integer-division

I have to calculate some simple mathematical expression, but when I do it in one row, the result always will be zero. But the correct result is obviously not zero. And its interesting but when I separate the parts of expression, I get the correct answer. Later I will divide with this result, so it should not be 0. The expression is like this: (X-X1)/(X2-X1) In this case the delta : 0 double delta = (x - x1) / (x2 - x1); But this way the delta will be correct: double top = x - x1; double bottom = x2 - x1; double delta = top/bottom; Do you have any idea, how could this happen? Mats Petersson

I have a function f(x)=(1^1)*(2^2)*(3^3)*.....(x^x) i have to calculate (f(x)/(f(x-r)*f(r)))modulo c i can calculate f(x) and (f(x-r)*f(r)). assume f(x) is a and f(x-r)*f(r) is b. c is some number that is very larger. `` so i how can calculate (a/b)%c your f(x) is just ᴨ (PI cumulative multiplication) squared it is hard to write it in here so i will deifine g(x0,x1) instead g(x0,x1)=x0*(x0+1)*(x0+2)*...*x1 so: f(x)=g(1,x)^2 computing h(x,r,c)=f(x)/(f(x-r)*f(r))%c when you rewrite it to g() you get: h(x,r,c)=((g(1,x)/(g(1,x-r)*g(1,r)))^2)%c now simplify (and lower magnitude) as much as you can

I was just reading: Efficiently dividing unsigned value by a power of two, rounding up and I was wondering what was the fastest way to do this in CUDA. Of course by "fast" I mean in terms of throughput (that question also addressed the case of subsequent calls depending on each other). For the lg() function mentioned in that question (base-2 logarithm of divisor), suppose we have: template <typename T> __device__ int find_first_set(T x); template <> __device__ int find_first_set<uint32_t>(uint32_t x) { return __ffs(x); } template <> __device__ int find_first_set<uint64_t>(uint64_t x) { return

In Java, if we divide byte s, short s or int s, we always get an int . If one of the operands is long , we'll get long . My question is - why does byte or short division not result in byte or short ? Why always int ? Apparently I'm not looking for the "because JLS says so" answer, I am asking about the technical rationale for this design decision in the Java language. Consider this code sample: byte byteA = 127; byte byteB = -128; short shortA = 32767; short shortB = -32768; int intA = 2147483647; int intB = - -2147483648; long longA = 9223372036854775807L; long longB = -9223372036854775808L;

This question already has an answer here: Why does integer division in C# return an integer and not a float? 7 answers If I divide 150 by 100, I should get 1.5. But I am getting 1.0 when I divided like I did below: double result = 150 / 100; Can anyone tell me how to get 1.5? try: double result = (double)150/100; When you are performing the division as before: double result = 150/100; The devision is first done as an Int and then it gets cast as a double hence you get 1.0, you need to have a double in the equation for it to divide as a double. Cast one of the ints to a floating point type. You