Re: Curious case of arithmetic expansion

[Pierre Gaston]

On Sun, Apr 23, 2017 at 3:28 PM, Florian Mayer <mayerflorian@me.com> wrote:

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...

 

This is indirection indeed, but in arithmetic evaluation it's not.

Quoting the manual:

"The value of a variable is evaluated as an  arithmetic  _expression_
 when  it  is  referenced, or when a variable which has been given the integer attribute using declare -i is assigned a value. "

Consider this:

foo=1+3

echo $foo

echo $((foo++))

echo $foo