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?

## Solution 1

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 * 10^{1}+ 2 * 10^{0}+ 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 * 2^{1}+ 0 * 2^{0}+ 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.

## Solution 2

`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)
{
...
}
```

## Solution 3

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))`

## Solution 4

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)
```

## Solution 5

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.

## Solution 6

Use `decimal`

. It doesn't have this "problem".

## Solution 7

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.

## Solution 8

`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.

## Solution 9

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.

## Solution 10

Official MS help, especially interested "Precision in Comparisons" part in context of the question. https://docs.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.");
```

## Solution 11

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)

## Solution 12

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`

.

## Solution 13

```
// 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.

## Solution 14

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.

## Solution 15

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")
```

## Solution 16

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
}
```

## Solution 17

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;
}
```

## Solution 18

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://docs.microsoft.com/en-us/dotnet/api/system.double.compareto?view=netcore-3.1