pow

C: i got different results with pow(10,2) and pow(10,j), j=2;

由 橙三吉。 提交于 2019-11-26 14:51:58
问题 this one prints 100: int j=2; int i= pow(10,2); printf("%d\n", i); and this one prints 99: int j=2; int i= pow(10,j); printf("%d\n", i); why? 回答1: What's going on is that you have a C implementation whose standard library has a very low quality implementation of pow which is returning inexact results even when the exact result is representable in the type ( double ). The call to pow(10,2) seems to producing the value just below 100.0 , which, when rounded to an integer, yields 99. The reason

What is happening here in pow function?

由 廉价感情. 提交于 2019-11-26 14:46:01
问题 I have seen various answer here that depicts Strange behavior of pow function in C. But I Have something different to ask here. In the below code I have initialized int x = pow(10,2) and int y = pow(10,n) (int n = 2) . In first case it when I print the result it shows 100 and in the other case it comes out to be 99 . I know that pow returns double and it gets truncated on storing in int , but I want to ask why the output comes to be different. CODE1 #include<stdio.h> #include<math.h> int main

pow() seems to be out by one here

由 ∥☆過路亽．° 提交于 2019-11-26 14:11:08
问题 What's going on here: #include <stdio.h> #include <math.h> int main(void) { printf("17^12 = %lf\n", pow(17, 12)); printf("17^13 = %lf\n", pow(17, 13)); printf("17^14 = %lf\n", pow(17, 14)); } I get this output: 17^12 = 582622237229761.000000 17^13 = 9904578032905936.000000 17^14 = 168377826559400928.000000 13 and 14 do not match with wolfram alpa cf: 12: 582622237229761.000000 582622237229761 13: 9904578032905936.000000 9904578032905937 14: 168377826559400928.000000 168377826559400929

return value of pow() gets rounded down if assigned to an integer

由 穿精又带淫゛_ 提交于 2019-11-26 12:33:14
I am using the pow function in C and storing the return value in an integer type. see the code snippet below: for (i = 0; i < 5; i++){ val = (int)pow(5, i); printf("%d, ", val); } here i , and val are integers and the output is 1, 5, 24, 124, 624 . I believe this is because a float 25 is treated as 24.99999... which gets rounded down to 24 on assignment to an integer. How can I by pass this if I still need to store the return value in an int ? Marcelo Cantos Add 0.5 before casting to int . If your system supports it, you can call the C99 round() function, but I prefer to avoid it for

BigInteger.pow(BigInteger)?

由 こ雲淡風輕ζ 提交于 2019-11-26 09:03:46
问题 I\'m playing with numbers in Java, and want to see how big a number I can make. It is my understanding that BigInteger can hold a number of infinite size, so long as my computer has enough Memory to hold such a number, correct? My problem is that BigInteger.pow accepts only an int, not another BigInteger, which means I can only use a number up to 2,147,483,647 as the exponent. Is it possible to use the BigInteger class as such? BigInteger.pow(BigInteger) Thanks. 回答1: You can write your own,

How Math.Pow (and so on) actually works

由 佐手、 提交于 2019-11-26 08:34:24
问题 So I was googling for a long time and i found almost nothing. I found some info about possible implementation of Math.Pow from this url, but they are inaccurate, for example this code public static double PowerA(double a, double b) { int tmp = (int)(BitConverter.DoubleToInt64Bits(a) >> 32); int tmp2 = (int)(b * (tmp - 1072632447) + 1072632447); return BitConverter.Int64BitsToDouble(((long)tmp2) << 32); } static void Main(string[] args) { double x = 12.53, y = 16.45; Console.WriteLine(Math.Pow

Why isn&#39;t `int pow(int base, int exponent)` in the standard C++ libraries?

由 微笑、不失礼 提交于 2019-11-26 07:25:48
问题 I feel like I must just be unable to find it. Is there any reason that the C++ pow function does not implement the \"power\" function for anything except float s and double s? I know the implementation is trivial, I just feel like I\'m doing work that should be in a standard library. A robust power function (i.e. handles overflow in some consistent, explicit way) is not fun to write. 回答1: As a matter of fact, it does. Since C++11 there is a templated implementation of pow(int, int) --- and

How to: pow(real, real) in x86

由 懵懂的女人 提交于 2019-11-26 06:31:33
问题 I\'m looking for the implementation of pow(real, real) in x86 Assembly. Also I\'d like to understand how the algorithm works. 回答1: Just compute it as 2^(y*log2(x)) . There is a x86 instruction FYL2X to compute y*log2(x) and a x86 instruction F2XM1 to do exponentiation. F2XM1 requires an argument in [-1,1] range, so you'd have to add some code in between to extract the integer part and the remainder, exponentiate the remainder, use FSCALE to scale the result by an appropriate power of 2. 回答2:

How to do a fractional power on BigDecimal in Java?

由 谁都会走 提交于 2019-11-26 04:27:33
问题 In my little project I need to do something like Math.pow(7777.66, 5555.44) only with VERY big numbers. I came across a few solutions: Use double - but the numbers are too big Use BigDecimal.pow but no support for fractional Use the X^(A+B)=X^A*X^B formula (B is the remainder of the second num), but again no support for big X or big A because I still convert to double Use some kind of Taylor series algorithm or something like that - I\'m not very good at math so this one is my last option if

How is Math.Pow() implemented in .NET Framework?

由 青春壹個敷衍的年華 提交于 2019-11-26 00:35:04
问题 I was looking for an efficient approach for calculating a b (say a = 2 and b = 50 ). To start things up, I decided to take a look at the implementation of Math.Pow() function. But in .NET Reflector, all I found was this: [MethodImpl(MethodImplOptions.InternalCall), SecuritySafeCritical] public static extern double Pow(double x, double y); What are some of the resources wherein I can see as what\'s going on inside when I call Math.Pow() function? 回答1: MethodImplOptions.InternalCall That means
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