I can't get this to divide into a decimal. It is rounding to value 0.
private void button24_Click(object sender, EventArgs e)
{
double x = 0;
x = 1 / 2;
ans.Text = x.ToString();
}
When I debug, x is zero before it is sent to the textbox 'ans.'
I tried..and string variable is still zero..
double x = 1/5;
string displayX = x.ToString("0.0000");
It's integer division and those are the expected outputs.
double x = 1.0 / 5; // this will not perform integer division
double x = 1/5; // this does (1/5 = 0).
double x = 1D / 5; // this will not because 1 is treated as a double
You can do one of the follow:
double x = 1;
double y = 1.5;
double ans = x / y;
Replace double x = 1/5 with double x = 1.0/5 and that should fix it. Because both numbers you're dividing are integers, it still processes it as an integer, rather than as a double. When you think through logically, it makes some sense - it does the division in whatever form those numbers are and then saves it to the variable; the variable type is inconsequential to the actual equation.
(I realize there are other answers already, but hopefully this will help you see why the issue exists.)
Related
Given two numbers X and Y, how can I compute how many X's can fit into Y?
Good old division to the rescue!
float x = 16;
float y = 12345;
float howMany = ((float)y)/x; //> 771.5625
int floor = (int)howMany; //> 771
int ceil = (int)(howMany+0.5f);//> 772
Alliteratively, since you leave us guessing, you may want:
int lenX = "16".Length;
int lenY = "12345".Length;
float howMany = (float)lenY/lenX; //> 2.5
There are two ways your question could be interpreted.
(1) How many fractional X's can fit into Y?
In mathematics, the answer to this question is Y / X. In a programming language, if X and Y are integer values, you'll have to take care to cast them to floating point values before performing the division.
When X and Y are integers
int X = 5;
int Y = 17;
double N = (double)Y / (double)X;
// N is 3.4
When X and Y are real numbers
double X = 2.5;
double Y = 11.5;
double N = Y / X;
// N is 4.6
(2) How many whole X's can fit into Y?
In mathematics, the answer to this question is ⌊Y / X⌋, the floor of Y divided by X. In a programming language, if X and Y are integer values, the / operator applied to them will typically perform integer division. Integer division discards the remainder of division, so you don't have to call any floor function.
When X and Y are integers
int X = 5;
int Y = 17;
int N = Y / X;
// N is 3
When X and Y are real numbers
double X = 2.5;
double Y = 11.5;
int N = (int)Math.Floor(Y / X);
// N is 4
in my ASP.NET project i did a survey page that uses Application to save the votes. I have a problem with the making of the percentages amount. I've tried many things. here is the problematic part of my code:
double x = (count / sum) ;
double f = (count1 / sum) ;
double g = (count2 / sum) ;
double h = (count3 / sum) ;
if (sum > 0)
{
a = (int)x * 100;
b = (int)f * 100;
c = (int)g * 100;
d = (int)h * 100;
}
I used breakpoints and figured out that the problem was in the double variables: the (count/sum) equals 0 anyway.
I'm assuming count and sum are integer types.
The result of division of 2 integers is a truncated integer.
You need to cast one side of the division to a double, then the result will be double
So
((double)count)/sum
What are the datatypes of count, count[1-3] and sum? If they are integral types, then integer division is performed. This
int x = 100 ;
int y = 300 ;
double z = x / y ;
yields the value 0.0 for z.
Try something like
double h = (double) ( count3 / sum ) ;
You might also want to move your test for sum > 0 up: as coded, if sum is zero, you'll throw a DivideByZeroException before you get to your test, thus rendering your test moot.
Your count and sum variables are probably integers. Cast one of them to double:
double x = count / (double)sum;
UPDATE:
Actually, if you want the percentage as an integer, you can skip the doubles altogether:
int a = 100 * count / sum;
Trying to create a Mandelbrot set, I have been trying to using 8 and 15 digit floating point variables in my program, and have run into an issue: the double approximates to 0. I tried unit testing and wrote this code to isolate the issue. Using the variable viewer, the h and w values both were on 0.0, as opposed to 0.00185185185185 and 0.0015625, yet when I just write double h = 0.0015625, it works.
Many thanks for assistance.
int apparentwidth = 3;
int apparentheight = 2;
int height = 1080;
int width = 1920;
double w = (apparentwidth / width);
double h = (apparentheight / height);
Console.WriteLine(w);
Console.WriteLine(h);
You're dividing two int variables, and the result is an int. You're storing the result in a double, but the scale (portion after the decimal) has already been lost.
Use double throughout:
double apparentwidth = 3;
double apparentheight = 2;
double height = 1080;
double width = 1920;
double w = (apparentwidth / width);
double h = (apparentheight / height);
Or cast one of the variables to a double when dividing:
double w = ((double)apparentwidth / width);
double h = ((double)apparentheight / height);
You are doing integer division on accident.
double w = (double)apparentwidth / (double)width;
double h = (double)apparentheight / (double)height;
Just to provide further explanation to the other answers:
When you do mathematical operations on ints, the result is an int, and this is achieved by effectively truncating the non-integer portion:
3 / 2 == 1
When an operation is performed involving at least one double, the int input is first converted to a double, so the result is a double:
1.0 / 4 == 0.25
4 * 0.25 == 1
And of course, if both inputs are double the result is double and no implicit conversion occurs:
1.0 / 4.0 == 0.25
I need to somehow get one number before floating point and value after that floating point. Example:
Before: 212.12345;
After: 2.12345
Any Ideas?
Assuming you have:
decimal x = 212.12345m;
you can use the modulo operator:
decimal result = x % 10;
Note that the number should be represented as a decimal if you care about the accurate value.
See also: Meaning of "%" operation in C# for the numeric type double
You can do like this:
public double GetFirst(double a)
{
double b = a / 10.0;
return (b - (int)b) * 10.0;
}
try this
double x = 1;
var y = x/10;
var z = (y % (Math.Floor(y))) * 10;
Try this code
string num = "15464612.12345";
string t = num.Split('.')[0];
num = t[t.Length-1].ToString() + "." + num.Split('.')[1];
my approach was to find the number 210, and substract it....
will work for any number as well as smaller then 10.
double f1 = 233.1234;
double f2 = f1 - (((int)f1 / 10) * 10);
The result of all of the division equations in the below for loop is 0. How can I get it to give me a decimal e.g.:
297 / 315 = 0.30793650793650793650793650793651
Code:
using System;
namespace TestDivide
{
class Program
{
static void Main(string[] args)
{
for (int i = 0; i <= 100; i++)
{
decimal result = i / 100;
long result2 = i / 100;
double result3 = i / 100;
float result4 = i / 100;
Console.WriteLine("{0}/{1}={2} ({3},{4},{5}, {6})", i, 100, i / 100, result, result2, result3, result4);
}
Console.ReadLine();
}
}
}
Answer:
Thanks Jon and everyone, this is what I wanted to do:
using System;
namespace TestDivide
{
class Program
{
static void Main(string[] args)
{
int maximum = 300;
for (int i = 0; i <= maximum; i++)
{
float percentage = (i / (float)maximum) * 100f;
Console.WriteLine("on #{0}, {1:#}% finished.", i, percentage);
}
Console.ReadLine();
}
}
}
You're using int/int, which does everything in integer arithmetic even if you're assigning to a decimal/double/float variable.
Force one of the operands to be of the type you want to use for the arithmetic.
for (int i = 0; i <= 100; i++)
{
decimal result = i / 100m;
long result2 = i / 100;
double result3 = i / 100d;
float result4 = i / 100f;
Console.WriteLine("{0}/{1}={2} ({3},{4},{5}, {6})",
i, 100, i / 100d, result, result2, result3, result4);
}
Results:
0/100=0 (0,0,0, 0)
1/100=0.01 (0.01,0,0.01, 0.01)
2/100=0.02 (0.02,0,0.02, 0.02)
3/100=0.03 (0.03,0,0.03, 0.03)
4/100=0.04 (0.04,0,0.04, 0.04)
5/100=0.05 (0.05,0,0.05, 0.05)
(etc)
Note that that isn't showing the exact value represented by the float or the double - you can't represent 0.01 exactly as a float or double, for example. The string formatting is effectively rounding the result. See my article on .NET floating binary point for more information as well as a class which will let you see the exact value of a double.
I haven't bothered using 100L for result2 because the result would always be the same.
Try
i / 100.0
because i is an int: i / 100 performs integer division, then the result, that is always 0, is casted to the target type. You need to specify at least one non-int literal in your expression:
i / 100.0
Because i is an integer and 100 is an integer...so you have an integer division
Try (decimal)i / 100.0 instead
No matter where you store it, an integer divided by an integer will always be an integer.
You need to force a floating point operation "double / double" instead of an "int / int"
double result = (double)297 / (double)315 ;
this is integer division whatever the type of variable you storing in,
so int / int = int
double result3 = ((double)i) / 100;
Because i is a int value and you divide by an integer so the result is an integer ! and so you need to divide by 100.0 to have an implicit cast in float or specify 100f or 100d
In my case I had only vars and no int
float div = (var1 - var2) / float.Parse(var1.ToString());