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3

1

This question already has an answer here:

• 
  Why does my shell script choke on whitespace or other special characters?
  
  4 answers
  
  


• 
  Security implications of forgetting to quote a variable in bash/POSIX shells
  
  3 answers
  
  


• 
  When is double-quoting necessary?
  
  1 answer
  
  

Is this a Bash bug?

$ mkdir test && cd test && echo "a" > "some.file"
test$ echo '*'
*
test$ TEST=$(echo '*')
test$ echo $TEST
some.file

Why is the second output the resolution of * into (all) filenames instead of just a literal * output? Looks like a bug in Bash?

Tried on Ubuntu 18.04, bash version 4.4.19(1)-release. Expect it will be the same on other OS'es.

bash shell quoting command-substitution variable-substitution

share|improve this question

edited Jun 3 at 18:00

Gilles

559k13411471657

asked Jun 3 at 0:16

RoelRoel

1267

marked as duplicate by muru, Scott, ilkkachu bash
Users with the  bash badge can single-handedly close bash questions as duplicates and reopen them as needed.

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Jun 3 at 19:08

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
  

  
  
  
  
  
  

  
  
  General procedure when finding bugs in Unix: 1) Check out Single Unix Specification online for relevant definitions and rationales. 2) Submit a bugreport to austingroupbugs.net 3) Submit bugreport to the developer (with a patch if possible).
  
  – DannyNiu
  Jun 3 at 1:10
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @DannyNiu: Your comment is not very practical.  Documents like POSIX and the Single Unix Specification do not consist of millions and millions of examples; they contain explanations (with, maybe, a few examples).  Those documents are of little help to somebody who doesn’t understand what is happening and why; reading them is like learning to drive by reading a roadmap. And this question isn’t asking what to do.  If you want to post a question, “What should I do if I’ve found a bug in a widely used software product?”, and post an answer, go ahead.
  
  – G-Man
  Jun 3 at 4:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @G-Man I did find DannyNiu's answer helpful in terms of the information provided, "just in case".
  
  – Roel
  Jun 3 at 4:28
  

  
  
  
  

add a comment | 

3

1

This question already has an answer here:

• 
  Why does my shell script choke on whitespace or other special characters?
  
  4 answers
  
  


• 
  Security implications of forgetting to quote a variable in bash/POSIX shells
  
  3 answers
  
  


• 
  When is double-quoting necessary?
  
  1 answer
  
  

Is this a Bash bug?

$ mkdir test && cd test && echo "a" > "some.file"
test$ echo '*'
*
test$ TEST=$(echo '*')
test$ echo $TEST
some.file

Why is the second output the resolution of * into (all) filenames instead of just a literal * output? Looks like a bug in Bash?

Tried on Ubuntu 18.04, bash version 4.4.19(1)-release. Expect it will be the same on other OS'es.

bash shell quoting command-substitution variable-substitution

share|improve this question

edited Jun 3 at 18:00

Gilles

559k13411471657

asked Jun 3 at 0:16

RoelRoel

1267

marked as duplicate by muru, Scott, ilkkachu bash
Users with the  bash badge can single-handedly close bash questions as duplicates and reopen them as needed.

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StackExchange.helpers.removeMessages();

);
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);
Jun 3 at 19:08

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
  

  
  
  
  
  
  

  
  
  General procedure when finding bugs in Unix: 1) Check out Single Unix Specification online for relevant definitions and rationales. 2) Submit a bugreport to austingroupbugs.net 3) Submit bugreport to the developer (with a patch if possible).
  
  – DannyNiu
  Jun 3 at 1:10
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @DannyNiu: Your comment is not very practical.  Documents like POSIX and the Single Unix Specification do not consist of millions and millions of examples; they contain explanations (with, maybe, a few examples).  Those documents are of little help to somebody who doesn’t understand what is happening and why; reading them is like learning to drive by reading a roadmap. And this question isn’t asking what to do.  If you want to post a question, “What should I do if I’ve found a bug in a widely used software product?”, and post an answer, go ahead.
  
  – G-Man
  Jun 3 at 4:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @G-Man I did find DannyNiu's answer helpful in terms of the information provided, "just in case".
  
  – Roel
  Jun 3 at 4:28
  

  
  
  
  

add a comment | 

This question already has an answer here:

• 
  Why does my shell script choke on whitespace or other special characters?
  
  4 answers
  
  


• 
  Security implications of forgetting to quote a variable in bash/POSIX shells
  
  3 answers
  
  


• 
  When is double-quoting necessary?
  
  1 answer
  
  

Is this a Bash bug?

$ mkdir test && cd test && echo "a" > "some.file"
test$ echo '*'
*
test$ TEST=$(echo '*')
test$ echo $TEST
some.file

Why is the second output the resolution of * into (all) filenames instead of just a literal * output? Looks like a bug in Bash?

Tried on Ubuntu 18.04, bash version 4.4.19(1)-release. Expect it will be the same on other OS'es.

bash shell quoting command-substitution variable-substitution

share|improve this question

edited Jun 3 at 18:00

Gilles

559k13411471657

asked Jun 3 at 0:16

RoelRoel

1267

$ mkdir test && cd test && echo "a" > "some.file"
test$ echo '*'
*
test$ TEST=$(echo '*')
test$ echo $TEST
some.file

share|improve this question

edited Jun 3 at 18:00

Gilles

559k13411471657

asked Jun 3 at 0:16

RoelRoel

1267

share|improve this question

edited Jun 3 at 18:00

Gilles

559k13411471657

asked Jun 3 at 0:16

RoelRoel

1267

edited Jun 3 at 18:00

Gilles

559k13411471657

edited Jun 3 at 18:00

Gilles

559k13411471657

asked Jun 3 at 0:16

RoelRoel

1267

asked Jun 3 at 0:16

RoelRoel

1267

1 Answer
1

active

oldest

votes

13

No, it is not a bug. You have shown that

echo '*'

will produce a literal *. Hence when you substitute this output, as per the following command

TEST=$(echo '*')

it will put * into the variable $TEST. Then when you

echo $TEST

the glob will expand here. You can verify this by running this last command, changing directories, then running it again.

You will get the * output if you say

echo "$TEST"

as explained here,
the double quotes allow the variable to be expanded
but prevent the glob from expanding.

share|improve this answer

edited Jun 3 at 4:37

G-Man

14.7k94175

answered Jun 3 at 0:22

SparhawkSparhawk

10.8k849102

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Aha, I see. Thank you. I missed the fact that the echo * does not have quotes and thus resolves. It's harder to see when you have it inside a complex script :) Makes sense now.
  
  – Roel
  Jun 3 at 1:36
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Roel FWIW, after a few years of experience you will start to treat $TEST instead of "$TEST" as a code smell and it becomes a bit easier to see. It's still not much of a smell though so it will still trip you up sometimes
  
  – slebetman
  Jun 3 at 9:24
  

  
  
  
  


• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  
  And this is why my general approach is, if it starts with a dollar, quote it :). Yes, you CAN learn the specific cases where you don't have to - but it's almost always harmless to quote them in cases where you don't have to, so you may as well just quote your variables everywhere!
  
  – Muzer
  Jun 3 at 10:15
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @slebetman: And for the rest of us without this instinct, shellcheck will do the complaining ( shellcheck.net ).
  
  – Piskvor
  Jun 3 at 12:42
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Roel many decades ago, I learned -- the hard way -- that 99.99999% of the time it's your fault, not a bug in the compiler or interpreter... :)
  
  – RonJohn
  Jun 3 at 14:56
  
  

  
  
  
  

 | 
show 1 more comment

13

No, it is not a bug. You have shown that

echo '*'

will produce a literal *. Hence when you substitute this output, as per the following command

TEST=$(echo '*')

it will put * into the variable $TEST. Then when you

echo $TEST

the glob will expand here. You can verify this by running this last command, changing directories, then running it again.

You will get the * output if you say

echo "$TEST"

as explained here,
the double quotes allow the variable to be expanded
but prevent the glob from expanding.

share|improve this answer

edited Jun 3 at 4:37

G-Man

14.7k94175

answered Jun 3 at 0:22

SparhawkSparhawk

10.8k849102

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Aha, I see. Thank you. I missed the fact that the echo * does not have quotes and thus resolves. It's harder to see when you have it inside a complex script :) Makes sense now.
  
  – Roel
  Jun 3 at 1:36
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Roel FWIW, after a few years of experience you will start to treat $TEST instead of "$TEST" as a code smell and it becomes a bit easier to see. It's still not much of a smell though so it will still trip you up sometimes
  
  – slebetman
  Jun 3 at 9:24
  

  
  
  
  


• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  
  And this is why my general approach is, if it starts with a dollar, quote it :). Yes, you CAN learn the specific cases where you don't have to - but it's almost always harmless to quote them in cases where you don't have to, so you may as well just quote your variables everywhere!
  
  – Muzer
  Jun 3 at 10:15
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @slebetman: And for the rest of us without this instinct, shellcheck will do the complaining ( shellcheck.net ).
  
  – Piskvor
  Jun 3 at 12:42
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Roel many decades ago, I learned -- the hard way -- that 99.99999% of the time it's your fault, not a bug in the compiler or interpreter... :)
  
  – RonJohn
  Jun 3 at 14:56
  
  

  
  
  
  

 | 
show 1 more comment

13

No, it is not a bug. You have shown that

echo '*'

will produce a literal *. Hence when you substitute this output, as per the following command

TEST=$(echo '*')

it will put * into the variable $TEST. Then when you

echo $TEST

the glob will expand here. You can verify this by running this last command, changing directories, then running it again.

You will get the * output if you say

echo "$TEST"

as explained here,
the double quotes allow the variable to be expanded
but prevent the glob from expanding.

share|improve this answer

edited Jun 3 at 4:37

G-Man

14.7k94175

answered Jun 3 at 0:22

SparhawkSparhawk

10.8k849102

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Aha, I see. Thank you. I missed the fact that the echo * does not have quotes and thus resolves. It's harder to see when you have it inside a complex script :) Makes sense now.
  
  – Roel
  Jun 3 at 1:36
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Roel FWIW, after a few years of experience you will start to treat $TEST instead of "$TEST" as a code smell and it becomes a bit easier to see. It's still not much of a smell though so it will still trip you up sometimes
  
  – slebetman
  Jun 3 at 9:24
  

  
  
  
  


• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  
  And this is why my general approach is, if it starts with a dollar, quote it :). Yes, you CAN learn the specific cases where you don't have to - but it's almost always harmless to quote them in cases where you don't have to, so you may as well just quote your variables everywhere!
  
  – Muzer
  Jun 3 at 10:15
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @slebetman: And for the rest of us without this instinct, shellcheck will do the complaining ( shellcheck.net ).
  
  – Piskvor
  Jun 3 at 12:42
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Roel many decades ago, I learned -- the hard way -- that 99.99999% of the time it's your fault, not a bug in the compiler or interpreter... :)
  
  – RonJohn
  Jun 3 at 14:56
  
  

  
  
  
  

 | 
show 1 more comment

No, it is not a bug. You have shown that

echo '*'

will produce a literal *. Hence when you substitute this output, as per the following command

TEST=$(echo '*')

it will put * into the variable $TEST. Then when you

echo $TEST

the glob will expand here. You can verify this by running this last command, changing directories, then running it again.

You will get the * output if you say

echo "$TEST"

as explained here,
the double quotes allow the variable to be expanded
but prevent the glob from expanding.

share|improve this answer

edited Jun 3 at 4:37

G-Man

14.7k94175

answered Jun 3 at 0:22

SparhawkSparhawk

10.8k849102

echo '*'

TEST=$(echo '*')

echo $TEST

echo "$TEST"

share|improve this answer

edited Jun 3 at 4:37

G-Man

14.7k94175

answered Jun 3 at 0:22

SparhawkSparhawk

10.8k849102

edited Jun 3 at 4:37

G-Man

14.7k94175

edited Jun 3 at 4:37

G-Man

14.7k94175

answered Jun 3 at 0:22

SparhawkSparhawk

10.8k849102

answered Jun 3 at 0:22

SparhawkSparhawk

10.8k849102