On Sun, Apr 23, 2017 at 4:07 PM, Florian Mayer <mayerflor...@me.com> wrote:
> It does not matter, how this construct in this particular context is > called. > The difference between $(()) and (()) is that $(()) actually expands to > something > whereas (()) just executes a C-like expression. In ((<expression>)) > <expression> can also > include assignments, as the bash manual that you properly cited, also > elaborates on. > You can do, for example, things like > $ foo=2 > $ ((foo+=100)) # fo is now 102 > $ ((++(foo++))) > or even > $ ((foo++, foo++, foo++, foo++, foo+=100)) > and (oh boy why) even > $ foo=(123 321) > $ ((foo[0]++, foo[1]—)) > > So I might have chosen the wrong subject text for this mail, > but again, it does not matter whether those constructs actually *expand* to > some string > or not. The side effects are what matter here. And in my opinion those are > not correct... > I understand what you want, I'm just explaining the result you get and why it doesn't do what you want. As it is, a variable can contain the name of another variable but it can also contain any arbitrary string that is a correct arithmetic expression. So if you have: foo=bar+14+baz baz='moo*2' moo=1 what shoud echo $((foo++)) do? Your case is only really a special case, arguably having only indirection instead of what we have would perhaps have been a better idea, but it is this way for so long that I doubt it will change. PS: you can perhaps use name reference instead moo=1; declare -n foo=bar bar=moo echo $((foo++)) echo $moo
