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Indirectly expand variables in shell

8

2

I need to indirectly reference a variable in the bash shell.

I basically want to what you can do in make by writing $($(var)).

I have tried using $$var which would be the most straight forward solution in bash but then I get this error:

bash: $$var: bad substitution

Is there a way to do this?

What I am trying to do is to iterate over all the arguments ($1, $2, $3, ...) to a program using an iteration variable and I cannot do this without indirection.

bash shell environment-variables variable arguments

share|improve this question

asked Sep 15 '15 at 15:37

wefwefa3wefwefa3

5152722

add a comment | 

8

2

I need to indirectly reference a variable in the bash shell.

I basically want to what you can do in make by writing $($(var)).

I have tried using $$var which would be the most straight forward solution in bash but then I get this error:

bash: $$var: bad substitution

Is there a way to do this?

What I am trying to do is to iterate over all the arguments ($1, $2, $3, ...) to a program using an iteration variable and I cannot do this without indirection.

bash shell environment-variables variable arguments

share|improve this question

asked Sep 15 '15 at 15:37

wefwefa3wefwefa3

5152722

add a comment | 

I need to indirectly reference a variable in the bash shell.

I basically want to what you can do in make by writing $($(var)).

I have tried using $$var which would be the most straight forward solution in bash but then I get this error:

bash: $$var: bad substitution

Is there a way to do this?

What I am trying to do is to iterate over all the arguments ($1, $2, $3, ...) to a program using an iteration variable and I cannot do this without indirection.

bash shell environment-variables variable arguments

share|improve this question

asked Sep 15 '15 at 15:37

wefwefa3wefwefa3

5152722

bash: $$var: bad substitution

share|improve this question

asked Sep 15 '15 at 15:37

wefwefa3wefwefa3

5152722

share|improve this question

asked Sep 15 '15 at 15:37

wefwefa3wefwefa3

5152722

asked Sep 15 '15 at 15:37

wefwefa3wefwefa3

5152722

asked Sep 15 '15 at 15:37

wefwefa3wefwefa3

5152722

2 Answers
2

active

oldest

votes

13

If you have var1=foo and foo=bar, you can get bar by saying $!var1. However, if you want to iterate over the positional parameters, it's almost certainly better to do

for i in "$@"; do
 # something
done

share|improve this answer

answered Sep 15 '15 at 15:43

Tom HuntTom Hunt

6,70521536

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Is there a way to add multiple levels of indirection?
  
  – wefwefa3
  Sep 17 '15 at 12:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Not that I can quickly find. The obvious guess of $!!var1 doesn't work. However, you can always do it manually, e.g. tmp=$!var1; echo $!tmp.
  
  – Tom Hunt
  Sep 17 '15 at 15:05
  
  

  
  
  
  

add a comment | 

2

Using /bin/bash:

foo=bar
test=foo
echo $!test

# prints -> bar

share|improve this answer

answered Feb 14 '17 at 5:53

Nick GrealyNick Grealy

1266

add a comment | 

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13

If you have var1=foo and foo=bar, you can get bar by saying $!var1. However, if you want to iterate over the positional parameters, it's almost certainly better to do

for i in "$@"; do
 # something
done

share|improve this answer

answered Sep 15 '15 at 15:43

Tom HuntTom Hunt

6,70521536

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Is there a way to add multiple levels of indirection?
  
  – wefwefa3
  Sep 17 '15 at 12:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Not that I can quickly find. The obvious guess of $!!var1 doesn't work. However, you can always do it manually, e.g. tmp=$!var1; echo $!tmp.
  
  – Tom Hunt
  Sep 17 '15 at 15:05
  
  

  
  
  
  

add a comment | 

13

If you have var1=foo and foo=bar, you can get bar by saying $!var1. However, if you want to iterate over the positional parameters, it's almost certainly better to do

for i in "$@"; do
 # something
done

share|improve this answer

answered Sep 15 '15 at 15:43

Tom HuntTom Hunt

6,70521536

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Is there a way to add multiple levels of indirection?
  
  – wefwefa3
  Sep 17 '15 at 12:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Not that I can quickly find. The obvious guess of $!!var1 doesn't work. However, you can always do it manually, e.g. tmp=$!var1; echo $!tmp.
  
  – Tom Hunt
  Sep 17 '15 at 15:05
  
  

  
  
  
  

add a comment | 

If you have var1=foo and foo=bar, you can get bar by saying $!var1. However, if you want to iterate over the positional parameters, it's almost certainly better to do

for i in "$@"; do
 # something
done

share|improve this answer

answered Sep 15 '15 at 15:43

Tom HuntTom Hunt

6,70521536

for i in "$@"; do
 # something
done

share|improve this answer

answered Sep 15 '15 at 15:43

Tom HuntTom Hunt

6,70521536

answered Sep 15 '15 at 15:43

Tom HuntTom Hunt

6,70521536

answered Sep 15 '15 at 15:43

Tom HuntTom Hunt

6,70521536

2

Using /bin/bash:

foo=bar
test=foo
echo $!test

# prints -> bar

share|improve this answer

answered Feb 14 '17 at 5:53

Nick GrealyNick Grealy

1266

add a comment | 

2

Using /bin/bash:

foo=bar
test=foo
echo $!test

# prints -> bar

share|improve this answer

answered Feb 14 '17 at 5:53

Nick GrealyNick Grealy

1266

add a comment | 

Using /bin/bash:

foo=bar
test=foo
echo $!test

# prints -> bar

share|improve this answer

answered Feb 14 '17 at 5:53

Nick GrealyNick Grealy

1266

foo=bar
test=foo
echo $!test

# prints -> bar

share|improve this answer

answered Feb 14 '17 at 5:53

Nick GrealyNick Grealy

1266

answered Feb 14 '17 at 5:53

Nick GrealyNick Grealy

1266

answered Feb 14 '17 at 5:53

Nick GrealyNick Grealy

1266