integer-division

The modulo in Python is confusing. In Python, % operator is calculating the remainder: >>> 9 % 5 4 However: >>> -9 % 5 1 Why is the result 1 ? and not -4 ? Because in python, the sign matches the denominator. >>> 9 % -5 -1 >>> -9 % 5 1 For an explanation of why it was implemented this way, read the blog post by Guido . -10 % 5 is 0, ie, -10 is evenly divided by 5. You ask why -9 % 5 is not -4, and the answer is that both 1 and -4 can be correct answers, it depends on what -9 divided by 5 is. Of course -9 divided by 5 is 1.8, but this is integer division, in Python 3 represented by //, so I'll

How to compute the integer division, 2 64 /n? Assuming: unsigned long is 64-bit We use a 64-bit CPU 1 < n < 2 64 If we do 18446744073709551616ul / n , we get warning: integer constant is too large for its type at compile time. This is because we cannot express 2 64 in a 64-bit CPU. Another way is the following: #define IS_POWER_OF_TWO(x) ((x & (x - 1)) == 0) unsigned long q = 18446744073709551615ul / n; if (IS_POWER_OF_TWO(n)) return q + 1; else return q; Is there any faster (CPU cycle) or cleaner (coding) implementation? phuclv's idea of using -n is clever, but can be made much simpler. As

I have the following variables: int first = 0; int end = 0; Declare in the public class. Within a method: double diff = end / first; double finaldiff = 1 - diff; The end variable on System.out.println is 527 , the first is 480 . Why is the answer for diff coming out as 1 ? It should be 1.097916667 , I thought using a double would enable me to calculate into decimals? Dividing two int s will get you an int , which is then implicitly converted to double . Cast one to a double before the divison: double diff = (double)end / first; 来源： https://stackoverflow.com/questions/10376322/java-division

i have an Integer value: Integer value = 56472201; Where the value could be positive or negative. When I divide the value by 1000000, I want this result in the form 56.472201 but instead it gives me just the quotient. How am I able to get both the quotient and remainder values? cast it to float and then do it: int i = 56472201; float j = ((float) i)/1000000.0 Edit: Due to precision(needed in your case), use double. Also as pointed by Konrad Rudolph, no need for explicit casting: double j = i / 1000000.0; If you divide an int by a double you will be left with a double result as illustrated by

I'm wondering if there really is no 128-bit division intrinsic function in Visual C++? There is a 64x64=128 bit multiplication intrinsic function called _umul128() , which nicely matches the MUL x64 assembler instruction. Naturally, I assumed there would be a 128/64=64 bit division intrinsic as well (modelling the DIV instruction), but to my amazement neither Visual C++ nor Intel C++ seem to have it, at least it's not listed in intrin.h. Can someone confirm that? I tried grep'ing for the function names in the compiler executable files, but couldn't find _umul128 in the first place, so I guess

1/2 gives 0 as it should. However, -1/2 gives -1 , but I want it to round towards 0 (i.e. I want -1/2 to be 0), regardless of whether it's positive or negative. What is the best way to do that? Do floating point division then convert to an int. No extra modules needed. >>> int(float(-1)/2) 0 >>> int(float(-3)/2) -1 >>> int(float(1)/2) 0 >>> int(float(3)/2) 1 dawg Python's default division of integers is return the floor (towards negative infinity) with no ability to change that. You can read the BDFL's reason why. To do 'round up' division, you would use: >>> a=1 >>> b=2 >>> (a+(-a%b))//b 1 >>