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Print name if parameter passed to function

Function to evaluate variables in BASHHow do I get the value of a named option of an already running process in Linux?Conditional execution block with || and parentheses problemShell valid function name charactersWhile loop with result from function - BASHGet specific result from functionBash function assign value to passed parameterFunction to iterate over arrayHow am I allowed to pass a void parameter through bash functions?Pass parameter to Bash function which will serve as a pattern to awk

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4

1

I have written a little function that exits if the value of the function argument is empty, I would like to be able to also print the name of the parameter (not the value!) if it is possible, my following implementation fails to print the name of the parameter.

function exitIfEmpty()

 if [ -z "$1" ]
 then
 echo "Exiting because $!1 is empty"
 exit 1
 fi

when called like so

exitIfEmpty someKey

should print

Exiting because someKey is empty

bash

share|improve this question

edited Mar 26 at 17:42

GAD3R

28.5k1959116

asked Mar 26 at 15:06

XerxesXerxes

1956

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Wait, when you give it "someKey", $1 won't be empty. If it's empty, there's nothing to print.
  
  – choroba
  Mar 26 at 15:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @choroba I want to print the parameter name if possible not its value.
  
  – Xerxes
  Mar 26 at 15:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  What do you mean by the name? Function arguments don't have names.
  
  – choroba
  Mar 26 at 15:32
  

  
  
  
  

add a comment | 

4

1

I have written a little function that exits if the value of the function argument is empty, I would like to be able to also print the name of the parameter (not the value!) if it is possible, my following implementation fails to print the name of the parameter.

function exitIfEmpty()

 if [ -z "$1" ]
 then
 echo "Exiting because $!1 is empty"
 exit 1
 fi

when called like so

exitIfEmpty someKey

should print

Exiting because someKey is empty

bash

share|improve this question

edited Mar 26 at 17:42

GAD3R

28.5k1959116

asked Mar 26 at 15:06

XerxesXerxes

1956

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Wait, when you give it "someKey", $1 won't be empty. If it's empty, there's nothing to print.
  
  – choroba
  Mar 26 at 15:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @choroba I want to print the parameter name if possible not its value.
  
  – Xerxes
  Mar 26 at 15:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  What do you mean by the name? Function arguments don't have names.
  
  – choroba
  Mar 26 at 15:32
  

  
  
  
  

add a comment | 

I have written a little function that exits if the value of the function argument is empty, I would like to be able to also print the name of the parameter (not the value!) if it is possible, my following implementation fails to print the name of the parameter.

function exitIfEmpty()

 if [ -z "$1" ]
 then
 echo "Exiting because $!1 is empty"
 exit 1
 fi

when called like so

exitIfEmpty someKey

should print

Exiting because someKey is empty

bash

share|improve this question

edited Mar 26 at 17:42

GAD3R

28.5k1959116

asked Mar 26 at 15:06

XerxesXerxes

1956

function exitIfEmpty()

 if [ -z "$1" ]
 then
 echo "Exiting because $!1 is empty"
 exit 1
 fi

Exiting because someKey is empty

share|improve this question

edited Mar 26 at 17:42

GAD3R

28.5k1959116

asked Mar 26 at 15:06

XerxesXerxes

1956

share|improve this question

edited Mar 26 at 17:42

GAD3R

28.5k1959116

asked Mar 26 at 15:06

XerxesXerxes

1956

edited Mar 26 at 17:42

GAD3R

28.5k1959116

edited Mar 26 at 17:42

GAD3R

28.5k1959116

asked Mar 26 at 15:06

XerxesXerxes

1956

asked Mar 26 at 15:06

XerxesXerxes

1956

3 Answers
3

active

oldest

votes

13

What gets passed to the function is just a string. If you run func somevar, what is passed is the string somevar. If you run func $somevar, what is passed is (the word-split) value of the variable somevar. Neither is a variable reference, a pointer or anything like that, they're just strings.

If you want to pass the name of a variable to a function, and then look at the value of that variable, you'll need to use a nameref (Bash 4.3 or later, IIRC), or an indirect reference $!var. $!var expands to the value of the variable whose name is stored in var.

So, you just have it the wrong way in the script, if you pass the name of a variable to function, use "$!1" to get the value of the variable named in $1, and plain "$1" to get the name.

E.g. this will print variable bar is empty, exiting, and exit the shell:

#!/bin/bash
exitIfEmpty() 
 if [ -z "$!1" ]; then
 echo "variable $1 is empty, exiting"
 exit 1
 fi

foo=x
unset bar
exitIfEmpty foo
exitIfEmpty bar

share|improve this answer

edited Mar 27 at 10:07

answered Mar 26 at 15:24

ilkkachuilkkachu

63.6k10104182

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Also: : "$!1:?Variable $1 is empty or unset".
  
  – Kusalananda♦
  Mar 27 at 10:16
  
  

  
  
  
  

add a comment | 

3

Pass the name as second argument

function exitIfEmpty()

 if [ -z "$1" ]
 then
 echo "Exiting because $2 is empty"
 exit 1
 fi


exitIfEmpty "$someKey" someKey

share|improve this answer

edited Mar 27 at 17:30

answered Mar 26 at 15:18

JShorthouseJShorthouse

54829

add a comment | 

2

echo "Exiting because $1 is empty"

should do the trick.

share|improve this answer

edited Apr 16 at 13:45

answered Mar 26 at 15:07

PankiPanki

1,050514

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Wouldn't this always print Exiting because is empty, if the test is against $1 also? That doesn't seem very useful.
  
  – ilkkachu
  Mar 26 at 15:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  ITYM echo 'Exit because $1 is empty' or echo "Exit because $1 is empty" -- single-quote strings don't do variable expansion; double-quoted strings require $ to be escaped
  
  – jimbobmcgee
  Mar 26 at 19:22
  

  
  
  
  

add a comment | 

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13

What gets passed to the function is just a string. If you run func somevar, what is passed is the string somevar. If you run func $somevar, what is passed is (the word-split) value of the variable somevar. Neither is a variable reference, a pointer or anything like that, they're just strings.

If you want to pass the name of a variable to a function, and then look at the value of that variable, you'll need to use a nameref (Bash 4.3 or later, IIRC), or an indirect reference $!var. $!var expands to the value of the variable whose name is stored in var.

So, you just have it the wrong way in the script, if you pass the name of a variable to function, use "$!1" to get the value of the variable named in $1, and plain "$1" to get the name.

E.g. this will print variable bar is empty, exiting, and exit the shell:

#!/bin/bash
exitIfEmpty() 
 if [ -z "$!1" ]; then
 echo "variable $1 is empty, exiting"
 exit 1
 fi

foo=x
unset bar
exitIfEmpty foo
exitIfEmpty bar

share|improve this answer

edited Mar 27 at 10:07

answered Mar 26 at 15:24

ilkkachuilkkachu

63.6k10104182

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Also: : "$!1:?Variable $1 is empty or unset".
  
  – Kusalananda♦
  Mar 27 at 10:16
  
  

  
  
  
  

add a comment | 

13

What gets passed to the function is just a string. If you run func somevar, what is passed is the string somevar. If you run func $somevar, what is passed is (the word-split) value of the variable somevar. Neither is a variable reference, a pointer or anything like that, they're just strings.

If you want to pass the name of a variable to a function, and then look at the value of that variable, you'll need to use a nameref (Bash 4.3 or later, IIRC), or an indirect reference $!var. $!var expands to the value of the variable whose name is stored in var.

So, you just have it the wrong way in the script, if you pass the name of a variable to function, use "$!1" to get the value of the variable named in $1, and plain "$1" to get the name.

E.g. this will print variable bar is empty, exiting, and exit the shell:

#!/bin/bash
exitIfEmpty() 
 if [ -z "$!1" ]; then
 echo "variable $1 is empty, exiting"
 exit 1
 fi

foo=x
unset bar
exitIfEmpty foo
exitIfEmpty bar

share|improve this answer

edited Mar 27 at 10:07

answered Mar 26 at 15:24

ilkkachuilkkachu

63.6k10104182

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Also: : "$!1:?Variable $1 is empty or unset".
  
  – Kusalananda♦
  Mar 27 at 10:16
  
  

  
  
  
  

add a comment | 

What gets passed to the function is just a string. If you run func somevar, what is passed is the string somevar. If you run func $somevar, what is passed is (the word-split) value of the variable somevar. Neither is a variable reference, a pointer or anything like that, they're just strings.

If you want to pass the name of a variable to a function, and then look at the value of that variable, you'll need to use a nameref (Bash 4.3 or later, IIRC), or an indirect reference $!var. $!var expands to the value of the variable whose name is stored in var.

So, you just have it the wrong way in the script, if you pass the name of a variable to function, use "$!1" to get the value of the variable named in $1, and plain "$1" to get the name.

E.g. this will print variable bar is empty, exiting, and exit the shell:

#!/bin/bash
exitIfEmpty() 
 if [ -z "$!1" ]; then
 echo "variable $1 is empty, exiting"
 exit 1
 fi

foo=x
unset bar
exitIfEmpty foo
exitIfEmpty bar

share|improve this answer

edited Mar 27 at 10:07

answered Mar 26 at 15:24

ilkkachuilkkachu

63.6k10104182

#!/bin/bash
exitIfEmpty() 
 if [ -z "$!1" ]; then
 echo "variable $1 is empty, exiting"
 exit 1
 fi

foo=x
unset bar
exitIfEmpty foo
exitIfEmpty bar

share|improve this answer

edited Mar 27 at 10:07

answered Mar 26 at 15:24

ilkkachuilkkachu

63.6k10104182

answered Mar 26 at 15:24

ilkkachuilkkachu

63.6k10104182

answered Mar 26 at 15:24

ilkkachuilkkachu

63.6k10104182

3

Pass the name as second argument

function exitIfEmpty()

 if [ -z "$1" ]
 then
 echo "Exiting because $2 is empty"
 exit 1
 fi


exitIfEmpty "$someKey" someKey

share|improve this answer

edited Mar 27 at 17:30

answered Mar 26 at 15:18

JShorthouseJShorthouse

54829

add a comment | 

3

Pass the name as second argument

function exitIfEmpty()

 if [ -z "$1" ]
 then
 echo "Exiting because $2 is empty"
 exit 1
 fi


exitIfEmpty "$someKey" someKey

share|improve this answer

edited Mar 27 at 17:30

answered Mar 26 at 15:18

JShorthouseJShorthouse

54829

add a comment | 

Pass the name as second argument

function exitIfEmpty()

 if [ -z "$1" ]
 then
 echo "Exiting because $2 is empty"
 exit 1
 fi


exitIfEmpty "$someKey" someKey

share|improve this answer

edited Mar 27 at 17:30

answered Mar 26 at 15:18

JShorthouseJShorthouse

54829

function exitIfEmpty()

 if [ -z "$1" ]
 then
 echo "Exiting because $2 is empty"
 exit 1
 fi


exitIfEmpty "$someKey" someKey

share|improve this answer

edited Mar 27 at 17:30

answered Mar 26 at 15:18

JShorthouseJShorthouse

54829

answered Mar 26 at 15:18

JShorthouseJShorthouse

54829

answered Mar 26 at 15:18

JShorthouseJShorthouse

54829

2

echo "Exiting because $1 is empty"

should do the trick.

share|improve this answer

edited Apr 16 at 13:45

answered Mar 26 at 15:07

PankiPanki

1,050514

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Wouldn't this always print Exiting because is empty, if the test is against $1 also? That doesn't seem very useful.
  
  – ilkkachu
  Mar 26 at 15:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  ITYM echo 'Exit because $1 is empty' or echo "Exit because $1 is empty" -- single-quote strings don't do variable expansion; double-quoted strings require $ to be escaped
  
  – jimbobmcgee
  Mar 26 at 19:22
  

  
  
  
  

add a comment | 

2

echo "Exiting because $1 is empty"

should do the trick.

share|improve this answer

edited Apr 16 at 13:45

answered Mar 26 at 15:07

PankiPanki

1,050514

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Wouldn't this always print Exiting because is empty, if the test is against $1 also? That doesn't seem very useful.
  
  – ilkkachu
  Mar 26 at 15:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  ITYM echo 'Exit because $1 is empty' or echo "Exit because $1 is empty" -- single-quote strings don't do variable expansion; double-quoted strings require $ to be escaped
  
  – jimbobmcgee
  Mar 26 at 19:22
  

  
  
  
  

add a comment | 

echo "Exiting because $1 is empty"

should do the trick.

share|improve this answer

edited Apr 16 at 13:45

answered Mar 26 at 15:07

PankiPanki

1,050514

echo "Exiting because $1 is empty"

share|improve this answer

edited Apr 16 at 13:45

answered Mar 26 at 15:07

PankiPanki

1,050514

answered Mar 26 at 15:07

PankiPanki

1,050514

answered Mar 26 at 15:07

PankiPanki

1,050514