What version of ksh are you referring to? I have installed
Homebrew's ksh93:
% ksh --version
version sh (AT&T Research) 93u+m/1.0.10 2024-08-01
In ksh, old style functions (f () { … }) have dynamic scoping and new style functions (function f { … }) have static scoping.
To my great surprise, with dynamic scoping all functions apparently share the same set of variables. In Zsh, when a nested function defines a local variable already defined in an enclosing function it gets its own variable that hides the existing one. In ksh, the already existing variable is apparently simply reused. The following quote from the man page confirms this understanding:
Functions defined with the name() syntax and functions defined with the function name syntax that are invoked with the .
special built-in are executed in the caller's environment and share all variables, options and traps with the caller.
The following function also seems to confirm it:
g1() {
typeset var=g1;
typeset -n ref;
g2() {
typeset var=g2;
ref=var;
echo "g2: var=$var";
echo "g2: ref=$ref";
}
g2;
echo "g1: var=$var";
echo "g1: ref=$ref";
}
% g1
g2: var=g2
g2: ref=g2
g1: var=g2
g1: ref=g2
Notice that in g1 var=g2 and not g1. I first thought that in g1 ref was still referring to the now dead var variable from g2 but no, there is simply a single var variable. The typeref in g2 doesn't create a new variable, it just updates the existing one. Apparently with dynamic scoping all ksh typeref statements are equivalent to Zsh's typeref -g statements.
Below is the same function using static scoping:
function h1 {
typeset var=h1;
typeset -n ref;
function h2 {
typeset var=h2;
ref=var;
echo "h2: var=$var";
echo "h2: ref=$ref";
}
h2;
echo "h1: var=$var";
echo "h1: ref=$ref";
}
% h1
h2: var=h2
h2: ref=var
h1: var=h1
named-refs.ksh[41]: h1: line 39: ref: no reference name
Here, what happens is that the ref=var in h2 doesn't initialize the ref defined in h1 because that ref isn't visible in h2 due to static scoping. Instead it creates a local string variable named ref, which leads to the output h2: ref=var. Then, in h1, the echo $ref fails because ref isn't initialized.
Let me know if I missed something but if all of this is correct, then all the different cases that we have discussed have no equivalent in ksh and thus "works like ksh" doesn't make much sense for them.
Philippe