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AppGameKit Classic Chat / determining if a number is a float?

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LiamBradbury
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Posted: 3rd Jul 2013 04:58
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I need to determine whether or not a number has decimal places, how would I do this within the code?

so if I'm counting from 1 to 10 and dividing each number by 2, I can say if the result is an integer or a float.

example:


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Ancient Lady
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Posted: 3rd Jul 2013 05:06 Edited at: 3rd Jul 2013 05:07
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Try this:


Rough code, but you can get the idea.

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easter bunny
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Posted: 3rd Jul 2013 05:06 Edited at: 3rd Jul 2013 05:15
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Try this:
'

edit: haha, posted at the same time
either way should work

edit2:
Here's couple of similarly useful functions I made a while back


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LiamBradbury
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Posted: 3rd Jul 2013 05:18
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Thanks guys! works great!

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LiamBradbury
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Posted: 3rd Jul 2013 07:18
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been playing around with this some more, Still not working.

My problem seems to be AppGameKit not recognizing floats properly.

This simple maths code should give me the answer 5.5 but instead returns 5



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LiamBradbury
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Posted: 3rd Jul 2013 07:20
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hmmm just tried this and it worked, weird?!?!



didn't realize they all had to be floats??

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easter bunny
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Posted: 3rd Jul 2013 07:28
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I'm pretty sure the inc command/statement only supports integers as well, I always use n#=n#+1 now for floats.

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Hodgey
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Posted: 3rd Jul 2013 07:40
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Quote: "a=11
b=2
c#=a/b

print c#"

This is an integer division and will always return an integer (as will any equation containing only integers). As you've found out, they need to be floats for it to work.

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Impetus73
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Posted: 3rd Jul 2013 11:27
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Seriously people? did you all forget about MOD ?



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JimHawkins
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Posted: 3rd Jul 2013 13:33
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Hmmm. So what if it's 1.0?

In any case, if the variable is declared correctly you should know whether it's a float or an integer.

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Impetus73
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Posted: 3rd Jul 2013 13:36
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Quote: "
I need to determine whether or not a number has decimal places
"


I guess .0 is the same as no decimal places.

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LiamBradbury
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Posted: 3rd Jul 2013 16:19 Edited at: 3rd Jul 2013 16:25
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Thanks for the help guys, I haven't yet tried the 'Mod' way, but MissJoJos way of "if round(a#)<>a#" worked fine

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Ancient Lady
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Posted: 3rd Jul 2013 16:38
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MissJoJo's method is the one most often used. I just wasn't thinking in those terms (having not remembered that there was a Round command in the AppGameKit set).

Impetus7's idea works as well, but only in the tightly defined code that you showed as your example.

If you don't know what is generating the value, the Round method will work.

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JimHawkins
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Posted: 3rd Jul 2013 16:50
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But a# IS a float, so what's the problem?

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Ancient Lady
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Posted: 3rd Jul 2013 17:00
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I think the issue is determining if there is a fractional component.

1.0 vs 1.1, for instance.

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Phaelax
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Posted: 3rd Jul 2013 18:01
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Marl
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Posted: 3rd Jul 2013 19:48
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or...

which is probably a tad faster

or if the source is not an int (similar to MissJoJo);

if floor(result#) <> ceil(result#)
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Ancient Lady
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Posted: 3rd Jul 2013 20:37
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I like the floor/ceil method.

It is good for the generic case where you have no knowledge about how the value was created.

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JimHawkins
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Posted: 3rd Jul 2013 22:46
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Floor(0 and Ceil() do NOT work for integer values.

Ceil(2.0) is 2.0
Floor(2.0) is 2.0

It would be necessary to add a small additional fractional part when floor() = ceil() and then test again.

I'd still like to know a circumstance when this is needed!

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Ancient Lady
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Posted: 3rd Jul 2013 22:56
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Well, in the case you show, the value (2.0) is not a float in the sense that there is not a non-zero value after the decimal point. Which is what I believe to be the goal of the initial poster.

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Phaelax
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Posted: 4th Jul 2013 00:33
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Marl, how does this:
if t mod 2 = 1

Check for a float value? It's just going to return 0 or 1 for even and odd numbers. Isn't mod just for integer division anyway?

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Marl
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Posted: 4th Jul 2013 00:45 Edited at: 4th Jul 2013 00:48
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Technically it doesn't.

(and technically, that was Impetus73's code - I was offering an optimization but put code instead of quote like a noob)

As AL pointed out, for this thread, the definition of "is a float" has been adapted to suite the OP's request - to determine if a number has any non-zero digits after the decimal place.

So the point of this particular instruction is - as you rightly point out - to determine if the source integer being halved is odd or even - Which would ultimately lead to whether decimal places become involved.

If working from an Int source, checking with Int instructions will be faster than converting and checking the float. If the result shows no decimal places then float maths may not be needed at all.
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LiamBradbury
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Posted: 4th Jul 2013 01:13
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Just to clarify I wanted to be able to read if there were any decimal places at all on the number, so for example 5/2 = 2.5, were as 6/2 = 3 , I made a program which searches for the highest prime number and in order to do so I needed to divide the number I was testing by all the numbers before it and if any of them divided perfectly into an integer value (such as 6/2=3) then that equals 1 prime factor. Prime numbers only have 2 prime factors, 1 and itself.

See this post for the full code

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Ancient Lady
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Posted: 4th Jul 2013 01:43
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I did the same program in Fortran way back when I was working in the computer room during my freshman year. We had a card reader that was hooked up to the mainframe several hundred miles away.

I let my program run for two minutes. And I got a huge ration of grief for using up so much time for something so trivial. Two minutes on a shared mainframe back in 1977 was very expensive. There were no PCs then and certainly no terminals for entering programs.

I had fun doing it and never wasted computer resources like that again.

Now, I can use all the time I want on any of my computers and no one complains.

In your case, since are working strictly with integers to begin with, simply use the Mod function with the two values. If it returns anything other than zero, then you have a fractional part if you do the division.

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Ancient Lady
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JimHawkins
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Posted: 4th Jul 2013 08:53
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Liam - Are you trying to find prime factors, or just generating primes?

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Hodgey
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Posted: 4th Jul 2013 10:18
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Quote: " I needed to divide the number I was testing by all the numbers before it"

I believe you only need to test up to the sqrt of the given number.

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LiamBradbury
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Posted: 4th Jul 2013 11:02 Edited at: 4th Jul 2013 11:07
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@jim hawkins, the program finds both, it determines if it is a prime number judging on how many prime factors the number has.

@hodgey Not sure, I'll have to look into it more, but I found that if I test up to half of the number being tested then I can make sure all prime factors are accounted for. I did this by dividing each number by 2 up to half of the testing number, if the result is a number requiring no decimal places (eg 2.0) then it is a prime factor.

example:



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LiamBradbury
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Posted: 4th Jul 2013 11:03
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@Ancient Lady - cool story! I wasn't even born then! haha 1986 I was born!

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Impetus73
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Posted: 4th Jul 2013 21:20
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My solution, solved the problem presented in the original post, what else do you want? haha

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