Why isn't my x parameter being read correctly?

Community Support
1

{y}[1;2] produces 2 as expected

{x}[1;2] produces error

'rank

I was expecting 1

2

i reckon the “find implicit args” expression looks something like:

q) f:{“c”$120+til 1+(“x”,“y”,“z”) ? x}

so it finds two args when invoked with “y” but only one arg with “x”:

q)f “y”

“xy”

q)f “x”

,“x”

can anyone point me to this in q.k? ?i’ve not yet tried to understand q.k?

3

Has it to do with lambdas being compiled?<o:p></o:p>

<o:p> </o:p>

From: personal-kdbplus@googlegroups.com [mailto:personal-kdbplus@googlegroups.com] On Behalf Of Don Nguyen
Sent: Monday, December 05, 2011 6:03 AM
To: personal-kdbplus@googlegroups.com
Subject: [personal kdb+] Why isn’t my x parameter being read correctly?<o:p></o:p>

<o:p> </o:p>

{y}[1;2] produces 2 as expected<o:p></o:p>

<o:p> </o:p>

{x}[1;2] produces error<o:p></o:p>

<o:p> </o:p>

'rank<o:p></o:p>

<o:p> </o:p>

I was expecting 1<o:p></o:p>

<o:p> </o:p>

–
Submitted via Google Groups

4

the problem is that?

{y} is dydadic - it takes two parameters: x and y
{x} is monadic - it takes one parameter: x

when you call a monadic with more than one parameter, you get a rank error.

just like?

q)sum[0;1]

'rank

q)sum[0]

0

q)sum[(0;1)]

1

5

On Dec 4, 2011, at 8:27 PM, Jack Andrews wrote:

> i reckon the “find implicit args” expression looks something like:
> q) f:{“c”$120+til 1+(“x”,“y”,“z”) ? x}
>
> so it finds two args when invoked with “y” but only one arg with “x”:
> q)f “y”
> “xy”
> q)f “x”
> ,“x”
>
> can anyone point me to this in q.k? i’ve not yet tried to understand
q.k

pretty sure it’s not in q.k

lots of syntax, especially stuff shared by k & q, is in the binary and
not available for us to read

it’s not the invocation, it’s the code of the function

{x} is monadic, {y} is dyadic, {z} is triadic

6

Thanks everyone for the insights