modulo

Is there some cool algorithm with bit wise operations? Often, the modulus and divide operations on a processor are the same thing. For instance, refer to http://jsimlo.sk/docs/cpu/index.php/div.html . This is the implementation of the divide instruction on Intel processors. Most of the time, modulus is just computed by dividing the two numbers. The quotient is stored in one register, and the remainder is stored in the other register. You would go after the remainder. If the divisor is known in advance (e.g. for code produced by a C compiler, this is a constant known at compile time) then

I apologize if this is a simple question but I'm having trouble grasping the concept of modulus division when the first number is smaller than the second number. For example when 1 % 4 my book says the remainder is 1. I don't understand how 1 is the remainder of 1 % 4. 1 / 4 is 0.25. Am I thinking about modulus division incorrectly? First, in Java, % is the remainder (not modulo) operator, which has slightly different semantics. That said, you need to think in terms of integer-only division, as if there were no fractional values. Think of it as storing items that cannot be divided: you can

I want to check, if a number is divisible by another number: for i = 1, 100 do if i % 2 == 0 then print( i .. " is divisible.") end end This should work without any problems, but with the Lua in my server the script doesn't run if there is a % in the script... I dont know whats the reason, so is there any "replacement" for that? So I could check the number divsibility? Thank you. It's not ideal, but according to the Lua 5.2 Reference Manual : a % b == a - math.floor(a/b)*b lhf Use math.fmod(x,y) which does what you want: Returns the remainder of the division of x by y that rounds the quotient

I wanted to use MOD function in SQL Server 2008R2 and followed this link but still got the message: 'MOD' is not a recognized built-in function name. DECLARE @m INT SET @m = MOD(321,11) SELECT @m Error: Msg 195, Level 15, State 10, Line 2 'MOD' is not a recognized built-in function name. Why I can't use this function from the link above? The MOD keyword only exists in the DAX language (tabular dimensional queries), not TSQL Use % instead. Ref: Modulo In TSQL, the modulo is done with a percent sign. SELECT 38 % 5 would give you the modulo 3 for your exact sample, it should be like this. DECLARE

Paraphrasing from in "Programming Pearls" book (about c language on older machines, since book is from the late 90's): Integer arithmetic operations ( + , - , * ) can take around 10 nano seconds whereas the % operator takes up to 100 nano seconds. Why there is that much difference? How does a modulus operator work internally? Is it same as division ( / ) in terms of time? The modulus/modulo operation is usually understood as the integer equivalent of the remainder operation - a side effect or counterpart to division. Except for some degenerate cases (where the divisor is a power of the

I'm brand new to Erlang. How do you do modulo (get the remainder of a division)? It's % in most C-like languages, but that designates a comment in Erlang. Several people answered with rem, which in most cases is fine. But I'm revisiting this because now I need to use negative numbers and rem gives you the remainder of a division, which is not the same as modulo for negative numbers. grifaton In Erlang, 5 rem 3. gives 2 , and -5 rem 3. gives -2 . If I understand your question, you would want -5 rem 3. to give 1 instead, since -5 = -2 * 3 + 1. Does this do what you want? mod(X,Y) when X > 0 -> X

I have a code in which i am computing x % 25. x always takes a positive value but its dynamic range is large. I found out that this particular code piece of computing a x % 25 is taking large cycles. I need to optimize it. Pre-computed lookup table is ruled out due to the possible large memory size of the table. As second approach i coded a fragment below(C code) - mod(a, b) { int r = a; while(r >= b) { r = r - b; } return r; } 1.) How can i optimize this code further for cycles(squeeze it to max)? 2.) Is there any entirely different optimized way to achieve x % 25( i know its not a common