, in python will the built-in divmod() function be better performance-wise than using the % and // operators (suppose of course one needs both the quotient and the remainder)? q, r = divmod(n, d) q, r = (n // d, n % d) 回答1: To measure is to know (all timings on a Macbook Pro 2.8Ghz i7): >>> import sys, timeit >>> sys.version_info sys.version_info(major=2, minor=7, micro=12, releaselevel='final

问题 Trying to figure out how to write a recursive predicate divide_by(X, D, I, R) that takes as input a positive integer X and a divisor D, and returns the answer as the whole number part I and the remainder part R, however, I can't seem to get my head around Prolog. How would I go about doing this? 回答1: There are predefined evaluable functors for this. (div)/2 and (mod)/2 always rounding down. Recommended by LIA-1, Knuth etc. (//)/2 and (rem)/2 rounding toward zero (actually, it's implementation

问题 I'm making a forum in PHP. I have to display all forum categories in a table, and to do so, I have used a while loop. However, I want to have only 3 td's in every table row. To loop through the categories, I'm using a while loop with the query, so I don't think I can use modulus here. 回答1: Why can't you use modulus? Just add a counter somewhere, and if it hits % 3 == 0 reset the counter and do your stuff. You might need to do some extra if's for first and last and stuff like that, but there

问题 Derived from this question : (Java) How does java do modulus calculations with negative numbers? Anywhere to force PHP to return positive 51? update Looking for a configuration setting to fix, instead hard-guessing Or other math function like bcmath? updated Not entire convinced by that java answer, as it does not take account of negative modulus -13+(-64) =? 回答1: If GMP is available, you can use gmp_mod Calculates n modulo d. The result is always non-negative, the sign of d is ignored.

问题 now I have one formula: int a = 53, x = 53, length = 62, result; result = (a + x) % length; but how to calculate reverse modulus to get the smallest "x" if I known result already (53 + x) % 62 = 44 //how to get x i mean what's the formula or logic to get x 回答1: private int ReverseModulus(int div, int a, int remainder) { if(remainder >= div) throw new ArgumentException("Remainder cannot be greater than or equal to divisor"); if(a < remainder) return remainder - a; return div + remainder - a; }

问题 Getting the modulus of a number can be easily done without the modulus operator or divisions, if your operand is a power of 2. In that case, the following formula holds: x % y = (x & (y − 1)) . This is often many performant in many architectures. Can the same be done for mod 31 ? int mod31(int a){ return a % 31; }; 回答1: Here are two ways to approach this problem. The first one using a common bit-twiddling technique, and if carefully optimized can beat hardware division. The other one