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DarkBASIC Discussion / Mod function(%)

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Caleb1994
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Posted: 2nd Sep 2009 15:29
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Simple very useful.

for those that don't know using mod on two numbers returns the remander after dividing the two numbers.

so 13 % 2 would return 1 because 2 * 6 == 12 and so the remander is 1.

i got 10 millaseconds on trying 1000000 % 20 so it's pretty good. here's the function



course you can't create operators so it's just a function but it works!

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Latch
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Posted: 2nd Sep 2009 16:10
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I think it can be simplified and sped up a bit:



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Caleb1994
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Posted: 2nd Sep 2009 19:18
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Nice! i didn't think that would work because of the float value that "num/modulus" would return! you have clearly proven me wrong good job!

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Libervurto
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Posted: 12th Sep 2009 01:41 Edited at: 12th Sep 2009 01:43
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dbc is quirky it will only return a float if the modulus (new word for me ) is a float, otherwise it will return an integer.

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TheComet
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Posted: 14th Sep 2009 14:38
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@Latch

If I used the numbers 13 and 2 with your function, 13-(13/2*2)=-1, since 13-(13/2=6.5~7*2=14)=-1. You'd have to add another 0-" at the start.



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Latch
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Posted: 14th Sep 2009 16:00
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@ThComet
No, I'm pretty sure the function would return 1 as it stands. num/modulus returns the interger portion of the division. 13/2 = 6.5 but the return value is 6. If it were mod(2,13), then 2/13 = 0.154 and the return value from 2/13 would be 0 . Integers divided by integers return a truncated integer in DBC. In some BASICs there is actually a TRUNC() (short for truncate) function that returns the integer portion of a floating point number. In DBC, INT() is the equivalent function to TRUNC. But in the case of one integer divided by another, DBC maintains the data type as integer.

mod(13,2)

value=13 - ((13/2) * 2 )
value=13 - ((6) *2 )
value= 13 - 12
value= 1

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TheComet
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Posted: 14th Sep 2009 19:27
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Ah, I thought 6.5 is rounded up to 7... Whoops, sorry.

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