. (dot) Does Not Work with mod?

Community Support
1

Hi,

I’m facing a strange error with mod and . (dot) operators. Please see the session log below:

===============SESSION=======================

D:\DEV>q

KDB+ 3.2 2015.03.05 Copyright (C) 1993-2015 Kx Systems

w32/ 4()core 4095MB Think flying-thinkpad 192.168.10.112 NONEXPIRE


Welcome to kdb+ 32bit edition

For support please see http://groups.google.com/d/forum/personal-kdbplus

Tutorials can be found at http://code.kx.com/wiki/Tutorials

To exit, type \

To remove this startup msg, edit q.q

q)mod

k){x-y*x div y}

q)mod . 192 2

'

q).[mod;192 2]

0

q)

===============SESSION=======================

Why is it that mod does not work with . (dot) operator (NOTE: the error message is empty)?

2

you need to wrap it in ()

q)(mod). 192 2

0

to make it a noun. Same would for for e.g. +

3

Yes, I figured that out after the original post.

But I’m still curious, though: Is this requirement due to the fact that mod was defined as a k function instead of a q one?

=============TEST=============

q)f:{x-y*x div y}

q)f

{x-y*x div y}

q)f . 1 2

1

q)mod

k){x-y*x div y}

q)mod . 1 2

'

q)(mod). 1 2

1

=============TEST=============

4

this should explain all you need…

https://groups.google.com/forum/#!searchin/personal-kdbplus/parentheses/personal-kdbplus/r60hWq7g8xw/xpALjkMip9IJ

http://www.kx.com/q/d/a/q1.htm#Verbs and Adverbs

http://code.kx.com/wiki/Reference/Grammar

HTH,

Sean