integer-division

问题 As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance. Closed 6 years ago . i have the program written however I have two problems and need assistance to correct. the problems are 1) i do not want the counter i

问题 This question already has answers here : C program to convert Fahrenheit to Celsius always prints zero (8 answers) Closed 3 years ago . Im not getting correct answer with the following code below. Can anyone debug this code? When I input radius = 5, the answer I get is 500.000000 whereas the original answer should be 523.80952. Can anyone please explain what's wrong here? Sphere volume formula =4/3(π x r^3) #include <stdio.h> int main() { float radius = 0; float volume; float pie = 0; printf(

问题 Trying to do something very simple and calculate the signup_flow rate by mailing. I have a column called signup_flow - each row has an integer value of either 1 or 0. Trying to calculate the rate of signup_flow per mailing. I am using the following query. SUM(signup_flow) and COUNT(signup_flow) return the correct values per mailing. When I try to do a simple calculated field (sum/count*100) it returns 0. This is my first time working with Postgres, does it not support calculations? SELECT

问题 Is there a way to compute the result of ((UINT_MAX+1)/x)*x-1 in C without resorting to unsigned long (where x is unsigned int )? (respective "without resorting to unsigned long long " depending on architecture.) 回答1: It is rather simple arithmetic: ((UINT_MAX + 1) / x) * x - 1 = ((UINT_MAX - x + x + 1) / x) * x - 1 = ((UINT_MAX - x + 1) / x + 1) * x - 1 = (UINT_MAX - x + 1) / x) * x + (x - 1) 回答2: With integer divisions we have the following equivalence (y/x)*x == y - y%x So we have ((UINT

问题 I have a 64-bit long that I want to round down to the nearest 10,000, so I am doing a simple: long myLong = 123456789 long rounded = (myLong / 10000) * 10000; //rounded = 123450000 This appears to do what I expect, but as I'm not 100% on the internals of how integer types get divided, I am just slightly concerned that there may be situations where this doesn't work as expected. Will this still work at very large numbers / edge cases? 回答1: Yes, it will work, so long as no result, intermediate

问题 Unlike in C, in Java is the result of x/y and x%y well-defined even for negative operands. Surprisingly, it's defined by rounding towards zero, and not by rounding down (i.e., towards negative infinity). Does anybody have taken any advantage of this definition? In most cases I just don't care, but sometimes I had to work around this, e.g., when computing an index using modulo array.length . This is no rant, I'm really interested if there are uses for this definition. 回答1: It's easier to

问题 I am trying to divide a 64 bits integral type to a 32 bits one, and I am using RVDS 4.1 as a tool-chain. Cortex-M0 does not have hardware divisor, so can I do the operation below? If so How? unsigned long int b = 2590202; unsigned long long int a = 953502716552001ULL; unsigned long long int result; result = a/b; 回答1: The compiler will compile the division operation as a call to a library subroutine that performs a software 64-bit division algorithm. You might have to also tell the compiler to

问题 This question already has answers here : Is floating point math broken? (31 answers) Closed 6 years ago . I am dividing a number( may be in amount format of XX.XX) by a integer. But when i do 6.6 / 6 or 3.3/3 , it gives 1.099999... , instead of 1.1. What i am doing is, I am dividing the amount equally among the number of people and if any remainder amount exists, I am adding that extra amount to one person among them. May be something of this kind, 104/6 will be divided as 17.85, 17.83, 17.83

问题 Can a scaled 64bit/32bit division performed by the hardware 128bit/64bit division instruction, such as: ; Entry arguments: Dividend in EAX, Divisor in EBX shl rax, 32 ;Scale up the Dividend by 2^32 xor rdx,rdx and rbx, 0xFFFFFFFF ;Clear any garbage that might have been in the upper half of RBX div rbx ; RAX = RDX:RAX / RBX ...be faster in some special cases than the scaled 64bit/32bit division performed by the hardware 64bit/32bit division instruction, such as: ; Entry arguments: Dividend in

问题 I am using this struct to represent 128bit integers: typedef struct { uint64_t low, high; } uint128; (Unless you can point me to a fast 128bit integer library I can not change that) Now I want to print such a value in base 10, using printf . I probably need division by 10 to do that, but no division is implemented yet. How can I do this? The solution does not have to be super efficient, as long as it works. EDIT: I like all solutions you came up with. You are awesome. 回答1: void printu128