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Error when running ((x++)) as root

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3

I make the following bash script in file temp.sh with the following content:

age=0;
((age++));

When I run it as a normal user, it runs fine.

But when i run it as root I get error:

./temp.sh: 4: ./temp.sh: age++: not found

Why is that?

root

share|improve this question

edited 4 hours ago

Hermann Ingjaldsson

asked 6 hours ago

Hermann IngjaldssonHermann Ingjaldsson

79741430

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Are you using the same shell and settings for both users ? What's the output of echo $SHELL as non-root and as root ?
  
  – Httqm
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  its /bin/bash both as normal user and as root. But Jesse answered.
  
  – Hermann Ingjaldsson
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  The exact output of ./temp.sh: 4: ./temp.sh: age++: not found is generated by dash running an script called as ./temp.sh. That seems to be your root shell.
  
  – Isaac
  2 hours ago
  

  
  
  
  

add a comment | 

3

I make the following bash script in file temp.sh with the following content:

age=0;
((age++));

When I run it as a normal user, it runs fine.

But when i run it as root I get error:

./temp.sh: 4: ./temp.sh: age++: not found

Why is that?

root

share|improve this question

edited 4 hours ago

Hermann Ingjaldsson

asked 6 hours ago

Hermann IngjaldssonHermann Ingjaldsson

79741430

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Are you using the same shell and settings for both users ? What's the output of echo $SHELL as non-root and as root ?
  
  – Httqm
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  its /bin/bash both as normal user and as root. But Jesse answered.
  
  – Hermann Ingjaldsson
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  The exact output of ./temp.sh: 4: ./temp.sh: age++: not found is generated by dash running an script called as ./temp.sh. That seems to be your root shell.
  
  – Isaac
  2 hours ago
  

  
  
  
  

add a comment | 

I make the following bash script in file temp.sh with the following content:

age=0;
((age++));

When I run it as a normal user, it runs fine.

But when i run it as root I get error:

./temp.sh: 4: ./temp.sh: age++: not found

Why is that?

root

share|improve this question

edited 4 hours ago

Hermann Ingjaldsson

asked 6 hours ago

Hermann IngjaldssonHermann Ingjaldsson

79741430

age=0;
((age++));

share|improve this question

edited 4 hours ago

Hermann Ingjaldsson

asked 6 hours ago

Hermann IngjaldssonHermann Ingjaldsson

79741430

share|improve this question

edited 4 hours ago

Hermann Ingjaldsson

asked 6 hours ago

Hermann IngjaldssonHermann Ingjaldsson

79741430

asked 6 hours ago

Hermann IngjaldssonHermann Ingjaldsson

79741430

asked 6 hours ago

Hermann IngjaldssonHermann Ingjaldsson

79741430

1 Answer
1

active

oldest

votes

8

In the absence of a hashbang, /bin/sh is likely being used. Some POSIX shells do support the ++ and -- operators, and ((...)) for arithmetic evaluations, but are not required to.

Since you have not included a hashbang in your example I will assume you are not using one and therefore your script is likely running in a POSIX shell that does not support said operator. Such a shell would interpret ((age++)) as the age++ command being run inside two nested sub-shells.

When you run it as a "normal" user it is likely being interpreted by bash or another shell that does support said operator and ((...)).

Related: Which shell interpreter runs a script with no shebang?

To fix this you can add a hashbang to your script:

#!/bin/bash
age=0
((age++))

Note: You do not need to terminate lines with ; in bash/shell.

---

To make your script portable to all POSIX shells you can use the following syntax:

age=$((age + 1))
age=$((age += 1))

share|improve this answer

edited 3 hours ago

answered 6 hours ago

Jesse_bJesse_b

15.5k33675

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  Or use the standard sh syntax: age=$((age + 1)), or : "$((age += 1))"
  
  – Stéphane Chazelas
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  The exact output of ./temp.sh: 4: ./temp.sh: age++: not found is generated by dash running an script called as ./temp.sh. That seems to be the root shell.
  
  – Isaac
  2 hours ago
  

  
  
  
  

add a comment | 

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8

In the absence of a hashbang, /bin/sh is likely being used. Some POSIX shells do support the ++ and -- operators, and ((...)) for arithmetic evaluations, but are not required to.

Since you have not included a hashbang in your example I will assume you are not using one and therefore your script is likely running in a POSIX shell that does not support said operator. Such a shell would interpret ((age++)) as the age++ command being run inside two nested sub-shells.

When you run it as a "normal" user it is likely being interpreted by bash or another shell that does support said operator and ((...)).

Related: Which shell interpreter runs a script with no shebang?

To fix this you can add a hashbang to your script:

#!/bin/bash
age=0
((age++))

Note: You do not need to terminate lines with ; in bash/shell.

---

To make your script portable to all POSIX shells you can use the following syntax:

age=$((age + 1))
age=$((age += 1))

share|improve this answer

edited 3 hours ago

answered 6 hours ago

Jesse_bJesse_b

15.5k33675

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  Or use the standard sh syntax: age=$((age + 1)), or : "$((age += 1))"
  
  – Stéphane Chazelas
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  The exact output of ./temp.sh: 4: ./temp.sh: age++: not found is generated by dash running an script called as ./temp.sh. That seems to be the root shell.
  
  – Isaac
  2 hours ago
  

  
  
  
  

add a comment | 

8

In the absence of a hashbang, /bin/sh is likely being used. Some POSIX shells do support the ++ and -- operators, and ((...)) for arithmetic evaluations, but are not required to.

Since you have not included a hashbang in your example I will assume you are not using one and therefore your script is likely running in a POSIX shell that does not support said operator. Such a shell would interpret ((age++)) as the age++ command being run inside two nested sub-shells.

When you run it as a "normal" user it is likely being interpreted by bash or another shell that does support said operator and ((...)).

Related: Which shell interpreter runs a script with no shebang?

To fix this you can add a hashbang to your script:

#!/bin/bash
age=0
((age++))

Note: You do not need to terminate lines with ; in bash/shell.

---

To make your script portable to all POSIX shells you can use the following syntax:

age=$((age + 1))
age=$((age += 1))

share|improve this answer

edited 3 hours ago

answered 6 hours ago

Jesse_bJesse_b

15.5k33675

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  Or use the standard sh syntax: age=$((age + 1)), or : "$((age += 1))"
  
  – Stéphane Chazelas
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  The exact output of ./temp.sh: 4: ./temp.sh: age++: not found is generated by dash running an script called as ./temp.sh. That seems to be the root shell.
  
  – Isaac
  2 hours ago
  

  
  
  
  

add a comment | 

In the absence of a hashbang, /bin/sh is likely being used. Some POSIX shells do support the ++ and -- operators, and ((...)) for arithmetic evaluations, but are not required to.

Since you have not included a hashbang in your example I will assume you are not using one and therefore your script is likely running in a POSIX shell that does not support said operator. Such a shell would interpret ((age++)) as the age++ command being run inside two nested sub-shells.

When you run it as a "normal" user it is likely being interpreted by bash or another shell that does support said operator and ((...)).

Related: Which shell interpreter runs a script with no shebang?

To fix this you can add a hashbang to your script:

#!/bin/bash
age=0
((age++))

Note: You do not need to terminate lines with ; in bash/shell.

---

To make your script portable to all POSIX shells you can use the following syntax:

age=$((age + 1))
age=$((age += 1))

share|improve this answer

edited 3 hours ago

answered 6 hours ago

Jesse_bJesse_b

15.5k33675

#!/bin/bash
age=0
((age++))

age=$((age + 1))
age=$((age += 1))

share|improve this answer

edited 3 hours ago

answered 6 hours ago

Jesse_bJesse_b

15.5k33675

answered 6 hours ago

Jesse_bJesse_b

15.5k33675

answered 6 hours ago

Jesse_bJesse_b

15.5k33675