What does the substitution ${!var_name+x} mean?

Multi tool use
Multi tool use












5















I found a script that has a function that checks if a variable is set but i don't understand it very well.



check_if_variable_is_set() {

    var_name=$1

    if [ -z "${!var_name+x}" ]; then

        false

    else

        true

    fi

}


What exactly happens with this substitution ?






bash shell-script variable-substitution







share|improve this question























  • related ${!FOO} and zsh.

    – αғsнιη
    10 hours ago
















5















I found a script that has a function that checks if a variable is set but i don't understand it very well.



check_if_variable_is_set() {

    var_name=$1

    if [ -z "${!var_name+x}" ]; then

        false

    else

        true

    fi

}


What exactly happens with this substitution ?






bash shell-script variable-substitution







share|improve this question























  • related ${!FOO} and zsh.

    – αғsнιη
    10 hours ago














5












5








5


1






I found a script that has a function that checks if a variable is set but i don't understand it very well.



check_if_variable_is_set() {

    var_name=$1

    if [ -z "${!var_name+x}" ]; then

        false

    else

        true

    fi

}


What exactly happens with this substitution ?






bash shell-script variable-substitution







share|improve this question














I found a script that has a function that checks if a variable is set but i don't understand it very well.



check_if_variable_is_set() {

    var_name=$1

    if [ -z "${!var_name+x}" ]; then

        false

    else

        true

    fi

}


What exactly happens with this substitution ?






bash shell-script variable-substitution




bash shell-script variable-substitution






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 11 hours ago









Karim ManaouilKarim Manaouil

1919




1919













  • related ${!FOO} and zsh.

    – αғsнιη
    10 hours ago



















  • related ${!FOO} and zsh.

    – αғsнιη
    10 hours ago

















related ${!FOO} and zsh.

– αғsнιη
10 hours ago





related ${!FOO} and zsh.

– αғsнιη
10 hours ago










1 Answer
1






active

oldest

votes


















11














In the bash shell, ${!var} is a variable indirection. It expands to the value of the variable whose name is kept in $var.



The variable expansion ${var+value} is a POSIX expansion that expands to value if the variable var is set (no matter if its value is empty or not).



Combining these, ${!var+x} would expand to x if the variable whose name is kept in $var is set.



Example:



$ foo=hello

$ var=foo






$ echo "${!var+$var is set, its value is ${!var}}"

foo is set, its value is hello






$ unset foo

$ echo "${!var+$var is set, its value is ${!var}}"


(empty line as output)




The function in the question could be shortened to



check_if_variable_is_set () { [ -n "${!1+x}" ]; }


or even:



check_if_variable_is_set () { [ -v "$1" ]; }


or even:



check_if_variable_is_set()[[ -v $1 ]]


Where -v is a bash test on a variable name which will be true if the named variable is set, and false otherwise.




POSIXly, it could be written:



check_if_variable_is_set() { eval '[ -n "${'"$1"'+x}" ]'; }


Note that all those are potential command injection vulnerabilities if the argument to that function could end up being under control of an attacker. Try for instance with check_if_variable_is_set 'a[$(id>&2)]'.



To guard against that, you may want to verify that the argument is a valid variable name first. For variables:



check_if_variable_is_set() {

  case $1 in

    ("" | *[![:alnum:]_]* | [![:alpha:]_]*) false;;

    (*)  eval '[ -n "${'"$1"'+x}" ]'

  esac

}


(note that [[:alpha:]] will check for alphabetical characters in your locale while some shells only accept alphabetical characters from the portable character set in their variable)






share|improve this answer


























  • There is nothing on earth more complete than this. You deserve a cookie for that.

    – Karim Manaouil
    8 hours ago











  • @KarimManaouil A third of that cookie goes to Stéphane though :-)

    – Kusalananda
    8 hours ago











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









11














In the bash shell, ${!var} is a variable indirection. It expands to the value of the variable whose name is kept in $var.



The variable expansion ${var+value} is a POSIX expansion that expands to value if the variable var is set (no matter if its value is empty or not).



Combining these, ${!var+x} would expand to x if the variable whose name is kept in $var is set.



Example:



$ foo=hello

$ var=foo






$ echo "${!var+$var is set, its value is ${!var}}"

foo is set, its value is hello






$ unset foo

$ echo "${!var+$var is set, its value is ${!var}}"


(empty line as output)




The function in the question could be shortened to



check_if_variable_is_set () { [ -n "${!1+x}" ]; }


or even:



check_if_variable_is_set () { [ -v "$1" ]; }


or even:



check_if_variable_is_set()[[ -v $1 ]]


Where -v is a bash test on a variable name which will be true if the named variable is set, and false otherwise.




POSIXly, it could be written:



check_if_variable_is_set() { eval '[ -n "${'"$1"'+x}" ]'; }


Note that all those are potential command injection vulnerabilities if the argument to that function could end up being under control of an attacker. Try for instance with check_if_variable_is_set 'a[$(id>&2)]'.



To guard against that, you may want to verify that the argument is a valid variable name first. For variables:



check_if_variable_is_set() {

  case $1 in

    ("" | *[![:alnum:]_]* | [![:alpha:]_]*) false;;

    (*)  eval '[ -n "${'"$1"'+x}" ]'

  esac

}


(note that [[:alpha:]] will check for alphabetical characters in your locale while some shells only accept alphabetical characters from the portable character set in their variable)






share|improve this answer


























  • There is nothing on earth more complete than this. You deserve a cookie for that.

    – Karim Manaouil
    8 hours ago











  • @KarimManaouil A third of that cookie goes to Stéphane though :-)

    – Kusalananda
    8 hours ago
















11














In the bash shell, ${!var} is a variable indirection. It expands to the value of the variable whose name is kept in $var.



The variable expansion ${var+value} is a POSIX expansion that expands to value if the variable var is set (no matter if its value is empty or not).



Combining these, ${!var+x} would expand to x if the variable whose name is kept in $var is set.



Example:



$ foo=hello

$ var=foo






$ echo "${!var+$var is set, its value is ${!var}}"

foo is set, its value is hello






$ unset foo

$ echo "${!var+$var is set, its value is ${!var}}"


(empty line as output)




The function in the question could be shortened to



check_if_variable_is_set () { [ -n "${!1+x}" ]; }


or even:



check_if_variable_is_set () { [ -v "$1" ]; }


or even:



check_if_variable_is_set()[[ -v $1 ]]


Where -v is a bash test on a variable name which will be true if the named variable is set, and false otherwise.




POSIXly, it could be written:



check_if_variable_is_set() { eval '[ -n "${'"$1"'+x}" ]'; }


Note that all those are potential command injection vulnerabilities if the argument to that function could end up being under control of an attacker. Try for instance with check_if_variable_is_set 'a[$(id>&2)]'.



To guard against that, you may want to verify that the argument is a valid variable name first. For variables:



check_if_variable_is_set() {

  case $1 in

    ("" | *[![:alnum:]_]* | [![:alpha:]_]*) false;;

    (*)  eval '[ -n "${'"$1"'+x}" ]'

  esac

}


(note that [[:alpha:]] will check for alphabetical characters in your locale while some shells only accept alphabetical characters from the portable character set in their variable)






share|improve this answer


























  • There is nothing on earth more complete than this. You deserve a cookie for that.

    – Karim Manaouil
    8 hours ago











  • @KarimManaouil A third of that cookie goes to Stéphane though :-)

    – Kusalananda
    8 hours ago














11












11








11







In the bash shell, ${!var} is a variable indirection. It expands to the value of the variable whose name is kept in $var.



The variable expansion ${var+value} is a POSIX expansion that expands to value if the variable var is set (no matter if its value is empty or not).



Combining these, ${!var+x} would expand to x if the variable whose name is kept in $var is set.



Example:



$ foo=hello

$ var=foo






$ echo "${!var+$var is set, its value is ${!var}}"

foo is set, its value is hello






$ unset foo

$ echo "${!var+$var is set, its value is ${!var}}"


(empty line as output)




The function in the question could be shortened to



check_if_variable_is_set () { [ -n "${!1+x}" ]; }


or even:



check_if_variable_is_set () { [ -v "$1" ]; }


or even:



check_if_variable_is_set()[[ -v $1 ]]


Where -v is a bash test on a variable name which will be true if the named variable is set, and false otherwise.




POSIXly, it could be written:



check_if_variable_is_set() { eval '[ -n "${'"$1"'+x}" ]'; }


Note that all those are potential command injection vulnerabilities if the argument to that function could end up being under control of an attacker. Try for instance with check_if_variable_is_set 'a[$(id>&2)]'.



To guard against that, you may want to verify that the argument is a valid variable name first. For variables:



check_if_variable_is_set() {

  case $1 in

    ("" | *[![:alnum:]_]* | [![:alpha:]_]*) false;;

    (*)  eval '[ -n "${'"$1"'+x}" ]'

  esac

}


(note that [[:alpha:]] will check for alphabetical characters in your locale while some shells only accept alphabetical characters from the portable character set in their variable)






share|improve this answer















In the bash shell, ${!var} is a variable indirection. It expands to the value of the variable whose name is kept in $var.



The variable expansion ${var+value} is a POSIX expansion that expands to value if the variable var is set (no matter if its value is empty or not).



Combining these, ${!var+x} would expand to x if the variable whose name is kept in $var is set.



Example:



$ foo=hello

$ var=foo






$ echo "${!var+$var is set, its value is ${!var}}"

foo is set, its value is hello






$ unset foo

$ echo "${!var+$var is set, its value is ${!var}}"


(empty line as output)




The function in the question could be shortened to



check_if_variable_is_set () { [ -n "${!1+x}" ]; }


or even:



check_if_variable_is_set () { [ -v "$1" ]; }


or even:



check_if_variable_is_set()[[ -v $1 ]]


Where -v is a bash test on a variable name which will be true if the named variable is set, and false otherwise.




POSIXly, it could be written:



check_if_variable_is_set() { eval '[ -n "${'"$1"'+x}" ]'; }


Note that all those are potential command injection vulnerabilities if the argument to that function could end up being under control of an attacker. Try for instance with check_if_variable_is_set 'a[$(id>&2)]'.



To guard against that, you may want to verify that the argument is a valid variable name first. For variables:



check_if_variable_is_set() {

  case $1 in

    ("" | *[![:alnum:]_]* | [![:alpha:]_]*) false;;

    (*)  eval '[ -n "${'"$1"'+x}" ]'

  esac

}


(note that [[:alpha:]] will check for alphabetical characters in your locale while some shells only accept alphabetical characters from the portable character set in their variable)







share|improve this answer














share|improve this answer



share|improve this answer








edited 9 hours ago









Stéphane Chazelas

306k57577931




306k57577931










answered 11 hours ago









KusalanandaKusalananda

131k17249408




131k17249408













  • There is nothing on earth more complete than this. You deserve a cookie for that.

    – Karim Manaouil
    8 hours ago











  • @KarimManaouil A third of that cookie goes to Stéphane though :-)

    – Kusalananda
    8 hours ago



















  • There is nothing on earth more complete than this. You deserve a cookie for that.

    – Karim Manaouil
    8 hours ago











  • @KarimManaouil A third of that cookie goes to Stéphane though :-)

    – Kusalananda
    8 hours ago

















There is nothing on earth more complete than this. You deserve a cookie for that.

– Karim Manaouil
8 hours ago





There is nothing on earth more complete than this. You deserve a cookie for that.

– Karim Manaouil
8 hours ago













@KarimManaouil A third of that cookie goes to Stéphane though :-)

– Kusalananda
8 hours ago





@KarimManaouil A third of that cookie goes to Stéphane though :-)

– Kusalananda
8 hours ago


















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