> And in my opinion those are not correct...
So in a scenario like the mentioned $ foo=bar; bar=moo; moo=123 the line $ ((foo++)); echo $foo $bar $moo should actually evaluate to „bar moo 124“ or at least say something like „error, can not execute side effects, because I can’t keep track of what I expanded in order to get a number" which it does not… I know that (()) probably was never intended for such a thing, but it’d be a nice-to-have addition, while being certainly doable :) > Am 23.04.2017 um 15:07 schrieb Florian Mayer <mayerflor...@me.com>: > > 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... > >> Am 23.04.2017 um 14:43 schrieb Pierre Gaston <pierre.gas...@gmail.com >> <mailto:pierre.gas...@gmail.com>>: >> >> >> >> On Sun, Apr 23, 2017 at 3:28 PM, Florian Mayer <mayerflor...@me.com >> <mailto:mayerflor...@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 >> >
