This question already has answers here:
Why does integer division in C# return an integer and not a float?
(8 answers)
Closed 6 years ago.
I want to calculate the average of two floating point numbers, but whatever the input, I am getting an integer returned.
What should I do to make this work?
public class Program
{
public static float Average(int a, int b)
{
return (a + b) / 2;
}
public static void Main(string[] args)
{
Console.WriteLine(Average(2, 1));
}
}
There're two problems with your code
Evident one - Integer division - e.g. 1 / 2 == 0 not 0.5 since result must be integer
Hidden one - Integer overflow - e.g. a + b can overflow int.MaxValue and you'll get negative result
The most accurate implementation is
public static float Average(int a, int b)
{
return 0.5f * a + 0.5f * b;
}
Tests:
Average(1, 2); // 1.5
Average(int.MaxValue, int.MaxValue); // some large positive value
The trick is to write the expression as 0.5 * a + 0.5 * b, which also obviates the potential for int overflow (acknowledge Dmitry Bychenko).
Currently your expression is evaluated in integer arithmetic, which means that any fractional part is discarded.
In setting one of the values in each term to a floating point literal, the entire expression is evaluated in floating point.
Finally, if you want the type of the expression to be a float, then use
0.5f * a + 0.5f * b
The f suffix is used to denote a float literal.
return (a + b) / 2F; tells the compiler to treat the number as a float, otherwise it will be treated as an int.
Use this:
public static float Average(int a, int b)
{
return (float)(a + b) / 2;
}
You can use:
(float)(a + b) / 2.0
This will return float
Sorry, if anyone has answered the same way (I did not read all answers)
Related
This question already has answers here:
Why I cannot the get percentage by using Int
(9 answers)
Closed 6 years ago.
When I enter, in the Windows Calculator utility, "15036/18218*100=" it returns 82.53375782193435
What I really want is 17.47 (100 - 82.53), but that's beside the point at the moment.
With this code:
// Example: thisQty == 3182; totalQty == 18218
private string GetPercentage(int thisQty, int totalQty)
{
int diff = totalQty - thisQty; // this equates to 15036
double prcntg = (diff/totalQty)*100; // this equates to 0.0 for some reason
return string.Format("{0}%", prcntg);
}
...I'm getting 0.0 for the prcntg value. Why? ISTM that this is the same operation that I'm doing by hand in the Calculator utility. Why doesn't it return 82.53375782193435?
The dividing of 2 ints will be an int even if the correct mathematical answer is with a fraction.
In order to have it keep the decimal part you must divide with a number of a type that holds the fraction part (like double or decimal):
Console.WriteLine(GetPercentage(3182, 18218));
private string GetPercentage(int thisQty, int totalQty)
{
int diff = totalQty - thisQty; // this equates to 15036
double prcntg = (diff / (double)totalQty) * 100;
return string.Format("{0}%", prcntg);
}
BTW - it doesn't matter if you cast to double the diff or the totalQty - for both it will do the / operation returning a double - which means keeping the fraction part
You are using an integer value, (which doesn't store factional part), so cast it to double, or use the parameter type as double (my recommendation). Your operation, 15036/18218 resolves to, 0.82 and in an integer value that is stored as 0... Where finally 0 * 100 is going to resolve to 0 anyways and that is where you get the result.
Try this instead,
private string GetPercentage(double thisQty, double totalQty)
{
double diff = totalQty - thisQty; // this equates to 15036
double prcntg = (diff/totalQty) * 100.0; // this equates to 0.0 for some reason
return string.Format("{0}%", prcntg);
}
This would have the fractional part too and you will get the result.
Based on Gilad Green's answer, here is what I ended up with, which gives the value I ultimately want, and also rounds the value to an integer:
private string GetPercentage(int thisQty, int totalQty)
{
int diff = totalQty - thisQty;
double prcntg = (diff / (double)totalQty) * 100;
prcntg = 100 - prcntg;
int roundedPercent = Convert.ToInt32(prcntg);
return string.Format("{0}%", roundedPercent);
}
I am attempting to implement the BesselK method from Boost (a C++ library).
The Boost method accepts two doubles and returns a double. (I have it implemented below as cyl_bessel_k .)
The equation I modeled this off of comes from Boosts documention:
http://www.boost.org/doc/libs/1_45_0/libs/math/doc/sf_and_dist/html/math_toolkit/special/bessel/mbessel.html
I have also been checking values against Wolfram:
http://www.wolframalpha.com/input/?i=BesselK%283%2C1%29
I am able to match output from the Boost method when passing a positive non-integer value for "v". However, when an integer is passed, my output is severely off. So,there is an obvious discontinuity issue. From reading up on this, it seems that this issue arises from passing a negative integer to the gamma function. Somehow reflection comes into play here with the Bessel_I method, but I'm nearing the end of my math skillset.
1.) What needs to happen to the bessel_i method with reflection to make this work?
2.) I'm currently doing a partial sum approach. Boost uses a continuous fraction approach. How can I modify this to account for convergence?
Any input is appreciated! Thank you!
static double cyl_bessel_k(double v, double x)
{
if (v > 0)
{
double iNegativeV = cyl_bessel_i(-v, x);
double iPositiveV = cyl_bessel_i(v, x);
double besselSecondKind = (Math.PI / 2) * ((iNegativeV - iPositiveV ) / (Math.Sin(Math.PI * v)));
return besselSecondKind;
}
else
{
//error handling
}
}
static double cyl_bessel_i(double v, double x)
{
if (x == 0)
{
return 0;
}
double summed = 0;
double a = Math.Pow((0.5d * x), v);
for (double k = 0; k < 10; k++) //how to account for convergence? 10 is arbitrary
{
double b = Math.Pow(0.25d * Math.Pow(x, 2), k);
double kFactorial = SpecialFunctions.Factorial((int)k); //comes from MathNet.Numerics (Nuget)
double gamma = SpecialFunctions.Gamma(v + k + 1); //comes from MathNet.Numerics
summed += b / (kFactorial * gamma);
}
return a * summed;
}
After lots of refactoring and trying things that didn't work, this is what I came up with. It's mostly Boost logic that has been adapted and translated into C#.
It's not perfect though (likely due to rounding, precision,etc). Any improvements are welcome! Max error is 0.0000001926% between true Bessel_K value from Wolfram and my adapted method. This is occurs when parameter 'v' is an integer. For my purposes, this was close enough.
Link to fiddle:
https://dotnetfiddle.net/QIYzK6
Hopefully it saves someone some headache.
This question already has answers here:
C# is rounding down divisions by itself
(10 answers)
C# double not working as expected [duplicate]
(1 answer)
Closed 7 years ago.
I have c# program that calculates percentage and returns int value, but it always returns 0.
I have been writing code for 16 constitutive hours so I appreciate if you find the mistakes within it.
I debugged my code and I found that the value is being passed correctly.
private int returnFlag(int carCapacity, int subscribers)
{
int percentage = (subscribers / carCapacity)*100;
return percentage;
}
What you're seeing is the result of operating on two integers, and losing the fractional portion.
This piece of code, when using the values 5 and 14, will truncate to 0:
(subscribers / carCapacity)
You need to cast one of the operands to a double or decimal:
private int returnFlag(int carCapacity, int subscribers)
{
decimal percentage = ((decimal)subscribers / carCapacity) * 100;
return (int)percentage;
}
The issue is that since you're performing math on int (read: integer) values, any fractions or remainders get thrown out. This can be seen by changing your code to
int percentage = (subscribers / carCapacity);
percentage *= 100;
Since (subscribers / carCapacity) results in less than one, the only possible number an int can hold is 0 - and 0 * 100 is 0.
You can fix this by converting to a more precise number, such as double, before performing operations:
private int returnFlag(int carCapacity, int subscribers)
{
double percentage = ((double)subscribers / (double)carCapacity) * 100.0;
return (int)percentage;
}
Integer types (int) don't work with fractions. Change the types you are working with in your division to decimal, double, single, or float.
let suppose dis.text = 2, prc.text = 100, I am using these codes.It Should be
net_prc.text = 98.But its giving me -100.Can anybody tell me why?,And how can i get correct
discounted percentage??
private void net_prcTabChanged(object sender, EventArgs e)
{
int d;
int di;
int i;
d = Convert.ToInt32(dis.Text);
i = Convert.ToInt32(prc.Text);
di = -((d / 100) * i) + i;
net_prc.Text = di.ToString();
}
Try (d / 100.0) to force it to use floating point arithmetic
di = -((d / 100) * i) + i;
All values in this statement are Integers. You are going to be computing arithmetic with decimal places, and you need to increase the precision of your variables to a double or a float. Instead, add a decimal place to one of the values in the equation. This will force all values into doubles.
This is a process called Arithmetic Promotion. It is where, at run time, the precision of every variable in an equation is increased to the size of the most precise variable.
Proper way to do it would be, changing the datatype of di to float
di = (d * 100) / i;
C# has an odd way of doing maths, because your numbers are cast as integers, you can only do integer math with them. you need to initially have them as float or as double so you can do float math or anything at all that requires a decimal place within the calculations.
Even dis.text = 1.5
private void net_prcTabChanged(object sender, EventArgs e)
{
double d;
double di;
double i;
d = Convert.ToDouble(dis.Text);
i = Convert.ToDouble(prc.Text);
di = -((d * 100.0) / i ) + i;
net_prc.Text = di.ToString();
}
Your division, d / 100, is a division of integers, and it returns an integer, probably 0 (zero). This is certainly the case with your example d = 2.
Addition: If you really want to do this with integers (rather than changing to decimal or double like many other answers recommend), consider changing the sub-expression
((d / 100) * i)
into
((d * i) / 100)
because it will give you a better precision to do the division as the last operation. With the numbers of your example, d=2 and i=100, the first sub-expression will give 0*100 or 0, while the changed sub-expression yields 200/100 which will be 2. However, you will not get rounding to nearest integer; instead you will get truncating (fractional part is discarded no matter if it's close to 1).
This question already has answers here:
Why returns C# Convert.ToDouble(5/100) 0.0 and not 0.05
(7 answers)
Closed 9 years ago.
I must be doing something dumb:
float ans = (i/3);
So why when i = 7 is ans coming out at 2.0?
i is an int
It's because the / operator is performing an integer division if both operands are integers. You could do this:
float ans = (i / 3.0f);
You need to make one of the operands a float, otherwise the calculation is done with integers first (which always results in an integer), before converting the result to a float.
float ans = ((float) i) / 3;
It's doing integer division because i is an int and 3 is an int. Try this:
float ans = ((float)i/3.0f);
use float ans = (i / 3.0) or float ans = (i / 3f) or float ans = ((float)i / 3). / does an integer division if both sides are of type integer.
Very simple: in C#, int / int = int.
What you're looking for is:
float ans = ((float)i/3);
Otherwise you're taking two integers and dividing them to find the number of whole times the divisor goes in to the dividend. (As mentioned, an int/int=int regardless of the destination type. And, to the compiler, "3" is another integer (unless you specify it as 3.0f))
I am assuming that you have this in a loop of some sort. You could specify your i variable as a float instead.
for (float i = 0; i < 10; i++)
{
float ans = (i/3);
// statements
}
Just another solution.