> what shoud echo $((foo++)) do? Give out an error message hopefully (and it does), because the expression (bar+14+moo*2)++ makes no semantical sense in terms of arithmetic expressions the bash uses, because ++ only can apply to locations and (bar+14+moo*2) evaluates to a number no matter the contents of bar and moo…
So lets assume that bash only executes things like ((<exp>++)) if and only if <exp> is a proper location (that is a string that is no number). We know that in this case only two things can happen. (1) <exp> refers directly to a number value or (2) <exp> refers to another string that is a location In case of (1) the result is obvious. In case of (2) my complain applies. > 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. I wouldn’t consider this a special case, because the syntax and semantics are almost identical to those of C (with the exception of the numerous exceptions like recursive expansions etc.). So bash should also implement the same lvalue, rvalue semantics that C does. Of course, with the appropriate adjustments needed to fit in the context of a string typed command line interpreter :) > would perhaps have been a better idea, but it is this way for so long that I > doubt it will change. Mhh… > PS: you can perhaps use name reference instead > moo=1; > declare -n foo=bar bar=moo > echo $((foo++)) > echo $moo Thank you. > Am 23.04.2017 um 15:19 schrieb Pierre Gaston <pierre.gas...@gmail.com>: > > > > On Sun, Apr 23, 2017 at 4:07 PM, Florian Mayer <mayerflor...@me.com > <mailto: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 > >
