double compare 0.2 != 0.2 How?

[Bassem]

I'm working on a real-time reservation graph. And it keeps giving me error even I tested each Method and each statement except this one, which I did this early morning.
What suppose these statements do?

MessageBox.Show(Convert.ToString((1.0 - 0.8) == 0.2));

I felt wrong about anything I didn't see, I tested this too

MessageBox.Show(Convert.ToString(((double)(1.0 - 0.8)).CompareTo(0.2)));

I waited for True from each, but both give false!!!

Help!!!!!!!!!!! !!
Please any help!!!!!!!!!

[artov]

Because 0.2 cannot be represented exact on floating point, value readed by the compiler and calculated from two double numbers (where 0.8 cannot also be repsesented exact), there has to be some differences.

For online help, check http://www.h-schmidt.net/FloatApplet/IEEE754.html or read wikipedia.

[Plater]

What error are you getting?
EDIT: NM I see

[Bassem]

Great I know the representation of a floating-point number. But What should I do?!
I'm comparing two double numbers and get wrong result.
Any idea. I can't step back for that, my application depends on that comparison.

[Plater]

Outside the box time.

Consider this:
[code=c#]
double a = 1.0;
double b = 0.8;
double c = 0.2;
MessageBox.Show ("" + ((a - b) + (a - c) == a));
[/code]
The result is true. Perhaps you can work out the routine to do this to fit your needs?

[Bassem]

Hello Plater,
I'm so sorry for wasting your time with such a question. Thanks for replying.
I tested your code and the result is True, as it should of course.
Could you please test mine, it also should give True, but I swear with VS2005 that it gives me False.

            double a = 1.0;
            double b = 0.2;
            double c = 0.8;

            MessageBox.Show("" + ((1.0 - 0.8) == 0.2));

I fell stupid to ask you this, please copy and test it.
Tell me why it gives False!
Again I've to say sorry for wasting your time.

Regards,
Bassem

[Plater]

No no, you are correct, yours will give false.
I speculate that since one side is a specified value and the other is computed they are treated differently.
That is why I was saying use my code, they are functionaly identical mathmatically.

If yours was a function like this:
[code=c#]
bool TestDouble(doub le a, double b, double c)
{
return ((a - b) == c);
//could also write it ((2*a - b - c)==a)
}
[/code]
You would call it with:
MessageBox.Show (TestDouble(1.0 ,0.8,0.2).ToStr ing());

But as you said, you will get "False" instead of the expected "True"

So if you instead used my function:
[code=c#]
bool TestDouble(doub le a, double b, double c)
{
return ((a - b) + (a - c) == a));
}
[/code]
You could again call it with:
MessageBox.Show (TestDouble(1.0 ,0.8,0.2).ToStr ing());

Only now you would get the expected "True"

[Bassem]

This is the most strange thing I ever seen in C#, Can you please give me a reference to read more. I didn't understand yet the difference.

[Curtis Rutland]

This is an interesting topic. I'm just commenting to subscribe and remind me to look at it later.

[IanWright83]

Sorry I didn't reply earlier. Account has been locked out and only just figured out how to create a new account with my existing email address.

Anyway, what you need to look at is Math.Round(). I believe if you do the following you'll get your answer.

MessageBox.Show(Math.Round(1.0 - 0.8, 1) == 0.2));

[Curtis Rutland]

I'm actually interested in the "why" of this problem. Why does a computed value differ from a literal in this case.

[IanWright83]

I'm actually interested in the "why" of this problem. Why does a computed value differ from a literal in this case.

Quite simply because it is floating point arithmetic. It's a well known common problem.

http://en.wikipedia.org/wiki/Floating_point

[r035198x]

The morale of the story is that never (ever) use == (or !=) for comparing floating point numbers. There are numbers that the computer cannot represent in binary exactly simply because they are recurring numbers. For those numbers, the exact value stored is dependent on the platform being used and the precision possible on those platforms.
e.g
(Requesting your forgiveness for polluting your esteemed forum with the dreadful Java codes)

System.out.println(0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1);

does not print 1.0 on my computer.

Here is What every computer scientist should know about floating point arithmetic

[IanWright83]

I'm kinda amazed that the other moderator/forum leaders didn't know this actually... You can't get very far with doubles without knowing about the rounding issues.

[r035198x]

I'm not too amazed. The forum administrative roles do not reflect on how much someone knows about subjects. They do know some pretty amazing things though which I don't know too. That's what makes forums amazing. Everyone gets to learn.