This question already has answers here:
Random Number Between 2 Double Numbers
(13 answers)
Closed 9 years ago.
Hi how can i generate a range of "double" numbers? For example how can i generate numbers between 2.50 and 7.20
There is a method in Random() class for "int" numbers Next(Int32, Int32) is there something similar for double?
You can write an extension method to Random for Random.NextDouble(double MinValue,double MaxValue) so that you can use it everywhere:
public static class RandomExtensions
{
public static double NextDouble(this Random RandGenerator, double MinValue, double MaxValue)
{
return RandGenerator.NextDouble() * (MaxValue - MinValue) + MinValue;
}
}
Get a value from 0 to 1, then multiply it by 7.20 - 2.50 and add 2.50.
double result = (random.NextDouble() * (7.2 - 2.5)) + 2.5;
Yes, it's called Random.NextDouble(). This returns a double between 0 and 1.
var value = lower + (random.NextDouble() * (upper - lower))
will return what you need.
Related
This question already has answers here:
Why do these division equations result in zero?
(10 answers)
Why does integer division in C# return an integer and not a float?
(8 answers)
Closed 4 years ago.
I'm calculating the percentage to add in a datagrid, but when I report the data it always returns me 0, what am I doing wrong?
if I have the following case: the variable quantidadeInstalada has the value of 10 and the goal has 20 the concluido would have to return me 50 but it returns me 0
private void AdicionarPessoa()
{
string Valida = ValidaPessoa();
if (Valida.Equals(""))
{
double concluido=0, falta=0;
int quantidadeInstalada = Convert.ToInt32(ttbQuantidade.Text);
int meta = Convert.ToInt32(ttbMetaPessoa.Text);
concluido = (quantidadeInstalada/meta)*100;
falta = 100-concluido;
MessageBox.Show(concluido.Text);
}
else
MessageBox.Show(Valida);
}
It's probably due to the precision of the int. Use a decimal or double instead.
When we use an integer, we lose precision.
Console.WriteLine(100 / 17); // 5
Console.WriteLine(100 / 17m); // 5.8823529411764705882352941176
Console.WriteLine(100 / 17d); // 5.88235294117647
Console.WriteLine(100 / 17f); // 5.882353
Since integers always round down, 0.99 as an integer is 0.
Note that for precision, the types of the inputs matters.
double output = input1 * input2;
For example:
double outputA = 9 / 10;
Console.WriteLine(outputA); // 0
double outputB = 9 / 10d;
Console.WriteLine(outputB); // 0.9
double outputC = 9d / 10;
Console.WriteLine(outputC); // 0.9
Here is a Fiddle.
It's because you're doing integer division. You can fix it by casting one of the variables to a double like this:
concluido = ((double)quantidadeInstalada/meta)*100;
When you're dividing int by int (as in "quantidadeInstalada/meta"), you'll get an int. Either use a fractional type (e.g. decimal, double) from the very beginning, as Shaun suggested, or (if the integral types have to stay), cast the values to a fractional type in the division expression (as itsme86 shown).
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);
}
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.
This question already has answers here:
Division returns zero
(8 answers)
Closed 8 years ago.
I'm trying to get Celsius to take the Fahrenheit temperature as an argument, then return the temperature in Celsius
class Exercise1
{
static void Main(string[] args)
{
double fahrenheit;
Console.WriteLine("Enter farenheit: ");
fahrenheit = double.Parse(Console.ReadLine());
double cels;
Exercise1 me = new Exercise1();
cels = me.Celsius(fahrenheit);
Console.WriteLine(cels);
}
public double Celsius(double fahrenheit)
{
double celsius;
celsius = 5 / 9 * (fahrenheit - 32);
return celsius;
}
Your problem is integer division in your Celsius function. This line
celsius = 5 / 9 * (fahrenheit - 32);
Will always be 0 because 5 / 9 will be divided as integers which will always give you 0. To force floating-point division, one if your integers must be a double. So do this:
celsius = 5.0 / 9 * (fahrenheit - 32);
Note the 5.0 will force floating-point division.
5 / 9 would be seen as integers and thus make the entire calculation integer maths in this case.
Meaning you are essentially getting 0 * (fahrenheit - 32).
Cast either or both 5 and 9 to a double to force floating point maths.
This question already has answers here:
How might I convert a double to the nearest integer value?
(8 answers)
Closed 9 years ago.
I'am producing decimal such as:
0.8235294117647058823529411765
0.1764705882352941176470588235
I'd like to multiply them by 10 then round them. If the second number after the dot is less than 5 then make it 0. Otherwise, make it 1. For the above examples, that would be:
8
2
The result should be put in a int.
I thought it is simple.
Math.Round(10 * your_decimal);
Reference: http://msdn.microsoft.com/en-us/library/3s2d3xkk.aspx
You can use Math.Round() if you want to keep it as a decimal or round to a certain precision.
decimal dec = 0.8235294117647058823529411765m;
decimal rounded = Math.Round(dec * 10); // 8m
decimal roundedToOne = Math.Round(dec * 10, 1); // 8.2m
FYI, an explicit conversion is defined for decimal to int so you can round down by casting to an int
int a = (int)(dec * 10); // 8
This could be combined with a condition to round the number up and down
decimal num = dec * 10;
int a = (int)num + (num % 0 < .5m ? 0 : 1); // 8