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Bash: What does “masking return values” mean?
2019 Community Moderator ElectionHow to name a file in the deepest level of a directory treeWhat does || mean in bash?Printf formatting with variable format - what does this var reference?What does $'r' mean?What is return code 1 for git-rebase? What are other values it could return?How does one parse $!i (and what does it mean)?Should I double quote these parameter expansions?“Make sure not to read and write the same file in the same pipeline”How to return a values from a method which prints its out putWhy not to export variables on the same line you assign them?
shellcheck generated the following warning
SC2155: Declare and assign separately to avoid masking return
values
For this line of code
local key_value=$(echo "$current_line" | mawk '/.+=.+/ print $1 ')
What does "masking return values" mean, and how does it pertain to the aforementioned warning?
bash shell-script shellcheck
asked 4 hours ago
InquisitorInquisitor
304
New contributor
Inquisitor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
shellcheck generated the following warning
SC2155: Declare and assign separately to avoid masking return
values
For this line of code
local key_value=$(echo "$current_line" | mawk '/.+=.+/ print $1 ')
What does "masking return values" mean, and how does it pertain to the aforementioned warning?
bash shell-script shellcheck
asked 4 hours ago
InquisitorInquisitor
304
New contributor
Inquisitor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
shellcheck generated the following warning
SC2155: Declare and assign separately to avoid masking return
values
For this line of code
local key_value=$(echo "$current_line" | mawk '/.+=.+/ print $1 ')
What does "masking return values" mean, and how does it pertain to the aforementioned warning?
bash shell-script shellcheck
asked 4 hours ago
InquisitorInquisitor
304
New contributor
Inquisitor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
shellcheck generated the following warning
SC2155: Declare and assign separately to avoid masking return
values
For this line of code
local key_value=$(echo "$current_line" | mawk '/.+=.+/ print $1 ')
What does "masking return values" mean, and how does it pertain to the aforementioned warning?
bash shell-script shellcheck
bash shell-script shellcheck
asked 4 hours ago
InquisitorInquisitor
304
New contributor
Inquisitor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
InquisitorInquisitor
304
New contributor
Inquisitor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
Inquisitor
asked 4 hours ago
InquisitorInquisitor
304
New contributor
Inquisitor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
InquisitorInquisitor
304
asked 4 hours ago
InquisitorInquisitor
304
304
New contributor
Inquisitor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Inquisitor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Inquisitor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
When you declare a variable as either local or exported that in itself is a command that will return success or not.
$ var=$(false)
$ echo $?
1
$ export var=$(false)
$ echo $?
0
So if you wanted to act on the return value of your command (echo "$current_line" | mawk '/.+=.+/ print $1 '), you would be unable to since it's going to exit with 0 as long as the local declaration succeeds (which is almost always will).
In order to avoid this it suggests declaring separately and then assigning:
local key_value
key_value=$(echo "$current_line" | mawk '/.+=.+/ print $1 ')
This is a shellcheck rule I frequently ignore and IMO is safe to ignore as long as you know you aren't trying to act on the return value of that variable declaration.
You can ignore it by adding the following to the top of your script (Below the hashbang of course):
# shellcheck disable=SC2155
answered 4 hours ago
Jesse_bJesse_b
13.4k23370
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
When you declare a variable as either local or exported that in itself is a command that will return success or not.
$ var=$(false)
$ echo $?
1
$ export var=$(false)
$ echo $?
0
So if you wanted to act on the return value of your command (echo "$current_line" | mawk '/.+=.+/ print $1 '), you would be unable to since it's going to exit with 0 as long as the local declaration succeeds (which is almost always will).
In order to avoid this it suggests declaring separately and then assigning:
local key_value
key_value=$(echo "$current_line" | mawk '/.+=.+/ print $1 ')
This is a shellcheck rule I frequently ignore and IMO is safe to ignore as long as you know you aren't trying to act on the return value of that variable declaration.
You can ignore it by adding the following to the top of your script (Below the hashbang of course):
# shellcheck disable=SC2155
answered 4 hours ago
Jesse_bJesse_b
13.4k23370
add a comment |
When you declare a variable as either local or exported that in itself is a command that will return success or not.
$ var=$(false)
$ echo $?
1
$ export var=$(false)
$ echo $?
0
So if you wanted to act on the return value of your command (echo "$current_line" | mawk '/.+=.+/ print $1 '), you would be unable to since it's going to exit with 0 as long as the local declaration succeeds (which is almost always will).
In order to avoid this it suggests declaring separately and then assigning:
local key_value
key_value=$(echo "$current_line" | mawk '/.+=.+/ print $1 ')
This is a shellcheck rule I frequently ignore and IMO is safe to ignore as long as you know you aren't trying to act on the return value of that variable declaration.
You can ignore it by adding the following to the top of your script (Below the hashbang of course):
# shellcheck disable=SC2155
answered 4 hours ago
Jesse_bJesse_b
13.4k23370
add a comment |
When you declare a variable as either local or exported that in itself is a command that will return success or not.
$ var=$(false)
$ echo $?
1
$ export var=$(false)
$ echo $?
0
So if you wanted to act on the return value of your command (echo "$current_line" | mawk '/.+=.+/ print $1 '), you would be unable to since it's going to exit with 0 as long as the local declaration succeeds (which is almost always will).
In order to avoid this it suggests declaring separately and then assigning:
local key_value
key_value=$(echo "$current_line" | mawk '/.+=.+/ print $1 ')
This is a shellcheck rule I frequently ignore and IMO is safe to ignore as long as you know you aren't trying to act on the return value of that variable declaration.
You can ignore it by adding the following to the top of your script (Below the hashbang of course):
# shellcheck disable=SC2155
answered 4 hours ago
Jesse_bJesse_b
13.4k23370
When you declare a variable as either local or exported that in itself is a command that will return success or not.
$ var=$(false)
$ echo $?
1
$ export var=$(false)
$ echo $?
0
So if you wanted to act on the return value of your command (echo "$current_line" | mawk '/.+=.+/ print $1 '), you would be unable to since it's going to exit with 0 as long as the local declaration succeeds (which is almost always will).
In order to avoid this it suggests declaring separately and then assigning:
local key_value
key_value=$(echo "$current_line" | mawk '/.+=.+/ print $1 ')
This is a shellcheck rule I frequently ignore and IMO is safe to ignore as long as you know you aren't trying to act on the return value of that variable declaration.
You can ignore it by adding the following to the top of your script (Below the hashbang of course):
# shellcheck disable=SC2155
answered 4 hours ago
Jesse_bJesse_b
13.4k23370
answered 4 hours ago
Jesse_bJesse_b
13.4k23370
answered 4 hours ago
Jesse_bJesse_b
13.4k23370
answered 4 hours ago
Jesse_bJesse_b
13.4k23370
13.4k23370
add a comment |
add a comment |
Inquisitor is a new contributor. Be nice, and check out our Code of Conduct.
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