integer-division

问题 This question I have tried to solve it but couldn't get any way. Any pointers would be appreciated. Regular subtraction way of doing division is not the intention here, ingenious way of using shifting operator to get this done is the intention. 回答1: Here's a solution heavily inspired by Hacker's Delight that really uses only bit shifts: def divu9(n): q = n - (n >> 3) q = q + (q >> 6) q = q + (q>>12) + (q>>24); q = q >> 3 r = n - (((q << 2) << 1) + q) return q + ((r + 7) >> 4) #return q + (r >

问题 Apologies for the confusing title. I'm not sure how to better describe what I'm trying to accomplish. I'm essentially trying to do the reverse of getting the high half of a 64-bit multiplication in C for platforms where int64_t divHi64(int64_t dividend, int64_t divisor) { return ((__int128)dividend << 64) / (__int128)divisor; } isn't possible due to lacking support for __int128 . 回答1: This can be done without a multi-word division Suppose we want to do ⌊2 64 × x ⁄ y ⌋ then we can transform

问题 When gcc sees multiplication or division of integer types that isn't supported in hardware, it generates call to special library function. http://gcc.gnu.org/onlinedocs/gccint/Integer-library-routines.html#Integer-library-routines According link above, long __divdi3 (long a, long b) used for division of long. However, here http://gcc.gnu.org/onlinedocs/gcc-3.3/gccint/Library-Calls.html divdi explained as "call for division of one signed double-word". When first source has cleary mapping of di

问题 If I divide by 0, I get either a ZeroDivisionError, Infinity or NaN depending on what is divided. ruby-1.9.2-p180 :018 > 0.0 / 0 => NaN ruby-1.9.2-p180 :020 > 3.0 / 0 => Infinity ruby-1.9.2-p180 :021 > 3 / 0 ZeroDivisionError: divided by 0 I understand that 0.0 / 0 is not an Infinity (in math terms), while 3.0 / 0 is but why then isn't 3 / 0 an Infinity? Why dividing an integer throws an exception but dividing a float doesn't? 回答1: In Ruby, not all numbers are created equal (pun intended).

问题 How can I tell the MSVC compiler to use the 64bit/32bit division operation to compute the result of the following function for the x86-64 target: #include <stdint.h> uint32_t ScaledDiv(uint32_t a, uint32_t b) { if (a > b) return ((uint64_t)b<<32) / a; //Yes, this must be casted because the result of b<<32 is undefined else return uint32_t(-1); } I would like the code, when the if statement is true, to compile to use the 64bit/32bit division operation, e.g. something like this: ; Assume

问题 I really don't think this is a precision problem, the answer should be about 0.226. Here's the exact code: val = I(i,j) bucketSize pos = val / bucketSize I is just a matrix I'm taking values from. Here is the output from MATLAB: val = 29 bucketSize = 128 pos = 0 What am I missing? 回答1: My guess would be that your matrix I is pixel data loaded from an image file, which will have values that are typically unsigned 8-bit integers. As already mentioned, converting both integer values to a double

问题 I have a class for fixed-point arithmetic, of which this is the salient portion: template <typename I, I S> struct fixed { I value; fixed(I i) : value(i * S) {} template <typename J, J T> fixed(const fixed<J, T> &fx) { if (S % T == 0) value = fx.value * (S / T); else if (T % S == 0) value = fx.value / (T / S); else value = S * fx.value / T; } static_assert(S >= 1, "Fixed-point scales must be at least 1."); }; On GCC 4.4.5, the following line of code: fixed<int, 8> f = fixed<int, 2>(1);

问题 This question already has an answer here : Integer division compared to floored quotient: why this surprising result? (1 answer) Closed 3 years ago . In Python 2.7 and 3.x , why does integer division give me a non-correct number when dividing by a number 0 < x < 1 ? Negative numbers -1 < x < 0 even work correctly: >>> 1//.1 9.0 >>> 1//-.1 -10.0 I understand that integer division with a negative (or positive) number rounds toward negative infinity, however I would have thought 1//.1 should

问题 I know that similar questions has been asked in the past, but I have implemented after a long process the algorithm to find the quotient correctly using the division by repeated subtraction method. But I am not able to find out the remainder from this approach. Is there any quick and easy way for finding out remainder in 64bit/64bit division on 32bit processor. To be more precise I am trying to implement ulldiv_t __aeabi_uldivmod( unsigned long long n, unsigned long long d) Referenced in this

问题 I know that similar questions has been asked in the past, but I have implemented after a long process the algorithm to find the quotient correctly using the division by repeated subtraction method. But I am not able to find out the remainder from this approach. Is there any quick and easy way for finding out remainder in 64bit/64bit division on 32bit processor. To be more precise I am trying to implement ulldiv_t __aeabi_uldivmod( unsigned long long n, unsigned long long d) Referenced in this