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How to protect bash function from being overridden?

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12

3

In the bash shell, we can define a function f with

f() echo Hello; 

and then redeclare/override it, without any error or warning messages, with

f() echo Bye; 

I believe there is a way to protect functions from being overridden in this way.

bash bash-functions

share|improve this question

edited Sep 27 at 9:52

kyb

asked Sep 26 at 7:23

kybkyb

30622 silver badges1515 bronze badges

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  the same as with variables, with typeset -r: typeset -rf f.
  
  – mosvy
  Sep 26 at 7:25
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  or readonly -f f
  
  – mosvy
  Sep 26 at 7:26
  

  
  
  
  

add a comment
 | 

12

3

In the bash shell, we can define a function f with

f() echo Hello; 

and then redeclare/override it, without any error or warning messages, with

f() echo Bye; 

I believe there is a way to protect functions from being overridden in this way.

bash bash-functions

share|improve this question

edited Sep 27 at 9:52

kyb

asked Sep 26 at 7:23

kybkyb

30622 silver badges1515 bronze badges

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  the same as with variables, with typeset -r: typeset -rf f.
  
  – mosvy
  Sep 26 at 7:25
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  or readonly -f f
  
  – mosvy
  Sep 26 at 7:26
  

  
  
  
  

add a comment
 | 

In the bash shell, we can define a function f with

f() echo Hello; 

and then redeclare/override it, without any error or warning messages, with

f() echo Bye; 

I believe there is a way to protect functions from being overridden in this way.

bash bash-functions

share|improve this question

edited Sep 27 at 9:52

kyb

asked Sep 26 at 7:23

kybkyb

30622 silver badges1515 bronze badges

f() echo Hello; 

f() echo Bye; 

share|improve this question

edited Sep 27 at 9:52

kyb

asked Sep 26 at 7:23

kybkyb

30622 silver badges1515 bronze badges

share|improve this question

edited Sep 27 at 9:52

kyb

asked Sep 26 at 7:23

kybkyb

30622 silver badges1515 bronze badges

asked Sep 26 at 7:23

kybkyb

30622 silver badges1515 bronze badges

asked Sep 26 at 7:23

kybkyb

30622 silver badges1515 bronze badges

1 Answer
1

active

oldest

votes

25

You may declare f as a read-only function using readonly -f f or declare -g -r -f f (readonly is equivalent to declare -g -r). It's the -f option to these built-in utilities that makes them act on f as the name of a function, rather than on the variable f.

$ f() echo Hello; 
$ readonly -f f
$ f() echo Bye; 
bash: f: readonly function
$ unset -f f
bash: unset: f: cannot unset: readonly function
$ f
Hello

As you can see, making the function read-only not only protects it from getting overridden, but also protects it from being unset (removed completely).

---

Currently (as of bash-5.0.11), trying to modify a read-only function would not terminate the shell if one is using the errexit shell option (set -e). Chet, the bash maintainer, says that this is an oversight and that it will be changed with the next release.

share|improve this answer

edited Oct 1 at 20:20

answered Sep 26 at 7:25

Kusalananda♦Kusalananda

175k2020 gold badges333333 silver badges539539 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Attempt to override function produces message bash: f: readonly function and non-zero status code, but does not exit if errexit option enabled.
  
  – kyb
  Sep 27 at 9:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @kyb I noticed this as well. I'm not certain it's a bug in bash, but I will ask on one of the bash mailing lists to be sure.
  
  – Kusalananda♦
  Sep 27 at 10:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  good, please update your answer when you will be sure about this behavior.
  
  – kyb
  Sep 27 at 14:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @kyb Both Stephane Chazelas and Greg Wooledge weighed in on that question, both with plausible explanations. Stephane suggests that bash only exits when set -e is in effect when POSIX requires it to (and readonly -f is not POSIX). Greg points out that the bash manual never mentions "failure in function declaration" as reason for errexit to trigger an exit (unless a function declaration counts as a compound command, which he's pretty sure it does not). The thread is ongoing here: lists.gnu.org/archive/html/help-bash/2019-09/msg00039.html
  
  – Kusalananda♦
  Sep 27 at 20:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @kyb I'm also noticing that you never say anything about errexit or set -e in your question.
  
  – Kusalananda♦
  Sep 27 at 20:58
  

  
  
  
  

 | 
show 2 more comments

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25

You may declare f as a read-only function using readonly -f f or declare -g -r -f f (readonly is equivalent to declare -g -r). It's the -f option to these built-in utilities that makes them act on f as the name of a function, rather than on the variable f.

$ f() echo Hello; 
$ readonly -f f
$ f() echo Bye; 
bash: f: readonly function
$ unset -f f
bash: unset: f: cannot unset: readonly function
$ f
Hello

As you can see, making the function read-only not only protects it from getting overridden, but also protects it from being unset (removed completely).

---

Currently (as of bash-5.0.11), trying to modify a read-only function would not terminate the shell if one is using the errexit shell option (set -e). Chet, the bash maintainer, says that this is an oversight and that it will be changed with the next release.

share|improve this answer

edited Oct 1 at 20:20

answered Sep 26 at 7:25

Kusalananda♦Kusalananda

175k2020 gold badges333333 silver badges539539 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Attempt to override function produces message bash: f: readonly function and non-zero status code, but does not exit if errexit option enabled.
  
  – kyb
  Sep 27 at 9:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @kyb I noticed this as well. I'm not certain it's a bug in bash, but I will ask on one of the bash mailing lists to be sure.
  
  – Kusalananda♦
  Sep 27 at 10:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  good, please update your answer when you will be sure about this behavior.
  
  – kyb
  Sep 27 at 14:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @kyb Both Stephane Chazelas and Greg Wooledge weighed in on that question, both with plausible explanations. Stephane suggests that bash only exits when set -e is in effect when POSIX requires it to (and readonly -f is not POSIX). Greg points out that the bash manual never mentions "failure in function declaration" as reason for errexit to trigger an exit (unless a function declaration counts as a compound command, which he's pretty sure it does not). The thread is ongoing here: lists.gnu.org/archive/html/help-bash/2019-09/msg00039.html
  
  – Kusalananda♦
  Sep 27 at 20:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @kyb I'm also noticing that you never say anything about errexit or set -e in your question.
  
  – Kusalananda♦
  Sep 27 at 20:58
  

  
  
  
  

 | 
show 2 more comments

25

You may declare f as a read-only function using readonly -f f or declare -g -r -f f (readonly is equivalent to declare -g -r). It's the -f option to these built-in utilities that makes them act on f as the name of a function, rather than on the variable f.

$ f() echo Hello; 
$ readonly -f f
$ f() echo Bye; 
bash: f: readonly function
$ unset -f f
bash: unset: f: cannot unset: readonly function
$ f
Hello

As you can see, making the function read-only not only protects it from getting overridden, but also protects it from being unset (removed completely).

---

Currently (as of bash-5.0.11), trying to modify a read-only function would not terminate the shell if one is using the errexit shell option (set -e). Chet, the bash maintainer, says that this is an oversight and that it will be changed with the next release.

share|improve this answer

edited Oct 1 at 20:20

answered Sep 26 at 7:25

Kusalananda♦Kusalananda

175k2020 gold badges333333 silver badges539539 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Attempt to override function produces message bash: f: readonly function and non-zero status code, but does not exit if errexit option enabled.
  
  – kyb
  Sep 27 at 9:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @kyb I noticed this as well. I'm not certain it's a bug in bash, but I will ask on one of the bash mailing lists to be sure.
  
  – Kusalananda♦
  Sep 27 at 10:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  good, please update your answer when you will be sure about this behavior.
  
  – kyb
  Sep 27 at 14:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @kyb Both Stephane Chazelas and Greg Wooledge weighed in on that question, both with plausible explanations. Stephane suggests that bash only exits when set -e is in effect when POSIX requires it to (and readonly -f is not POSIX). Greg points out that the bash manual never mentions "failure in function declaration" as reason for errexit to trigger an exit (unless a function declaration counts as a compound command, which he's pretty sure it does not). The thread is ongoing here: lists.gnu.org/archive/html/help-bash/2019-09/msg00039.html
  
  – Kusalananda♦
  Sep 27 at 20:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @kyb I'm also noticing that you never say anything about errexit or set -e in your question.
  
  – Kusalananda♦
  Sep 27 at 20:58
  

  
  
  
  

 | 
show 2 more comments

You may declare f as a read-only function using readonly -f f or declare -g -r -f f (readonly is equivalent to declare -g -r). It's the -f option to these built-in utilities that makes them act on f as the name of a function, rather than on the variable f.

$ f() echo Hello; 
$ readonly -f f
$ f() echo Bye; 
bash: f: readonly function
$ unset -f f
bash: unset: f: cannot unset: readonly function
$ f
Hello

As you can see, making the function read-only not only protects it from getting overridden, but also protects it from being unset (removed completely).

---

Currently (as of bash-5.0.11), trying to modify a read-only function would not terminate the shell if one is using the errexit shell option (set -e). Chet, the bash maintainer, says that this is an oversight and that it will be changed with the next release.

share|improve this answer

edited Oct 1 at 20:20

answered Sep 26 at 7:25

Kusalananda♦Kusalananda

175k2020 gold badges333333 silver badges539539 bronze badges

$ f() echo Hello; 
$ readonly -f f
$ f() echo Bye; 
bash: f: readonly function
$ unset -f f
bash: unset: f: cannot unset: readonly function
$ f
Hello

share|improve this answer

edited Oct 1 at 20:20

answered Sep 26 at 7:25

Kusalananda♦Kusalananda

175k2020 gold badges333333 silver badges539539 bronze badges

answered Sep 26 at 7:25

Kusalananda♦Kusalananda

175k2020 gold badges333333 silver badges539539 bronze badges

answered Sep 26 at 7:25

Kusalananda♦Kusalananda

175k2020 gold badges333333 silver badges539539 bronze badges