What I’m saying is, that if bash does recursively apply expansion
mechanisms on the identifiers until it can retrieve a number,
it should do it symmetrically. That is,
it should remember what chain of expansion had been necessary for
a particular number to appear at the end of the expansion.
So instead of
124 moo 123
The echo command should produce
bar moo 124
(The expansion chain here was foo->bar->moo->123)
> It's because it's not really indirection, rather the content of the variable
> is evaluated:
No it is really indirection. Bash even has a special (and very limited) syntax
for that.
Consider
$ foo=bar; bar=moo
You can get the string „moo“ through foo by using
$ echo ${!foo}
$ echo ${!!foo} # or something else does not work, though...
>
>
> It prints "124 moo 123" no?
>
> It's because it's not really indirection, rather the content of the variable
> is evaluated:
>
> $ foo=1+2;echo $((foo))
> 3
>
> In the last case it evaluates the value of foo, 123, and then increment the
> value of foo and foo becomes 124
>
