modulus

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. 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 is no reason not to use a modulo with a while. $i=0; while(guard()){ if($i % 3 == 0){ //ploing } $i++ } This

I originally tried this, however the % operator isn't defined for float64. func main(){ var a float64 a = 1.23 if a%1 == 0{ fmt.Println("yay") }else{ fmt.Println("you fail") } } Assuming your numbers will fit into an int64 , you can just compare the float value with a converted integer value to see if they're the same: if a == float64(int64(a)) { fmt.Println("yay") } else { fmt.Println("you fail") } Otherwise you can use the math.Trunc function detailed here , with something like: if a == math.Trunc(a) { fmt.Println("yay") } else { fmt.Println("you fail") } That one should work within the

I want to find all muliples of a number in PHP. I'm using something like this if($count != 20 ) to work out if $count is not equal to 20. But I also need this script to check if $count is not equal to 20, 40, 60, 80, 100, 120, 140, 160 etc. Any ideas? I think i need to use the modulus symbol ( % ), but I don't know. if ($count % 20 != 0) if ($count % 20 != 0) { // $count is not a multiple of 20 } marianboda If you don't want zero to be excluded: if ($count % 20 != 0 || $count == 0) You can do it like so: if($count % 20 != 0) 来源： https://stackoverflow.com/questions/2809864/find-multiples-of-a

There is a column in options that hold an integer. I want to select the row only if that value % 2 = 1. I know this can be done in 2 queries but is it possible to do it in 1? MySQL, SQL Server, PostgreSQL, SQLite support using the percent sign as the modulus: WHERE column % 2 = 1 For Oracle, you have to use the MOD function : WHERE MOD(column, 2) = 1 At least some versions of SQL (Oracle, Informix, DB2, ISO Standard) support: WHERE MOD(value, 2) = 1 MySQL supports '%' as the modulus operator: WHERE value % 2 = 1 select * from table where value % 2 = 1 works fine in mysql. Note: Disregard this

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 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; } e.g. : // (53 + x) % 62 = 44 var res = ReverseModulus(62,53,44); // res = 53 // (2 + x) % 8 = 3 var res =

I can write the program int a = 3; int b = 4; Console.WriteLine(a % b); The answer I get is 3. How does 3 mod 4 = 3??? I can't figure out how this is getting computed this way. NullUserException Because the remainder of 3 / 4 = 3. http://en.wikipedia.org/wiki/Modulo_operator If you can't figure out why the remainder is 3, we've got some more serious problems here. I wasn't quite sure what to expect, but I couldn't figure out how the remainder was 3. So you have 3 cookies, and you want to divide them equally between 4 people. Because there are more people than cookies, nobody gets a cookie (

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; }; rici 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 substitutes a multiply for the divide, similar to the optimization performed by gcc , and is far and away the