It just follows the parsing rules.
Think about it: before 6!:2 can be executed, you
have to know whether it is being invoked dyadically
or monadically. So you have to see whether its left
operand is a noun or not. But you can't just look at
] : that might be part of ]`] which is a noun. So
you have to go one more token, to the aa, before you
can decide to execute 6!:2.
Nouns are processed by value. Therefore, the value
of aa put on the stack is the value before executing 6!:2.
aa=. '1'
smoutput aa [ [ 6!:2 'smoutput aa=. ''2'' [ wait 1'
2
2
Henry Rich
> -----Original Message-----
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Devon
> McCormick
> Sent: Friday, October 17, 2008 10:50 PM
> To: J-programming forum
> Subject: [Jprogramming] Order of evaluation oddity
>
> Something I never noticed before:
>
> aa=. '1'
> smoutput aa [ 6!:2 'smoutput aa=. ''2'' [ wait 1'
> 2
> 1
> aa
> 2
>
> The leftmost "smoutput" gets the value of "a" before it's
> re-assigned within
> the "6!:2" statement?
>
> --
> Devon McCormick, CFA
> ^me^ at acm.
> org is my
> preferred e-mail
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