Thank you. I got the verb versus pro-verb problem; I guess it can be surprising 
to newcomers, but it is logical.

However, I don't understand why sending a verb such as f=: g to an adverb which 
(locally) redefines g allows f to see the local definition of g, whereas 
sending a verb such as h=: 3 : 'd' to an adverb which redefines d does not 
allow h to see the local definition of d. Is it because h is run in its own 
locale, while f is not? Would that mean that f=: 3 : 'g y' would not be able to 
see local g, while f=: g would?

Keep in mind that, although I did complete the locales lab a littlie while ago, 
I program as a hobby and thus have little to no experience whatsoever with OOP 
and locales.

Louis

> On 07 Nov 2016, at 03:17, Henry Rich <[email protected]> wrote:
> 
> I didn't read all this, but I sympathize with your perplexity. If you give an 
> individual testcase that produces a result you don't expect, and give a 
> thorough justification of why you expect something different, veterans here 
> will explain what's happening. More probably, you will discover the reason 
> yourself as you try to create the justification.
> 
> Just to pick up on one thing I did see: The value of (g =: ]) is (]), a 
> primitive with no name, which is not the same as (g), a name.
> 
> Henry Rich
> 
>> On 11/5/2016 12:38 PM, Louis de Forcrand wrote:
>> 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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