division

问题 I'm trying to diagnose and fix a bug which boils down to X/Y yielding an unstable result when X and Y are small: In this case, both cx and patharea increase smoothly. Their ratio is a smooth asymptote at high numbers, but erratic for "small" numbers. The obvious first thought is that we're reaching the limit of floating point accuracy, but the actual numbers themselves are nowhere near it. ActionScript "Number" types are IEE 754 double-precision floats, so should have 15 decimal digits of

问题 I was reading the section on C portability in the book C Traps and Pitfalls by Andrew Koening.. On an integer divison q = a/b; r = a%b; If a is a negative number, apparently the reminder r can be a negative or positive number, while satisfying the property q * b + r == a Normally I would expect r to be negative if dividend a is negative. And that is what I see in a intel machine with gcc. I am just curious have you ever seen a machine that would return a positive reminder when the dividend is

问题 I am battling to understand why my division is not working, below is my current code, which simply takes in two single digits and attempts to divide them: STDIN equ 0 SYS_READ equ 0 STDOUT equ 1 SYS_WRITE equ 1 segment .data num1 dq 0 num2 dq 0 quot dq 0 rem dq 0 segment .text global _start _start: mov rax, SYS_READ mov rdi, STDIN mov rsi, num1 mov rdx, 2 syscall mov rax, SYS_READ mov rdi, STDIN mov rsi, num2 mov rdx, 2 syscall mov rax, [num1] sub rax, '0' mov rbx, [num2] sub rbx, '0' xor rdx

问题 When doing a division im getting a rounded answer? double div; div = 25/8; printf("%lf",div); When i do this prints out 3.0000 why dont i get 3.125 ? 回答1: Because you are doing an integer division, try with: div = 25.0/8; or div = (double)25/8; Typing 25.0 means a double literal. You could also use 25.f for a float literal. Both of these trigger floating point division. 回答2: Either typecast explicitly to double data type or change numerator to get desired value i.e. 25.o or use floating point

问题 I am making a calculator in assembly language to be executed on an x86 processor. Basically, my calculator asks the user to enter two numbers and then to indicate which operation (addition, subtraction, multiplication and division) want to do with them. My calculator adds, subtracts and multiplies correctly but is unable to divide . In making a division, I always get 1 as the result. Then I leave my application code complete: section .data ; Messages msg1 db 10,'-Calculator-',10,0 lmsg1 equ $

问题 I need to divide two large integers WITHOUT using Biginteger since the Numbers can't be stored inside a primitive type , since I need to do it char by char from the strings I am given,I have already created a class called BigNumber, with this class I can: Add multiply compare two strings with large integers inside Now I only need to implement the Dividing method but I can't get my head around how to do it with two strings instead of one String and an Int, here's what I got so far, it works if

问题 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. 回答1: 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

问题 This question already has answers here : Closed 7 years ago . Possible Duplicate: Floating point arithmetic not producing exact results in Java I was doing this simple division but I get a very strange output: double a = 60/(1.2-1.1); a => 600.0000000000008 When it should be 600. thanks in advance 回答1: In IEEE-754 binary double, we need to consider 1.1 and 1.2 in the binary representation: 1.2 = 0b1.001100110011001100110011001100110011001100110011001100110011... 1.1 = 0b1

问题 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" 回答1: if you really want to use loop to divide two numbers, you can write it like code below int c=0;

问题 Based on the answers/comments of this question i wrote a performance test with gcc 4.9.2 (MinGW64) to estimate which way of multiple integer division is faster, as following: #include <emmintrin.h> // SSE2 static unsigned short x[8] = {0, 55, 2, 62003, 786, 5555, 123, 32111}; // Dividend __attribute__((noinline)) static void test_div_x86(unsigned i){ for(; i; --i) x[0] /= i, x[1] /= i, x[2] /= i, x[3] /= i, x[4] /= i, x[5] /= i, x[6] /= i, x[7] /= i; } __attribute__((noinline)) static void