Re: unexpected unmodified variable

[Ray Andrews <rayandrews@eastlink.ca>]

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On 2022-10-06 21:36, Bart Schaefer wrote:
> Yes, that's documented. Interpreting the argument of "return" counts
> as a math operation, so the math value of a "functions -M" function
> really is always the last math operation.

Bart, can you point me to something where I can read up on this 
'functions -M', I was just using it on faith and because things broke 
without it.  I've never used that before and never seemed to miss it.  
In the manual I read:

By default the function is implemented by a shell function of the same 
name; if
shellfn  is specified it gives the name of the corresponding shell func‐ 
tion
while mathfn remains the name used in  arithmetical expressions.   The name
of  the  function in $0 is mathfn (not shellfn as would usually be the 
case),
provided the option FUNCTION_ARGZERO is in  effect.   The positional
parameters  in  the shell function correspond to the arguments of the 
mathe‐
matical function call.  The result of the last arithmetical expression 
eval‐
uated inside the shell function (even if it is a form that normally only 
re‐
turns a status) gives the result of the mathematical function.

... which might just be the most impenetrable bit of computer bafflegab 
I've ever read, I can't get past the first sentence. You have to be 
Peter to understand that.


> I think you must have typo'd something somewhere.  Add "functions -t
> func2" so you can see a trace of what's executing.

No typo, everything is copied and pasted without edit:

The file 'test1':

    #!/usr/bin/zsh

    functions -M func2
    func2 ()
    {
    functions -t func2
    echo func2 here!
    count=2
    #return 3
    }
    func1 ()
    {
    functions -t func2
    local count=1

    var=$(( func2() ))

    # Count = 1, no effect:
    #var=$( func2 )

    echo var is: $var, count is $count
    }


The run:

0 /aWorking/Zsh/Source/Wk 3 $ . test1; func1
+func2:2:>functions -t func2
+func2:3:>echo func2 'here!'
func2 here!
+func2:4:>count=2
var is: 0, count is 2                          <<< With 'return 3' 
commented out.

0 /aWorking/Zsh/Source/Wk 3 $ . test1; func1
+func2:2:>functions -t func2
+func2:3:>echo func2 'here!'
func2 here!
+func2:4:>count=2
+func2:5:>return 3
var is: 3, count is 2                         <<< With 'return 3' active.

0 /aWorking/Zsh/Source/Wk 3 $ . test1; func1
+func2:2:>functions -t func2
+func2:3:>echo func2 'here!'
func2 here!
+func2:4:>count=2
var is: 3, count is 2                        <<< With 'return 3' 
commented out again.

> this case, doesn't matter.  What matters is that $(( func2() )) runs
> in the current shell, not in a subshell.

Right, I understand that.  So long as there's math involved, there's no 
subshell.

BTW, if I modify func2:

func2 ()
{
functions -t func2
echo func2 here!
#integer count=2
#return 3
}

... the same pattern results:

'# integer count=2':  and count is unchanged in func1, var = 0.

'integer count=2':  and count is unchanged in func1 but var = 2. I force 
it to be a math op.  The 'integer count' seems to be a new variable so 
the original is unchanged, that's clear.

'# integer count=2':  and var is '2'.  func2 retains the previous return 
value even thou the active line is commented out (again). Unless I'm 
making some kind of mistake, it seems buggy to me, the rv of func2 is 
floating unchanged somewhere.  When I explicitly change it, it changes, 
but then the new value floats there too. I've done it a dozen times, I 
don't think I'm wrong.  I have to start a new shell to erase that rv.  
Or is this some bizarre local issue here?

Tho I do wonder why '$(( ))' is 'local' and '$( )' spawns a subshell, 
dunno, maybe there's a good reason for that but one might wish for some 
method of asking '$( )' to run 'local' too, so that my original issue 
wouldn't come up.  I appreciate that this kind of thing is so deep in 
the way the shell runs that it might be next to impossible even if 
anyone but me was interested in it. Still: "setopt NO_SUBSHELL" might be 
very useful.




>
> In case it wasn't obvious, $(( func2 )) treats func2 as the name of a
> variable, not a function, so it doesn't run func2 at all.

Yes, that's clear.  '$(( func2() ))' is a math operation -- grab the 
return value, which is whatever the last math operation was. '$(( func2 
))' is ... basically nothing.


>

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