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Re: "typeset -p" and no_GLOBAL_EXPORT, other misc.



2024-03-11 21:13:32 -0700, Bart Schaefer:
[...]
> To even start fixing this mess, we'd have to explain what a function
> considers "local" and what "global".  Are all "inherited" scopes
> global, so local is the current scope only?  Gaah.
[...]

IMO, typeset -g should only mean: don't make it local, do not
instantiate a new variable in the current scope.

And we certainly don't want bash's typeset -g behaviour where it
means *the outermost scope*, which in shells with dynamic
scoping is essentially what "global" means, but it's not
particularly useful to single it out.

For instance, in bash:

bash-5.3$ integer() { typeset -gi -- "$@"; }
bash-5.3$ integer i=1+1
bash-5.3$ echo $i
2

fine, but:

bash-5.3$ f() { local i; integer i=2+2; echo "$i"; }
bash-5.3$ f

bash-5.3$ echo $i
4

That "integer" changed the type of the global (outer-most) i
variable instead of that of its caller.

IOW, most of the usages of "typeset -g" like with that "integer"
above is when you want to change attributes of a variable (set
the type with typeset which bash renamed to declare) without
wanting to make it local, like export or readonly do in
Bourne-like shells (except zsh for the latter), not change the
type of the variable in the outermost scope which I can't think
why you would want to.

In mksh or yash, readonly var is like typeset -gr var.

In bash, readonly var is like mksh/yash's typeset -gr var, but
unlike its own typeset -gr:

~$ bash -c 'b() { local a=1; a; echo $a; }; a() { readonly a=2; echo $a; }; a=0; b; echo $a'
2
2
0
~$ bash -c 'b() { local a=1; a; echo $a; }; a() { typeset -gr a=2; echo $a; }; a=0; b; echo $a'
1
1
2

~$ mksh -c 'b() { local a=1; a; echo $a; }; a() { readonly a=2; echo $a; }; a=0; b; echo $a'
2
2
0
~$ mksh -c 'b() { local a=1; a; echo $a; }; a() { typeset -gr a=2; echo $a; }; a=0; b; echo $a'
2
2
0


In zsh, readonly var, when not emulating other shells is more
like typeset -r:

~$ zsh -c 'b() { local a=1; a; echo $a; }; a() { typeset -gr a=2; echo $a; }; a=0; b; echo $a'
2
2
0
~$ zsh -c 'b() { local a=1; a; echo $a; }; a() { readonly a=2; echo $a; }; a=0; b; echo $a'
2
1
0
~$ zsh -c 'b() { local a=1; a; echo $a; }; a() { typeset -r a=2; echo $a; }; a=0; b; echo $a'
2
1
0

ksh93 does static scoping, so things are different there anyway,
there's only a global scope and one local scope not a stack of
them, so -g can mean "global" without ambiguity and functions
cannot access the variables of their caller unless they're
exported.

~$ ksh -c 'function b { typeset a=1; a; echo $a; }; function a { typeset -gr a=2; echo $a; }; a=0; b; echo $a'
2
1
2
~$ ksh -c 'function b { typeset a=1; a; echo $a; }; function a { readonly a=2; echo $a; }; a=0; b; echo $a'
2
1
2
~$ ksh -c 'function b { typeset -x a=1; a; echo $a; }; function a { readonly a=2; echo $a; }; a=0; b; echo $a'
2
1
0
~$ ksh -c 'function b { typeset -x a=1; a; echo $a; a=3; }; function a { readonly a=2; echo $a; }; a=0; b; echo $a'
2
1
0

readonly is still not equivalent to typeset -gr as it affects
the variable in the scope where it currently exists (whether
local, global or exported from an ancestor), not only in the
global scope:

$ ksh -c 'function a { typeset a=1; typeset -gr a; echo $a; }; a=0; a; echo $a; a=2'
1
0
ksh: a: is read only
$ ksh -c 'function a { typeset a=1; readonly a; echo $a; }; a=0; a; echo $a; a=2'
1
0

-- 
Stephane




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