Why does ! operator with set -e always succeed even for truthy return values?
I use a system where scripts utilize set -e:
#!/bin/sh
set -e
echo 1
true
echo 2
true
echo 3
false
echo notreached
If I do this:
#!/bin/sh
set -e
echo 1
true
echo 2
true
echo 3
! false
echo reached
...the last line is reached. However, if I do this:
#!/bin/sh
set -e
echo 1
true
echo 2
true
echo 3
! true
echo reached
...the last line is reached even though ! true should be false.
More testing:
user@linux:~$ ! false && echo ok
ok
user@linux:~$ ! true && echo ok
user@linux:~$
...so in these cases, the ! operator works correctly.
However, it does not work properly with set -e. Why?
Related: http://mywiki.wooledge.org/BashFAQ/105 ...although that does not answer my question.
bash shell-script
asked 1 hour ago
juhistjuhist
1333
add a comment |
I use a system where scripts utilize set -e:
#!/bin/sh
set -e
echo 1
true
echo 2
true
echo 3
false
echo notreached
If I do this:
#!/bin/sh
set -e
echo 1
true
echo 2
true
echo 3
! false
echo reached
...the last line is reached. However, if I do this:
#!/bin/sh
set -e
echo 1
true
echo 2
true
echo 3
! true
echo reached
...the last line is reached even though ! true should be false.
More testing:
user@linux:~$ ! false && echo ok
ok
user@linux:~$ ! true && echo ok
user@linux:~$
...so in these cases, the ! operator works correctly.
However, it does not work properly with set -e. Why?
Related: http://mywiki.wooledge.org/BashFAQ/105 ...although that does not answer my question.
bash shell-script
asked 1 hour ago
juhistjuhist
1333
add a comment |
I use a system where scripts utilize set -e:
#!/bin/sh
set -e
echo 1
true
echo 2
true
echo 3
false
echo notreached
If I do this:
#!/bin/sh
set -e
echo 1
true
echo 2
true
echo 3
! false
echo reached
...the last line is reached. However, if I do this:
#!/bin/sh
set -e
echo 1
true
echo 2
true
echo 3
! true
echo reached
...the last line is reached even though ! true should be false.
More testing:
user@linux:~$ ! false && echo ok
ok
user@linux:~$ ! true && echo ok
user@linux:~$
...so in these cases, the ! operator works correctly.
However, it does not work properly with set -e. Why?
Related: http://mywiki.wooledge.org/BashFAQ/105 ...although that does not answer my question.
bash shell-script
asked 1 hour ago
juhistjuhist
1333
I use a system where scripts utilize set -e:
#!/bin/sh
set -e
echo 1
true
echo 2
true
echo 3
false
echo notreached
If I do this:
#!/bin/sh
set -e
echo 1
true
echo 2
true
echo 3
! false
echo reached
...the last line is reached. However, if I do this:
#!/bin/sh
set -e
echo 1
true
echo 2
true
echo 3
! true
echo reached
...the last line is reached even though ! true should be false.
More testing:
user@linux:~$ ! false && echo ok
ok
user@linux:~$ ! true && echo ok
user@linux:~$
...so in these cases, the ! operator works correctly.
However, it does not work properly with set -e. Why?
Related: http://mywiki.wooledge.org/BashFAQ/105 ...although that does not answer my question.
bash shell-script
bash shell-script
asked 1 hour ago
juhistjuhist
1333
asked 1 hour ago
juhistjuhist
1333
asked 1 hour ago
juhistjuhist
1333
asked 1 hour ago
juhistjuhist
1333
asked 1 hour ago
juhistjuhist
1333
1333
add a comment |
add a comment |
1 Answer
1
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oldest
votes
Because that's what the standard says:
-e
When this option is on, when any command fails, the shell immediately shall exit, [...], with the following exceptions:
- The -e setting shall be ignored when executing the compound list following the while, until, if, or elif reserved word, a pipeline
beginning with the ! reserved word, or any command of an AND-OR list
other than the last.
http://pubs.opengroup.org/onlinepubs/9699919799.2018edition/utilities/V3_chap02.html#set
Bash's manual also points this out, with somewhat different wording.
I think that entry in BashFAQ doesn't even try to spell out the specifics, the phrase "These rules are extremely convoluted" makes that pretty clear. Though the linked pages seem to have more than one would want to know.
answered 1 hour ago
ilkkachuilkkachu
57k785158
add a comment |
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1 Answer
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1 Answer
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votes
Because that's what the standard says:
-e
When this option is on, when any command fails, the shell immediately shall exit, [...], with the following exceptions:
- The -e setting shall be ignored when executing the compound list following the while, until, if, or elif reserved word, a pipeline
beginning with the ! reserved word, or any command of an AND-OR list
other than the last.
http://pubs.opengroup.org/onlinepubs/9699919799.2018edition/utilities/V3_chap02.html#set
Bash's manual also points this out, with somewhat different wording.
I think that entry in BashFAQ doesn't even try to spell out the specifics, the phrase "These rules are extremely convoluted" makes that pretty clear. Though the linked pages seem to have more than one would want to know.
answered 1 hour ago
ilkkachuilkkachu
57k785158
add a comment |
Because that's what the standard says:
-e
When this option is on, when any command fails, the shell immediately shall exit, [...], with the following exceptions:
- The -e setting shall be ignored when executing the compound list following the while, until, if, or elif reserved word, a pipeline
beginning with the ! reserved word, or any command of an AND-OR list
other than the last.
http://pubs.opengroup.org/onlinepubs/9699919799.2018edition/utilities/V3_chap02.html#set
Bash's manual also points this out, with somewhat different wording.
I think that entry in BashFAQ doesn't even try to spell out the specifics, the phrase "These rules are extremely convoluted" makes that pretty clear. Though the linked pages seem to have more than one would want to know.
answered 1 hour ago
ilkkachuilkkachu
57k785158
add a comment |
Because that's what the standard says:
-e
When this option is on, when any command fails, the shell immediately shall exit, [...], with the following exceptions:
- The -e setting shall be ignored when executing the compound list following the while, until, if, or elif reserved word, a pipeline
beginning with the ! reserved word, or any command of an AND-OR list
other than the last.
http://pubs.opengroup.org/onlinepubs/9699919799.2018edition/utilities/V3_chap02.html#set
Bash's manual also points this out, with somewhat different wording.
I think that entry in BashFAQ doesn't even try to spell out the specifics, the phrase "These rules are extremely convoluted" makes that pretty clear. Though the linked pages seem to have more than one would want to know.
answered 1 hour ago
ilkkachuilkkachu
57k785158
Because that's what the standard says:
-e
When this option is on, when any command fails, the shell immediately shall exit, [...], with the following exceptions:
- The -e setting shall be ignored when executing the compound list following the while, until, if, or elif reserved word, a pipeline
beginning with the ! reserved word, or any command of an AND-OR list
other than the last.
http://pubs.opengroup.org/onlinepubs/9699919799.2018edition/utilities/V3_chap02.html#set
Bash's manual also points this out, with somewhat different wording.
I think that entry in BashFAQ doesn't even try to spell out the specifics, the phrase "These rules are extremely convoluted" makes that pretty clear. Though the linked pages seem to have more than one would want to know.
answered 1 hour ago
ilkkachuilkkachu
57k785158
edited 1 hour ago
answered 1 hour ago
ilkkachuilkkachu
57k785158
answered 1 hour ago
ilkkachuilkkachu
57k785158
answered 1 hour ago
ilkkachuilkkachu
57k785158
57k785158
add a comment |
add a comment |
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