REOPEN: Perform arithmetic expansion inside parameter expansion? [duplicate]
Clash Royale CLAN TAG#URR8PPP
up vote
-2
down vote
favorite
This question already has an answer here:
Double and triple substitution in bash and zsh
5 answers
Please reopen. This is not a duplicate, because here I am asking why it doesn't work, not just for workaround.
Bash manual says that
tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution (done in a left-to-right fashion)
If I understand http://unix.stackexchange.com/a/270324/674 correctly, "left-to-right" means that "brace expansion; tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution" have the same priority.
So is it possible to use arithmetic expansion inside parameter expansion? (i.e. one level recursion)
If no, why can't arithmetic expansion work inside parameter expansion, given that "tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution (done in a left-to-right fashion)"?
If yes, how?
For example,
$ set hello world
$ echo $2
world
$ echo $$((1+1))
bash: $$((1+1)): bad substitution
I hope to
- first expand
$((1+1))in$$((1+1))to2. and - then
$2toworld.
Thanks.
bash
asked Sep 26 '17 at 23:12
Tim
23k66225408
marked as duplicate by αғsýιη, Stephen Rauch, SatÃ…Â Katsura, Kusalananda bash
Users with the  bash badge can single-handedly close bash questions as duplicates and reopen them as needed.
StackExchange.ready(function()
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();
);
);
);
Sep 27 '17 at 6:53
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
 |Â
show 3 more comments
up vote
-2
down vote
favorite
This question already has an answer here:
Double and triple substitution in bash and zsh
5 answers
Please reopen. This is not a duplicate, because here I am asking why it doesn't work, not just for workaround.
Bash manual says that
tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution (done in a left-to-right fashion)
If I understand http://unix.stackexchange.com/a/270324/674 correctly, "left-to-right" means that "brace expansion; tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution" have the same priority.
So is it possible to use arithmetic expansion inside parameter expansion? (i.e. one level recursion)
If no, why can't arithmetic expansion work inside parameter expansion, given that "tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution (done in a left-to-right fashion)"?
If yes, how?
For example,
$ set hello world
$ echo $2
world
$ echo $$((1+1))
bash: $$((1+1)): bad substitution
I hope to
- first expand
$((1+1))in$$((1+1))to2. and - then
$2toworld.
Thanks.
bash
asked Sep 26 '17 at 23:12
Tim
23k66225408
marked as duplicate by αғsýιη, Stephen Rauch, SatÃ…Â Katsura, Kusalananda bash
Users with the  bash badge can single-handedly close bash questions as duplicates and reopen them as needed.
StackExchange.ready(function()
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();
);
);
);
Sep 27 '17 at 6:53
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
Because that's not left-to-right. What you're hoping for is not left-to-right. The documentation says expansion is done left-to-right. You want something else, I don't know what to call it; maybe "from-inside-to-outside" parsing? Please read the documentation and clear the English words in it like "left-to-right" so you can answer your own questions. (This question contains its own answer.)
– Wildcard
Sep 27 '17 at 0:40
What the manual tries to say is: parameter expansion is done before arithmetic expansion, so it works the other way around:set 6 7; echo $(($1*$2))will work as expected.
– Philippos
Sep 27 '17 at 6:53
3
The whole sentence of your citation is: "The order of expansions is: brace expansion; tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution (done in a left-to-right fashion); word splitting; and pathname expansion.
– Philippos
Sep 27 '17 at 10:44
3
@Tim they don't (and can't) have the same priority. Parameter expansion is done before arithmetic expansion, so$(( 1 + $a:-5 ))will work but$ $((1 + 1))won't. To solve your requirement you would needset 4 5; b=$((1+1)); echo $!b
– roaima
Sep 27 '17 at 12:14
1
Just to add more confusion in Tim's head,$@:$((1+1)):1would work though ;-) Hint:$@:1+1:1would work too.
– xhienne
Sep 27 '17 at 12:51
 |Â
show 3 more comments
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
This question already has an answer here:
Double and triple substitution in bash and zsh
5 answers
Please reopen. This is not a duplicate, because here I am asking why it doesn't work, not just for workaround.
Bash manual says that
tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution (done in a left-to-right fashion)
If I understand http://unix.stackexchange.com/a/270324/674 correctly, "left-to-right" means that "brace expansion; tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution" have the same priority.
So is it possible to use arithmetic expansion inside parameter expansion? (i.e. one level recursion)
If no, why can't arithmetic expansion work inside parameter expansion, given that "tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution (done in a left-to-right fashion)"?
If yes, how?
For example,
$ set hello world
$ echo $2
world
$ echo $$((1+1))
bash: $$((1+1)): bad substitution
I hope to
- first expand
$((1+1))in$$((1+1))to2. and - then
$2toworld.
Thanks.
bash
asked Sep 26 '17 at 23:12
Tim
23k66225408
This question already has an answer here:
Double and triple substitution in bash and zsh
5 answers
Please reopen. This is not a duplicate, because here I am asking why it doesn't work, not just for workaround.
Bash manual says that
tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution (done in a left-to-right fashion)
If I understand http://unix.stackexchange.com/a/270324/674 correctly, "left-to-right" means that "brace expansion; tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution" have the same priority.
So is it possible to use arithmetic expansion inside parameter expansion? (i.e. one level recursion)
If no, why can't arithmetic expansion work inside parameter expansion, given that "tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution (done in a left-to-right fashion)"?
If yes, how?
For example,
$ set hello world
$ echo $2
world
$ echo $$((1+1))
bash: $$((1+1)): bad substitution
I hope to
- first expand
$((1+1))in$$((1+1))to2. and - then
$2toworld.
Thanks.
This question already has an answer here:
Double and triple substitution in bash and zsh
5 answers
bash
bash
asked Sep 26 '17 at 23:12
Tim
23k66225408
asked Sep 26 '17 at 23:12
Tim
23k66225408
edited Sep 27 '17 at 12:20
asked Sep 26 '17 at 23:12
Tim
23k66225408
asked Sep 26 '17 at 23:12
Tim
23k66225408
asked Sep 26 '17 at 23:12
Tim
23k66225408
23k66225408
marked as duplicate by αғsýιη, Stephen Rauch, SatÃ…Â Katsura, Kusalananda bash
Users with the  bash badge can single-handedly close bash questions as duplicates and reopen them as needed.
StackExchange.ready(function()
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();
);
);
);
Sep 27 '17 at 6:53
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by αғsýιη, Stephen Rauch, SatÃ…Â Katsura, Kusalananda bash
Users with the  bash badge can single-handedly close bash questions as duplicates and reopen them as needed.
StackExchange.ready(function()
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();
);
);
);
Sep 27 '17 at 6:53
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
Because that's not left-to-right. What you're hoping for is not left-to-right. The documentation says expansion is done left-to-right. You want something else, I don't know what to call it; maybe "from-inside-to-outside" parsing? Please read the documentation and clear the English words in it like "left-to-right" so you can answer your own questions. (This question contains its own answer.)
– Wildcard
Sep 27 '17 at 0:40
What the manual tries to say is: parameter expansion is done before arithmetic expansion, so it works the other way around:set 6 7; echo $(($1*$2))will work as expected.
– Philippos
Sep 27 '17 at 6:53
3
The whole sentence of your citation is: "The order of expansions is: brace expansion; tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution (done in a left-to-right fashion); word splitting; and pathname expansion.
– Philippos
Sep 27 '17 at 10:44
3
@Tim they don't (and can't) have the same priority. Parameter expansion is done before arithmetic expansion, so$(( 1 + $a:-5 ))will work but$ $((1 + 1))won't. To solve your requirement you would needset 4 5; b=$((1+1)); echo $!b
– roaima
Sep 27 '17 at 12:14
1
Just to add more confusion in Tim's head,$@:$((1+1)):1would work though ;-) Hint:$@:1+1:1would work too.
– xhienne
Sep 27 '17 at 12:51
 |Â
show 3 more comments
2
Because that's not left-to-right. What you're hoping for is not left-to-right. The documentation says expansion is done left-to-right. You want something else, I don't know what to call it; maybe "from-inside-to-outside" parsing? Please read the documentation and clear the English words in it like "left-to-right" so you can answer your own questions. (This question contains its own answer.)
– Wildcard
Sep 27 '17 at 0:40
What the manual tries to say is: parameter expansion is done before arithmetic expansion, so it works the other way around:set 6 7; echo $(($1*$2))will work as expected.
– Philippos
Sep 27 '17 at 6:53
3
The whole sentence of your citation is: "The order of expansions is: brace expansion; tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution (done in a left-to-right fashion); word splitting; and pathname expansion.
– Philippos
Sep 27 '17 at 10:44
3
@Tim they don't (and can't) have the same priority. Parameter expansion is done before arithmetic expansion, so$(( 1 + $a:-5 ))will work but$ $((1 + 1))won't. To solve your requirement you would needset 4 5; b=$((1+1)); echo $!b
– roaima
Sep 27 '17 at 12:14
1
Just to add more confusion in Tim's head,$@:$((1+1)):1would work though ;-) Hint:$@:1+1:1would work too.
– xhienne
Sep 27 '17 at 12:51
2
2
Because that's not left-to-right. What you're hoping for is not left-to-right. The documentation says expansion is done left-to-right. You want something else, I don't know what to call it; maybe "from-inside-to-outside" parsing? Please read the documentation and clear the English words in it like "left-to-right" so you can answer your own questions. (This question contains its own answer.)
– Wildcard
Sep 27 '17 at 0:40
Because that's not left-to-right. What you're hoping for is not left-to-right. The documentation says expansion is done left-to-right. You want something else, I don't know what to call it; maybe "from-inside-to-outside" parsing? Please read the documentation and clear the English words in it like "left-to-right" so you can answer your own questions. (This question contains its own answer.)
– Wildcard
Sep 27 '17 at 0:40
What the manual tries to say is: parameter expansion is done before arithmetic expansion, so it works the other way around:
set 6 7; echo $(($1*$2)) will work as expected.– Philippos
Sep 27 '17 at 6:53
What the manual tries to say is: parameter expansion is done before arithmetic expansion, so it works the other way around:
set 6 7; echo $(($1*$2)) will work as expected.– Philippos
Sep 27 '17 at 6:53
3
3
The whole sentence of your citation is: "The order of expansions is: brace expansion; tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution (done in a left-to-right fashion); word splitting; and pathname expansion.
– Philippos
Sep 27 '17 at 10:44
The whole sentence of your citation is: "The order of expansions is: brace expansion; tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution (done in a left-to-right fashion); word splitting; and pathname expansion.
– Philippos
Sep 27 '17 at 10:44
3
3
@Tim they don't (and can't) have the same priority. Parameter expansion is done before arithmetic expansion, so
$(( 1 + $a:-5 )) will work but $ $((1 + 1)) won't. To solve your requirement you would need set 4 5; b=$((1+1)); echo $!b– roaima
Sep 27 '17 at 12:14
@Tim they don't (and can't) have the same priority. Parameter expansion is done before arithmetic expansion, so
$(( 1 + $a:-5 )) will work but $ $((1 + 1)) won't. To solve your requirement you would need set 4 5; b=$((1+1)); echo $!b– roaima
Sep 27 '17 at 12:14
1
1
Just to add more confusion in Tim's head,
$@:$((1+1)):1 would work though ;-) Hint: $@:1+1:1 would work too.– xhienne
Sep 27 '17 at 12:51
Just to add more confusion in Tim's head,
$@:$((1+1)):1 would work though ;-) Hint: $@:1+1:1 would work too.– xhienne
Sep 27 '17 at 12:51
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
3
down vote
You need either eval or indirection for this:
eval echo $$((1+1))
index=$((1+1))
echo $!index
answered Sep 26 '17 at 23:25
Hauke Laging
53.7k1282130
Thanks. I understand the two ways you mentioned. Why can't arithmetic expansion work inside parameter expansion, given that "tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution (done in a left-to-right fashion)"?
– Tim
Sep 26 '17 at 23:42
2
@Tim, because that wouldn't be left-to-right.
– Wildcard
Sep 27 '17 at 0:37
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
You need either eval or indirection for this:
eval echo $$((1+1))
index=$((1+1))
echo $!index
answered Sep 26 '17 at 23:25
Hauke Laging
53.7k1282130
Thanks. I understand the two ways you mentioned. Why can't arithmetic expansion work inside parameter expansion, given that "tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution (done in a left-to-right fashion)"?
– Tim
Sep 26 '17 at 23:42
2
@Tim, because that wouldn't be left-to-right.
– Wildcard
Sep 27 '17 at 0:37
add a comment |Â
up vote
3
down vote
You need either eval or indirection for this:
eval echo $$((1+1))
index=$((1+1))
echo $!index
answered Sep 26 '17 at 23:25
Hauke Laging
53.7k1282130
Thanks. I understand the two ways you mentioned. Why can't arithmetic expansion work inside parameter expansion, given that "tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution (done in a left-to-right fashion)"?
– Tim
Sep 26 '17 at 23:42
2
@Tim, because that wouldn't be left-to-right.
– Wildcard
Sep 27 '17 at 0:37
add a comment |Â
up vote
3
down vote
up vote
3
down vote
You need either eval or indirection for this:
eval echo $$((1+1))
index=$((1+1))
echo $!index
answered Sep 26 '17 at 23:25
Hauke Laging
53.7k1282130
You need either eval or indirection for this:
eval echo $$((1+1))
index=$((1+1))
echo $!index
answered Sep 26 '17 at 23:25
Hauke Laging
53.7k1282130
answered Sep 26 '17 at 23:25
Hauke Laging
53.7k1282130
answered Sep 26 '17 at 23:25
Hauke Laging
53.7k1282130
answered Sep 26 '17 at 23:25
Hauke Laging
53.7k1282130
53.7k1282130
Thanks. I understand the two ways you mentioned. Why can't arithmetic expansion work inside parameter expansion, given that "tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution (done in a left-to-right fashion)"?
– Tim
Sep 26 '17 at 23:42
2
@Tim, because that wouldn't be left-to-right.
– Wildcard
Sep 27 '17 at 0:37
add a comment |Â
Thanks. I understand the two ways you mentioned. Why can't arithmetic expansion work inside parameter expansion, given that "tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution (done in a left-to-right fashion)"?
– Tim
Sep 26 '17 at 23:42
2
@Tim, because that wouldn't be left-to-right.
– Wildcard
Sep 27 '17 at 0:37
Thanks. I understand the two ways you mentioned. Why can't arithmetic expansion work inside parameter expansion, given that "tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution (done in a left-to-right fashion)"?
– Tim
Sep 26 '17 at 23:42
Thanks. I understand the two ways you mentioned. Why can't arithmetic expansion work inside parameter expansion, given that "tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution (done in a left-to-right fashion)"?
– Tim
Sep 26 '17 at 23:42
2
2
@Tim, because that wouldn't be left-to-right.
– Wildcard
Sep 27 '17 at 0:37
@Tim, because that wouldn't be left-to-right.
– Wildcard
Sep 27 '17 at 0:37
add a comment |Â
2
Because that's not left-to-right. What you're hoping for is not left-to-right. The documentation says expansion is done left-to-right. You want something else, I don't know what to call it; maybe "from-inside-to-outside" parsing? Please read the documentation and clear the English words in it like "left-to-right" so you can answer your own questions. (This question contains its own answer.)
– Wildcard
Sep 27 '17 at 0:40
What the manual tries to say is: parameter expansion is done before arithmetic expansion, so it works the other way around:
set 6 7; echo $(($1*$2))will work as expected.– Philippos
Sep 27 '17 at 6:53
3
The whole sentence of your citation is: "The order of expansions is: brace expansion; tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution (done in a left-to-right fashion); word splitting; and pathname expansion.
– Philippos
Sep 27 '17 at 10:44
3
@Tim they don't (and can't) have the same priority. Parameter expansion is done before arithmetic expansion, so
$(( 1 + $a:-5 ))will work but$ $((1 + 1))won't. To solve your requirement you would needset 4 5; b=$((1+1)); echo $!b– roaima
Sep 27 '17 at 12:14
1
Just to add more confusion in Tim's head,
$@:$((1+1)):1would work though ;-) Hint:$@:1+1:1would work too.– xhienne
Sep 27 '17 at 12:51