The mod() function returns the integer remainder of a division


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Blackadder
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Oh, and by the way, this is not at all a complaint, I applaude this discrepancy to the manual. The fraction part of a division can easily be computed by


  fpart(a/b)


while mirroring the "normal" mod() behavior with a mod function as described in the manual is a tad more irky.


Bye, Malte.


Dave
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Inquisit's mod() function computes the integer remainder of a division


... as any well-behaved mod function should do. But indeed, you are correct in that the documentation is wrong.


~Dave


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Just a quick note: other than stated in the help, Inquisit's mod() function computes the integer remainder of a division.

Examples

mod(1,2)  = 1
mod(2,9)  = 2
mod(11,4) = 3


I haven't tested what mod(3.14,2) would yield but I assume it to be 1.14.

Bye, Malte


GO

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