I must say this is very confusing.
Shouldn’t any function that
a) makes no use of any foreigns or ?,
b) uses only local definition =. and
c) does not access any global variables (or has shadowed them with a local
definition)
return the same result every time it is run with identical (same value)
arguments?
a=: 1 : 0
g=. +:
u y
)
] a 5
5
(u=: ]) a 5
5
(g=: ]) a 5
5
g a 5
10
u a 5
|stack error: u
| u y
That is _very_ misleading. I understand that in the first 3 cases ] is passed
by value
and in the other two by name, but I still see this as strange behaviour. When
the adverb’s
argument is u, wether u has a value or not in the calling locale, gives a stack
error. Using g,
no matter the value of g, gives 10. Using any other undefined name gives a
value error.
Finally, using an anonymous function or proverb which isn’t g or u yields the
correct answer.
That’s 3 different results or errors for basically one input _value_ (not
counting the correct
value error).
The worst one IMHO is (g=: ]) a 5 versus g a 5. Finding such an error must be a
quite
painful experience. Note that this happens here as well:
a=: 1 : 0
u=. +:
g y
)
g=: u=: ]
g a 5
5
u a 5
5
g=: u
u=: ]
g a 5
10
u a 5
10
I guess this means that function arguments can access variables which are local
to the
calling adverb. So can functions that use a value referred to by a global
variable see
this value change if they are supplied as an argument to an adverb? Apparently
not:
g=: 3 : 'global'
global=: 5
a=: 1 : 0
global=. 10
u y
)
g 0
5
g a 0
5
This is how I would have expected it to work in the first place. Why then can a
function
argument to an adverb (unexpectedly) access a verb which is local to the adverb,
while it (as expected) cannot access a local variable?
Louis
> On 05 Nov 2016, at 15:33, Raul Miller <[email protected]> wrote:
>
> On Sat, Nov 5, 2016 at 10:09 AM (EST), I wrote:
>> ... Now that I think this through, I think I can do a passable job of that.)
>
> Actually, no, I do not.
>
> Extracting the locale from a name is straightforward:
>
> conamed=:3 :0
> 'a b'=. _2{.I.'_'='_',y
> if. (0=a)+.1~:#;:y do. NB. implied
> 18!:5 ''
> elseif. b=a+1 do. NB. indirect
> ".b}.y
> elseif. do. NB. direct
> <}:a}.y
> end.
> )
>
> And, getting the name of an adverb argument is trivial:
>
> ;u`''
>
> However, this does not address the issue faced by jtrace, which is
> finding the locale that jtrace was invoked from.
>
> As a workaround, I would recommend assuming that that locale was the
> base locale. This potentially reduces the utility of the jtrace
> facility, but it seems like the right balance of simplicity.
> Specifically, I am recommending changing line 8 of executet_jtrace_
> and executep_jtrace_ from
>
> ". 't_z=. ', ; t_x
> and
> ". 't_z=. ',t_x=. '(',(;:^:_1 t_x),')'
>
> to
>
> do_base_ 't_z=. ', ; t_x
> and
> do_base_ 't_z=. ',t_x=. '(',(;:^:_1 t_x),')'
>
> Does this make sense?
>
> Thanks,
>
> --
> Raul
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