I'm currently writing the Change Return Program on The Big List of Projects and I encountered a logic error in my code. Here is the snippet in question:
if(change / 1 >= 1) // checks if when divided the value is higher than 1, signaling that there is at least a dollar in change
{
double temp1 = change % 1; // declared as double for casting
temp1 = change - temp1; // and used to remove excess, non-dollar value
dollarNum = (int) temp1;
change -= (double)dollarNum;
}
Where change is a double that before this has a value of 10.01. However, after this snippet runs, the value becomes 0.00999999999999979, rather than 0.01 as I want it. Am I assuming something is wrong with casting here? Or is it something else entirely?
Also, note that I tried reformatting the last line to "change = change - dollarNum and tried swapping out dollarNum with temp1 to avoid casting in that line, but neither fixed the issue.
If I formatted anything wrong please let me know and I'll surely fix it as soon as I can.
A double is not suitable for this kind of calculation because it does not store the value exactly (internally it uses binary fractions instead of decimal ones).
Use 'the 'decimal' type instead (https://learn.microsoft.com/en-us/dotnet/csharp/language-reference/keywords/decimal):
decimal change = 10.1M;
if(change / 1 >= 1) //checks if when divided the value is higher than 1, signalling that there is at least a dollar in change
{
decimal temp1 = change % 1; //declared as double for casting
temp1 = change - temp1; //and used to remove excess, non-dollar value
int dollarNum = (int)temp1;
change -= (decimal)dollarNum;
}
I need to convert my value 2.8634 to 2.8. I tried the following ,
var no = Math.Round(2.8634,2,MidpointRounding.AwayFromZero)
I'm getting 2.87.
Suggest me some ideas how to convert.
Thanks
This might do the trick for you
decimal dsd = 2.8634m;
var no = Math.Truncate(dsd * 10) / 10;
Math.Truncate calculates the integral part of a specified decimal number. The number is rounded to the nearest integer towards zero.
You can also have a look on the difference between Math.Floor, Math.Ceiling, Math.Truncate, Math.Round with an amazing explanation.
Use this one.Hope this will work for you.
var no = Math.Round(2.8634,1,MidpointRounding.AwayFromZero)
It's a tad more cryptic (but more efficient) than calling a Math method, but you can simply multiply the value by 10, cast to an integer (which effectively truncates the decimal portion), and then divide by 10.0 (or 10d/10f, all just to ensure we don't get integer division) to get back the value you are after. I.e.:
float val = 2.8634;
val = ((int)(val * 10)) / 10.0;
I decided to re-create my question:
decimal dTotal = 0m;
foreach (DictionaryEntry item in _totals)
{
if (!string.IsNullOrEmpty(item.Value.ToString()))
{
dTotal += Convert.ToDecimal(item.Value);
}
}
Console.WriteLine(dTotal / 3600m);
Console.WriteLine(decimal.Round(dTotal / 3600m, 2));
Console.WriteLine(decimal.Divide(dTotal, 3600m));
The above code returns:
579.99722222222222222222222222
580.00
579.99722222222222222222222222
So, that is where my issues are coming from, I really need it to just display the 579.99; but any round, be it decimal.Round or Math.Round still return 580; even the string formats for {0:F} return 580.00.
How can i properly do this?
New answer (to new question)
Okay, so you've got a value of 579.99722222222222222222222222 - and you're asking that to be rounded to two decimal places. Isn't 580.00 the natural answer? It's closer to the original value than 579.99 is. It sounds like you essentially want flooring behaviour, but with a given number of digits. For that, you can use:
var floored = Math.Floor(original * 100) / 100;
In this case, you can do both in one step:
var hours = Math.Floor(dTotal / 36) / 100;
... which is equivalent to
var hours = Math.Floor((dTotal / 3600) * 100) / 100;
Original answer (to original question)
Sounds like you've probably got payTotal in an inappropriate form to start with:
using System;
class Test
{
static void Main()
{
decimal pay = 2087975.7m;
decimal time = pay / 3600;
Console.WriteLine(time); // Prints 579.99325
}
}
This is the problem:
var payTotal = 2087975.7;
That's assigning payTotal to a double variable. The value you've actually got is 2087975.69999999995343387126922607421875, which isn't what you wanted. Any time you find yourself casting from double to decimal or vice versa, you should be worried: chances are you've used the wrong type somewhere. Currency values should absolutely be stored in decimal rather than double (and there are various other Stack Overflow questions talking about when to use which).
See my two articles on floating point for more info:
Binary floating point in .NET
Decimal floating point in .NET
(Once you've got correct results, formatting them is a different matter of course, but that shouldn't be too bad...)
I want to calculate the percentage. But the compiler is giving an error that the input string is not in a correct format. Can some one elaborate what i am missing here?
private double per()
{
double a = Convert.ToDouble(tbEnglish.Text+tbUrdu.Text+tbPhysics.Text+tbChemistry.Text+tbMaths.Text);
double d = 500;
double lblResult = (a / d)*100;
return lblResult;
}
You're concatenating the strings and then trying to convert that one result into a double. So for results of 75.6, 92.1, 56.3 78.2 and 72.3 you'd end up trying to parse "75.692.156.378.272.3".
Parse each value and then sum them.
However, I would strongly recommend that you use decimal for this instead of double. You should also consider using TryParse instead of Parse so that you can handle user input errors gracefully. Here's the solution sticking with Parse:
public decimal AveragePercentage()
{
decimal sum = decimal.Parse(tbEnglish.Text) +
decimal.Parse(tbUrdu.Text) +
decimal.Parse(tbPhysics.Text) +
decimal.Parse(tbChemistry.Text) +
decimal.Parse(tbMaths.Text);
return sum / 5m;
}
Out of interest, in your original code why are you dividing by 500 and then multiplying by 100? Why not just divide by 5 (as mine does now that I've noticed what was going on)?
As a side note, it's very important to differentiate between compile-time errors and execution-time errors. It wasn't the compiler saying that the input string wasn't in the correct format - it was the Convert.ToDouble method, at execution time. In this case it was relatively obvious, but in other situations we could have been chasing our tails for a while trying to find a compile-time problem when it was actually failing at execution time.
I don't have Visual Studio available to me here on my Linux box, but I think you're better off with code like this.
private double per()
{
double a = Convert.ToDouble(tbEnglish.Text);
a += Convert.ToDouble(tbPhysics.Text);
a += Convert.ToDouble(tbChemistry.Text);
a += Convert.ToDouble(tbMaths.Text);
double d = 500;
double lblResult = (a / d)*100;
return lblResult;
}
In your example, you end up building a string that will look like: "75.692.156.372.3", which cannot be parsed into a double.
You need to convert all the TextBox.Text values into Decimals before using the + operator.
I've a double variable called x.
In the code, x gets assigned a value of 0.1 and I check it in an 'if' statement comparing x and 0.1
if (x==0.1)
{
----
}
Unfortunately it does not enter the if statement
Should I use Double or double?
What's the reason behind this? Can you suggest a solution for this?
It's a standard problem due to how the computer stores floating point values. Search here for "floating point problem" and you'll find tons of information.
In short – a float/double can't store 0.1 precisely. It will always be a little off.
You can try using the decimal type which stores numbers in decimal notation. Thus 0.1 will be representable precisely.
You wanted to know the reason:
Float/double are stored as binary fractions, not decimal fractions. To illustrate:
12.34 in decimal notation (what we use) means
1 * 101 + 2 * 100 + 3 * 10-1 + 4 * 10-2
The computer stores floating point numbers in the same way, except it uses base 2: 10.01 means
1 * 21 + 0 * 20 + 0 * 2-1 + 1 * 2-2
Now, you probably know that there are some numbers that cannot be represented fully with our decimal notation. For example, 1/3 in decimal notation is 0.3333333…. The same thing happens in binary notation, except that the numbers that cannot be represented precisely are different. Among them is the number 1/10. In binary notation that is 0.000110011001100….
Since the binary notation cannot store it precisely, it is stored in a rounded-off way. Hence your problem.
double and Double are the same (double is an alias for Double) and can be used interchangeably.
The problem with comparing a double with another value is that doubles are approximate values, not exact values. So when you set x to 0.1 it may in reality be stored as 0.100000001 or something like that.
Instead of checking for equality, you should check that the difference is less than a defined minimum difference (tolerance). Something like:
if (Math.Abs(x - 0.1) < 0.0000001)
{
...
}
You need a combination of Math.Abs on X-Y and a value to compare with.
You can use following Extension method approach
public static class DoubleExtensions
{
const double _3 = 0.001;
const double _4 = 0.0001;
const double _5 = 0.00001;
const double _6 = 0.000001;
const double _7 = 0.0000001;
public static bool Equals3DigitPrecision(this double left, double right)
{
return Math.Abs(left - right) < _3;
}
public static bool Equals4DigitPrecision(this double left, double right)
{
return Math.Abs(left - right) < _4;
}
...
Since you rarely call methods on double except ToString I believe its pretty safe extension.
Then you can compare x and y like
if(x.Equals4DigitPrecision(y))
Comparing floating point number can't always be done precisely because of rounding. To compare
(x == .1)
the computer really compares
(x - .1) vs 0
Result of sybtraction can not always be represeted precisely because of how floating point number are represented on the machine. Therefore you get some nonzero value and the condition evaluates to false.
To overcome this compare
Math.Abs(x- .1) vs some very small threshold ( like 1E-9)
From the documentation:
Precision in Comparisons
The Equals method should be used with caution, because two apparently equivalent values can be unequal due to the differing precision of the two values. The following example reports that the Double value .3333 and the Double returned by dividing 1 by 3 are unequal.
...
Rather than comparing for equality, one recommended technique involves defining an acceptable margin of difference between two values (such as .01% of one of the values). If the absolute value of the difference between the two values is less than or equal to that margin, the difference is likely to be due to differences in precision and, therefore, the values are likely to be equal. The following example uses this technique to compare .33333 and 1/3, the two Double values that the previous code example found to be unequal.
So if you really need a double, you should use the techique described on the documentation.
If you can, change it to a decimal. It' will be slower, but you won't have this type of problem.
Use decimal. It doesn't have this "problem".
Exact comparison of floating point values is know to not always work due to the rounding and internal representation issue.
Try imprecise comparison:
if (x >= 0.099 && x <= 0.101)
{
}
The other alternative is to use the decimal data type.
double (lowercase) is just an alias for System.Double, so they are identical.
For the reason, see Binary floating point and .NET.
In short: a double is not an exact type and a minute difference between "x" and "0.1" will throw it off.
Double (called float in some languages) is fraut with problems due to rounding issues, it's good only if you need approximate values.
The Decimal data type does what you want.
For reference decimal and Decimal are the same in .NET C#, as are the double and Double types, they both refer to the same type (decimal and double are very different though, as you've seen).
Beware that the Decimal data type has some costs associated with it, so use it with caution if you're looking at loops etc.
Official MS help, especially interested "Precision in Comparisons" part in context of the question.
https://learn.microsoft.com/en-us/dotnet/api/system.double.equals
// Initialize two doubles with apparently identical values
double double1 = .333333;
double double2 = (double) 1/3;
// Define the tolerance for variation in their values
double difference = Math.Abs(double1 * .00001);
// Compare the values
// The output to the console indicates that the two values are equal
if (Math.Abs(double1 - double2) <= difference)
Console.WriteLine("double1 and double2 are equal.");
else
Console.WriteLine("double1 and double2 are unequal.");
1) Should i use Double or double???
Double and double is the same thing. double is just a C# keyword working as alias for the class System.Double
The most common thing is to use the aliases! The same for string (System.String), int(System.Int32)
Also see Built-In Types Table (C# Reference)
Taking a tip from the Java code base, try using .CompareTo and test for the zero comparison. This assumes the .CompareTo function takes in to account floating point equality in an accurate manner. For instance,
System.Math.PI.CompareTo(System.Math.PI) == 0
This predicate should return true.
// number of digits to be compared
public int n = 12
// n+1 because b/a tends to 1 with n leading digits
public double MyEpsilon { get; } = Math.Pow(10, -(n+1));
public bool IsEqual(double a, double b)
{
// Avoiding division by zero
if (Math.Abs(a)<= double.Epsilon || Math.Abs(b) <= double.Epsilon)
return Math.Abs(a - b) <= double.Epsilon;
// Comparison
return Math.Abs(1.0 - a / b) <= MyEpsilon;
}
Explanation
The main comparison function done using division a/b which should go toward 1. But why division? it simply puts one number as reference defines the second one. For example
a = 0.00000012345
b = 0.00000012346
a/b = 0.999919002
b/a = 1.000081004
(a/b)-1 = 8.099789405475458e-5
1-(b/a) = 8.100445524503848e-5
or
a=12345*10^8
b=12346*10^8
a/b = 0.999919002
b/a = 1.000081004
(a/b)-1 = 8.099789405475458e-5
1-(b/a) = 8.100445524503848e-5
by division we get rid of trailing or leading zeros (or relatively small numbers) that pollute our judgement of number precision. In the example, the comparison is of order 10^-5, and we have 4 number accuracy, because of that in the beginning code I wrote comparison with 10^(n+1) where n is number accuracy.
Adding onto Valentin Kuzub's answer above:
we could use a single method that supports providing nth precision number:
public static bool EqualsNthDigitPrecision(this double value, double compareTo, int precisionPoint) =>
Math.Abs(value - compareTo) < Math.Pow(10, -Math.Abs(precisionPoint));
Note: This method is built for simplicity without added bulk and not with performance in mind.
As a general rule:
Double representation is good enough in most cases but can miserably fail in some situations. Use decimal values if you need complete precision (as in financial applications).
Most problems with doubles doesn't come from direct comparison, it use to be a result of the accumulation of several math operations which exponentially disturb the value due to rounding and fractional errors (especially with multiplications and divisions).
Check your logic, if the code is:
x = 0.1
if (x == 0.1)
it should not fail, it's to simple to fail, if X value is calculated by more complex means or operations it's quite possible the ToString method used by the debugger is using an smart rounding, maybe you can do the same (if that's too risky go back to using decimal):
if (x.ToString() == "0.1")
Floating point number representations are notoriously inaccurate because of the way floats are stored internally. E.g. x may actually be 0.0999999999 or 0.100000001 and your condition will fail. If you want to determine if floats are equal you need to specify whether they're equal to within a certain tolerance.
I.e.:
if(Math.Abs(x - 0.1) < tol) {
// Do something
}
My extensions method for double comparison:
public static bool IsEqual(this double value1, double value2, int precision = 2)
{
var dif = Math.Abs(Math.Round(value1, precision) - Math.Round(value2, precision));
while (precision > 0)
{
dif *= 10;
precision--;
}
return dif < 1;
}
To compare floating point, double or float types, use the specific method of CSharp:
if (double1.CompareTo(double2) > 0)
{
// double1 is greater than double2
}
if (double1.CompareTo(double2) < 0)
{
// double1 is less than double2
}
if (double1.CompareTo(double2) == 0)
{
// double1 equals double2
}
https://learn.microsoft.com/en-us/dotnet/api/system.double.compareto?view=netcore-3.1