pow

After upgrading to Java 1.8.0_20 our test system reported errors, but the code was not changed. I found out, that Math.pow() called with exactly the same input parameters yields different results upon repreated calls. In Java 1.8.0_11 it behaves as expected and returns always the same value, but with Java 1.8.0_20 and above it sometimes returns slightly different values. This is similar to the question Math.pow yields different result depending on java version , but different because the results of pow() differ within one VM. The following JUint test fails when run under Java 1.8.0_20 and

Let's say I have the following code snippet: int i; double value; for(i = 0; i < CONSTANT; i++) { value = (double)pow(2, i); } Trying to compile this code yields an "undefined reference to `pow'" error. Including or excluding math.h makes no difference, since it ends up being included anyway. Raising 2.0 to a hardcoded power works okay, but everything fails if I substitute the exponent by an expression that contains i . What am I doing wrong? Thanks. It's a very interesting behavior, and a good learning example. To solve your problem, add -lm to your gcc command line (provided you're using gcc

I'm programming for a while now(beginner), and recursive functions are a somewhat abstract concept for me. I would not say I'm stuck, program works fine, I'm just wondering if the function itself could be written without the pow function in the code (but still doing exactly what the problem suggests) Problem: http://prntscr.com/30hxg9 My solution: #include<stdio.h> #include<math.h> int power(int, int); int main(void) { int x, n; printf("Enter a number and power you wish to raise it to: "); scanf_s("%d %d", &x, &n); printf("Result: %d\n", power(n, x)); return 0; } int power(int x, int n) { if

I read that the pow(double, double) function is defined in "math.h" but I can't find its declaration. Does anybody know where this function declared? And where is it implemented in C? Reference: http://publications.gbdirect.co.uk/c_book/chapter9/maths_functions.html Quite often, an include file such as <math.h> will include other header files that actually declare the functions you would expect to see in <math.h> . The idea is that the program gets what it expects when it includes <math.h> , even if the actual function definitions are in some other header file. Finding the implementation of a

While doing my homework I noticed something really strange that I just can't figure out why. int x = 5; cout << pow(x, 2); The result is 25. That's fine. But if I write the same program like this: int x = 5; int y = pow(x, 2); cout << y; The result is 24! When x is 2, 3, 4, 6, 7, 8 no problem, but with 5, 10, 11, 13 etc. result is 1 lower than it should be. Same thing with if(). for (int x = 1; x <= 20 ; x++) { if (x * x == pow(x, 2)) cout << x << endl; } It prints out numbers 1, 2, 3, 4, 6, 8, 12, 16. std::pow() returns a floating point number. If the result is for instance 24.99999999 and

This question already has an answer here: Why pow(10,5) = 9,999 in C++ 8 answers I wrote this code and first time of loop result is 99. Why is result 99, when it should be 100? #include <iostream> #include<math.h> using namespace std; int main () { int skt = 0; int sk[3]; int nsk = 3; sk[0]=1; sk[1]=2; sk[2]=8; for (int i=0; i<nsk; i++) { skt = skt + (sk[i]*pow(10.0,nsk-i-1)); cout <<" "<<skt<<endl; } } the result of this code 99 119 127 ,but if I use library cmath it is correct answer #include <iostream> #include<cmath> using namespace std; int main () { int skt = 0; int sk[3]; int nsk = 3;

Our professor said that you can't calculate a b if a<0 using pow() because pow() uses natural logarithms to calculate it (a b =e b ln a ) and since it's undefined for negative numbers it can't be calculated. I tried it and it works as long as b is an integer. I have searched through math.h and further files, but was unable to find how the function is defined and what it uses to calculate. I also tried searching the internet, but without any success. There are similar questions on Stack Overflow right here and here (for C#). (the last one is good, but I was unable to find sourcecode.) So the

So, If i would like to calculate the value of 6^8 mod 5 using the pow function, what should I put in a line?? In the assumption that You don't need to import it first I know that pow is used like pow (x, y) = pow (6, 8) = 6^8 and My guess is mod.pow(6,8) Thank you! It's simple: pow takes an optional 3rd argument for the modulus. From the docs : pow(x, y[, z]) Return x to the power y; if z is present, return x to the power y, modulo z (computed more efficiently than pow(x, y) % z) . The two-argument form pow(x, y) is equivalent to using the power operator: x**y . So you want: pow(6, 8, 5) Not