Bash: What does “masking return values” mean?
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 $1 }')
What does "masking return values" mean, and how does it pertain to the aforementioned warning?
bash shell-script shellcheck
New contributor
add a comment |
shellcheck
generated the following warning
SC2155: Declare and assign separately to avoid masking return
values
For this line of code
local key_value=$(echo "$current_line" | mawk '/.+=.+/ {print $1 }')
What does "masking return values" mean, and how does it pertain to the aforementioned warning?
bash shell-script shellcheck
New contributor
add a comment |
shellcheck
generated the following warning
SC2155: Declare and assign separately to avoid masking return
values
For this line of code
local key_value=$(echo "$current_line" | mawk '/.+=.+/ {print $1 }')
What does "masking return values" mean, and how does it pertain to the aforementioned warning?
bash shell-script shellcheck
New contributor
shellcheck
generated the following warning
SC2155: Declare and assign separately to avoid masking return
values
For this line of code
local key_value=$(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
New contributor
New contributor
edited 4 hours ago
Inquisitor
New contributor
asked 5 hours ago
InquisitorInquisitor
304
304
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add a comment |
add a comment |
1 Answer
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votes
When you declare a variable as either local
or export
ed 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
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 export
ed 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
add a comment |
When you declare a variable as either local
or export
ed 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
add a comment |
When you declare a variable as either local
or export
ed 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
When you declare a variable as either local
or export
ed 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 5 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.
Inquisitor is a new contributor. Be nice, and check out our Code of Conduct.
Inquisitor is a new contributor. Be nice, and check out our Code of Conduct.
Inquisitor is a new contributor. Be nice, and check out our Code of Conduct.
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