division

I read in a paper that division and multiplication of a number by a power of 2 is a trivial process. I have searched a lot of internet for the explanation but doesn't get it. Can any one explain in easy words what does this actually meant to say. Vlad It is trivial from the bit operations perspective. Multiplying by 2 is equivalent to a shift left by 1 bit, division is a right shift. similarly it is the same trivial to multiply and divide by any power of 2. int a = 123; // or in binary format: a = 0b01111011; assert (a * 2) == (a << 1); // 2 = 2^1, (a << 1) = 11110110 assert (a / 2) == (a >> 1

I have value A of type DECIMAL(19,8) - the scale is 8 , so the number of decimal digits that will be stored to the right of the decimal point is 8 . Now, I am dividing A on B , where B is BIGINT . For, example: SELECT CAST(3 AS DECIMAL(19, 8)) / CAST(27 AS BIGINT) -- 0.111111111111111111111111111 ,CAST(300 AS DECIMAL(19, 8)) / CAST(27 AS BIGINT) -- 11.111111111111111111111111111 ,CAST(75003 AS DECIMAL(19, 8)) / CAST(13664400 AS BIGINT) -- 0.005488934750153684025643277 the output values are with length: 29 , 30 , 29 respectively. Could anyone tell why the length of the value for the three

Using bcdiv, i can't divide with small float using scientific notation : Working code : bcscale(30); $a = '1' ; $b = '0.00000001'; $result = bcdiv($a, $b); var_dump($result); Results in : string(20) "100000000.0000000000" Non-working code : bcscale(30); $a = '1' ; $b = '1e-8'; $result = bcdiv($a, $b); var_dump($result); Results in : Warning: bcdiv() [function.bcdiv]: Division by zero in C:\wamp\www\utilitaires\test_bcdiv.php on line XX NULL How can i do this division properly, with the less precision loss ? That is because, actually, bcmath doesn't support scientific notation. It isn't

hi i'm doing a java activity that will divide the two given numbers without using the "/" operator. I want to use a loop statement. System.out.print("Enter Divident: "); int ans1 = Integer.parseInt(in.readLine()); System.out.print("Enter Divisor: "); int ans2 = Integer.parseInt(in.readLine()); The output is: Enter Dividend: 25 Enter Divisor 5 5 How can solve this without using this "ans1/ans2" if you really want to use loop to divide two numbers, you can write it like code below int c=0; while(ans1 >= ans2){ ans1 -= ans2; c++; } after loop c equals quotient and ans1 equals reminder of division

I'm having an issue trying to divide two decimals and then display the result. Annoyingly this is only happening on our server, and it appears to work perfectly fine if I run the code locally. This is the code that I am trying to run decimal dOne = -966.96M; decimal dTwo = 2300M; decimal dResult = Decimal.Round((dOne / dTwo), 28, MidpointRounding.AwayFromZero); The resulting number (as generated from windows calculator) is -0.43346086956521739130434782608696 This always results in an overflow exception: System.OverflowException: Value was either too large or too small for a Decimal. at System

Output of the below code confusing me. Why NaN sometimes and Infinity other times ? public static void main (String[] args) { double a = 0.0; double b = 1.0; int c = 0; System.out.println(a/0.0); System.out.println(a/0); System.out.println(b/0.0); System.out.println(b/0); System.out.println(c/0.0); System.out.println(c/0); } Outputs is: NaN NaN Infinity Infinity NaN Exception in thread "main" java.lang.ArithmeticException: / by zero What is the deciding factor here ? This is because of The IEEE Standard for Floating-Point Arithmetic (IEEE 754) which is a technical standard for floating-point

does the following integer arithmetic property hold? (m/n)/l == m/(n*l) At first I thought I knew answer (does not hold), but now am not sure. Does it hold for all numbers or only for certain conditions, i.e. n > l ? the question pertains to computer arithmetic, namely q = n/m, q*m != n , ignoring overflow. Case1 assume m = kn+b (b<n), left = (m/n)/l = ((kn+b)/n)/l = (k+b/n)/l = k/l (b/n=0, because b<n) right = (kn+b)/(n*l) = k/l + b/(n*l) = k/l (b/(n*l)=0, because b<n) => left = right Case2 assume m = kn, left = (m/n)/l = (kn/n)/l = k/l right = kn/(n*l) = k/l => left = right So, (m/n)/l == m/