Re: saved from prince of eval

[Bart Schaefer <schaefer@brasslantern.com>]

On Nov 8,  9:07am, Ray Andrews wrote:
}
} I'm trying to get some mileage out of the '(e)' flag, but it frustrates me

If you haven't already, you should read through the "Rules" section
under "Parameter Expansion".  "man zshexpn" and search for "Rules".

In particular the last sentence of the #1 rule "Nested Substitution":

     ... Note  that,  unless  the  '(P)'
     flag is present, the flags and any subscripts apply directly to
     the value of the nested substitution; for example,  the  expan-
     sion ${${foo}} behaves exactly the same as ${foo}.

Then note that (e) isn't applied until nearly the end of the procedure,
at rule #10.

So in this expression:

} 1:    $ foo="${(e)${array}[${top}, ${bottom}]}"

First ${array} expands, and then [$[top},${bottom}] applies to that
value -- which isn't an array, it's a string, so the subscripts are
extracting a range of characters from that string.  Finally that
substring is re-evaluated (but probably is nothing interesting).

What you needed was

    foo="${(e)${(P)array}[${top}, ${bottom}]}"

though as previously explained that doesn't work as you want for an
associative $array.
 
} If I insert the literal name of the array in place of "${array}" 
} everything is fine.

In "${(e)the_literal_array[${top}, ${bottom}]}" the subscript applies
directly the array (because there is no nested expression).

} But this works:
} 
} 2:    $ bar='\$${array}[${top}, ${bottom}]'
} 3:    $ foo="${(e)$(print -R "${(e)${bar}}")}"

This describes that you've got at least three levels of unexpanded
references:

- whatever array is named by $array contains unexpanded references
  in the array elements
- bar contains unexpanded references to $array, $top, and $bottom
- expanding \$ in $bar becomes an unexpanded reference to the array
  named by $array with the subscript expanded from $top and $bottom

I have no idea why you want to put yourself in that situation, but if
you have three unexpanded levels then you need three re-evaluations:

- the inner (e) flag
- $(...), which you can think of as:  eval "..." | read
- the outer (e) flag
 
} I'm not sure how to interpret it tho. Is 'print' doing the work here, or 
} is print a bystander as a nested use of '(e)' works?

In this case the "print" is doing nothing except provide the standard
output to be captured from $(...).

You can replace the $(...) with another ${(e)...} to get your third
needed re-evaluation, and you don't need the inermost ${ } around bar:

    foo="${(e)${(e)${(e)bar}}}"

Again it's pretty ridiculous that you're doing anything like this in
the first place.

}       $ foo='stranger'
}       $ bar='echo howdy $foo'
}       $ eval baz=$bar; echo $baz
}     zsh: command not found: howdy << Ferkrissakes

Well, think about it a bit harder.  Or better yet, "setopt xtrace" and
watch what is happening.

torch% foo='stranger'
torch% bar='echo howdy $foo'
torch% setopt xtrace
torch% eval baz=$bar; echo $baz
+zsh:4> eval 'baz=echo howdy $foo'
+(eval):1> baz=echo howdy stranger
zsh: command not found: howdy

The assignment becomes "baz=echo" which is prefixed to the command
"howdy stranger".