Noob question, probably easy answer...

Lodeclaw

A colleague of mine is learning C#, as am I. I am ahead of him, so he asked me to help him figure out why his tiny test program isn't working. For the life of me I can't figure it out. Here's the code:

private void button1\_Click(object sender, EventArgs e)
{
    int tf = int.Parse(textBox1.Text);

    int tc = 5 / 9 \* (tf - 32);
    label1.Text = tc.ToString();
}

This is supposed to convert the temperature in Fahrenheit (tf) to the temperature in celcius (tc). As far as I can tell it should work, but the value tc always comes out as 0. What am I overlooking? X|

Lost User

Lodeclaw wrote:

int tc = 5 / 9 * (tf - 32);

Integer devision, a common mistake for beginners, it happens to all of us. 5/9 is .5555 which is 0. Convert to doubles and try it then, should be fine after that.

Lodeclaw

It's the same result.

Lost User

No it isn't... I used same variable names...

        double tf = double.Parse(textBox1.Text);

        double tc = (5.0 / 9.0) \* (tf - 32);
        label1.Text = tc.ToString();

Kristian Sixhoj

double tf = double.Parse(textBox1.Text);

double tc = 5 / 9 * (tf - 32);
label1.Text = tc.ToString();

If that doesn't work, something's really wrong here. :rolleyes:

Kristian Sixhoej "You can't undo the past... but you can certainly not repeat it." - Bruce Willis

Lost User

Kristian Sixhoej wrote:

If that doesn't work, something's really wrong here.

Untrue, I think it will still do integer devision on 5 / 9, you have to explicitly put .0 on either to do double devision. Weird, I know, I just noticed that behavior on a test.

Lodeclaw

Oh, I see the problem. We're not entering the values in the calculation as doubles. (5.0 rather than 5) Thanks, Eliott.

Lost User

Anytime

Kristian Sixhoj

EliottA wrote:

you have to explicitly put .0 on either to do double devision.

Didn't knew that.

EliottA wrote:

Weird, I know, I just noticed that behavior on a test.

IIRC I have never had that problem when doing double divisions. But I probably don't recall correct. ;P

Kristian Sixhoej "You can't undo the past... but you can certainly not repeat it." - Bruce Willis

Lost User

Remember to change your int.Parse and datatype to double and double.Parse otherwise entering 10.0 in that textbox and parsing it to an int might either through an exception or truncate the double value (I don't know which one) in either case, duplicating the original error!!

Lost User

I didn't know it either, and I never ran into a problem like that before, I never had to implicitly add a .0 unless it was in java.. even then I vaguely recall something about automatic upgrading datatype or something, I don't know. I completely forget, just remember seeing integer devision smack me in the face a few times in my first programming course.

Rob Philpott

Being a bit anal about this, I'd do this:

double tf = double.Parse(textBox1.Text);
double tc = 5d / 9d * (tf - 32d);
label1.Text = tc.ToString();

Regards, Rob Philpott.

Lodeclaw

Oh, that's fascinating! Placing a 'd' after a whole number will really make it a double?

Lost User

Yep, although I never really have ever used it. I'd rather put 5.0 or something of the sort.

musefan

will 5.0 def making it a double as opposed to float? Im not trying to correct, seroius question

My opinion is... If someone has already posted an answer, dont post the SAME answer

Lost User

From my recollection yes it will, the default is always double unless explicitly indicated otherwise.

musefan

;)

My opinion is... If someone has already posted an answer, dont post the SAME answer

Guffa

EliottA wrote:

I never ran into a problem like that before

That's probably because it's not that usual to divide two literal values. Usually one of the operands is a double variable, then the compiler will cast the other operand to double also.

Despite everything, the person most likely to be fooling you next is yourself.