Understanding 'self' - problem with .z.s[]

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1

I wrote a function for permutation:

q) perm:{[a] $[2 = count a ; (a;reverse a);raze each[{ (first
x) ,/: perm[x _ 0]} ] each[{x rotate y}[;a]] til count a ]}
and
q) perm 1 2 3
works fine.

However, when i replace the recursive call to ‘perm’ with ‘.z.s’, it
gives error 'stack.

q) perm:{[a] $[2 = count a ; (a;reverse a);raze each[{ (first
x) ,/: .z.s[x _ 0]} ] each[{x rotate y}[;a]] til count a ]}
q)perm 1 2 3

GIVES ERROR:

'stack
@
{ (first x) ,/: .z.s[x _ 0]}
`int$()

Can anyone explain the reason for the same ?
Thanks

2

.z.s in the place you’ve put it is not equals perm, but rather nested lambda it’s lodged in. So, you’re blowing up the call stack by never exiting recursion.

Cheers,
Oleg

3

Yeah, .z.s is the self call for the function??

{ (first x) ,/: .z.s[x _ 0]}

and does not call perm as intended.

Example :?

f:{show `Hi;{$[x<0;x;[show x;.z.s[x-1]]]}}

Output : f 4

`Hi

4

3

2

1

0

-1


Cheers !

Sayantan.