Strange rounding

[puter]

doing a simple calculation using C#. I am setting distance = speedConv / 3600.

speedConv is between 0 and 60. When I print out all the values speedConv, for example is 30...but distance is always 0. I have calculated several values and they are in the area of .005, which should be well within the range of a double.

Does anyone know of any reason this value would be getting rounded to 0?


Thanks

[v0id]

How do you print it? Your problem could be that you're using a function or a format, which only allows you to print integers, and that will therefore result in the missing digits.

[justto_doit]

Hello,
May I know what is the data type of

distance
speedConv

[puter]

distance and speedConv are both doubles.

I know the printing method works because when I print out another double (MPG) it prints out as a number similar to 50.22342123

So, that tells me that there should 1) be plenty of decimal places and 2) not be a problem with the printing.

It's almost like it's taking a number that is less than 1 and rounding it to zero for unknown reasons. Oh, and this is a debug print.

[justto_doit]

its funny~
I have do something similar but I cannot get this error

e.g.
double distance;
double speedConv;
for (speedConv = 0; speedConv < 30; speedConv++)
{
distance = speedConv / 3600;
Debug.WriteLine(distance);
}

or may you point out where is the difference between us?

[puter]

Originally Posted by justto_doit

its funny~
I have do something similar but I cannot get this error

e.g.
double distance;
double speedConv;
for (speedConv = 0; speedConv < 30; speedConv++)
{
distance = speedConv / 3600;
Debug.WriteLine(distance);
}

or may you point out where is the difference between us?

only difference is that yours is counting up in order.

Right now I'm generating random data to test my code, so they are never the same number.

That should not affect the test.

What error are you getting? I don't get an error, I just get a distance of 0.

[puter]

I figured it out, but I think it's stupid.

apparently if you divide a double by an int literal C# thinks it would be smart to just change that thar double output to an int.

dunno why the developers thought that was a good idea, but changing it to 3600.0 fixed it.

There wasn't even a warning for the loss of precision.

[Xav]

If you divide the answer using two integer values, then of course the answer will be an integer. You're right, changing it to 3600.0 does the job.

[puter]

I'm not using two integer values tho.

I'm dividing a double by an integer.

it's stupid that 1) the compile goes to the less precise result and 2) does not produce a warning about loss of precision.

[Xav]

It's known as an "implicit" cast - there is no potential loss of precision, because by passing in 3600 and not 3600.0, you are effectively telling C# you only want that level of accuracy in your answer.