integer-division

I understand how to do it for powers of 2 so that's not my question. For example, if I want to find 5% of a number using a bit shift instead of an integer divide, how would i calculate that? So instead of (x * 20 / 19), I could do (x * 100 >> 11). Now this isn't right but it's close and I arrived at it using trial and error. How would I determine the most possible precise shift to use? High Performance Mark Best approach is to let the compiler do it for you. You simply write a/b in your language of choice, and the compiler generates the bit twiddling. EDIT (I hope you don't mind, i'm adding

In C, is there a difference between integer division a/b and floor(a/b) where both a and b are integers? More specifically what happens during both processes? Pete Becker a/b does integer division. If either a or b is negative, the result depends on the compiler (rounding can go toward zero or toward negative infinity in pre-C99; in C99+, the rounding goes toward 0). The result has type int . floor(a/b) does the same division, converts the result to double, discards the (nonexistent) fractional part, and returns the result as a double. ouah floor returns a double while a / b where both a and b

I was writing this code in C when I encountered the following problem. #include <stdio.h> int main() { int i=2; int j=3; int k,l; float a,b; k=i/j*j; l=j/i*i; a=i/j*j; b=j/i*i; printf("%d %d %f %f\n",k,l,a,b); return 0; } Can anyone tell me why the code is returning zero for the first and third variables ( k and a )? What I think you are experiencing is integer arithmetic . You correctly suppose l and b to be 2, but incorrectly assume that k and a will be 3 because it's the same operation. But it's not, it's integer arithmetic (rather than floating-point arithmetic). So when you do i / j

I have a 128-bit number stored as 2 64-bit numbers ("Hi" and "Lo"). I need only to divide it by a 32-bit number. How could I do it, using the native 64-bit operations from CPU? (Please, note that I DO NOT need an arbitrary precision library. Just need to know how to make this simple division using native operations. Thank you). If you are storing the value (128-bits) using the largest possible native representation your architecture can handle (64-bits) you will have problems handling the intermediate results of the division (as you already found :) ). But you always can use a SMALLER

I'm working on Cortex-A8 and Cortex-A9 in particular. I know that some architectures don't come with integer division, but what is the best way to do it other than convert to float, divide, convert to integer? Or is that indeed the best solution? Cheers! = ) The compiler normally includes a divide in its library, gcclib for example I have extracted them from gcc and use them directly: https://github.com/dwelch67/stm32vld/ then stm32f4d/adventure/gcclib going to float and back is probably not the best solution. you can try it and see how fast it is...This is a multiply but could as easily make

Questions arise when I type in these expressions to Python 3.3.0 -10 // 3 # -4 -10 % 3 # 2 10 // -3 # -4 10 % -3 # -2 -10 // -3 # 3 It appears as though it takes the approximate floating point (-3.33)? and rounds down either way in integer division but in the modulo operation it does something totally different. It seems like it returns the remainder +/-1 and only switches the sign depending on where the negative operand is. I am utterly confused, even after looking over other answers on this site! I hope someone can clearly explain this too me! The book says hint: recall this magic formula a

int main (void) { int fahrenheit; // fahrenheit stands for fahrenheit double c; // c stands for celsius printf("Enter your fahrenheit, we'll covnvert it into celsius! "); scanf("%f", &fahrenheit); c = 5/9 * (fahrenheit - 32); printf("Here is your %f in celsius!.\n"); return (0); } I've followed the code through break points and when it takes in my input the calculations are off, but the formula is correct. Some sort of logic error I can't put my finger on. Please help! The scanf call uses the wrong format string. You are reading an int so you need it to be: scanf("%d", &fahrenheit); The