Re: Curious case of arithmetic expansion

[Florian Mayer]

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