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Print name if parameter passed to function

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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 yesterday

GAD3R

27.5k1858114

asked yesterday

XerxesXerxes

1956

New contributor

Xerxes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Wait, when you give it "someKey", $1 won't be empty. If it's empty, there's nothing to print.
  
  – choroba
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @choroba I want to print the parameter name if possible not its value.
  
  – Xerxes
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  What do you mean by the name? Function arguments don't have names.
  
  – choroba
  yesterday
  

  
  
  
  

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 yesterday

GAD3R

27.5k1858114

asked yesterday

XerxesXerxes

1956

New contributor

Xerxes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Wait, when you give it "someKey", $1 won't be empty. If it's empty, there's nothing to print.
  
  – choroba
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @choroba I want to print the parameter name if possible not its value.
  
  – Xerxes
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  What do you mean by the name? Function arguments don't have names.
  
  – choroba
  yesterday
  

  
  
  
  

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 yesterday

GAD3R

27.5k1858114

asked yesterday

XerxesXerxes

1956

New contributor

Xerxes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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 yesterday

GAD3R

27.5k1858114

asked yesterday

XerxesXerxes

1956

New contributor

Xerxes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited yesterday

GAD3R

27.5k1858114

asked yesterday

XerxesXerxes

1956

New contributor

Xerxes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited yesterday

GAD3R

27.5k1858114

edited yesterday

GAD3R

27.5k1858114

asked yesterday

XerxesXerxes

1956

New contributor

Xerxes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked yesterday

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 15 hours ago

answered yesterday

ilkkachuilkkachu

62.8k10103180

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Also: : "$!1:?Variable $1 is empty or unset".
  
  – Kusalananda♦
  15 hours ago
  
  

  
  
  
  

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 8 hours ago

answered yesterday

JShorthouseJShorthouse

50728

New contributor

JShorthouse 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 | 

2

echo "Exiting because $1 is empty"

should do the trick.

share|improve this answer

answered yesterday

PankiPanki

838412

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Wouldn't this always print Exiting because is empty, if the test is against $1 also? That doesn't seem very useful.
  
  – ilkkachu
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  yesterday
  

  
  
  
  

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 15 hours ago

answered yesterday

ilkkachuilkkachu

62.8k10103180

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Also: : "$!1:?Variable $1 is empty or unset".
  
  – Kusalananda♦
  15 hours ago
  
  

  
  
  
  

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 15 hours ago

answered yesterday

ilkkachuilkkachu

62.8k10103180

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Also: : "$!1:?Variable $1 is empty or unset".
  
  – Kusalananda♦
  15 hours ago
  
  

  
  
  
  

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 15 hours ago

answered yesterday

ilkkachuilkkachu

62.8k10103180

#!/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 15 hours ago

answered yesterday

ilkkachuilkkachu

62.8k10103180

answered yesterday

ilkkachuilkkachu

62.8k10103180

answered yesterday

ilkkachuilkkachu

62.8k10103180

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 8 hours ago

answered yesterday

JShorthouseJShorthouse

50728

New contributor

JShorthouse 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 | 

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 8 hours ago

answered yesterday

JShorthouseJShorthouse

50728

New contributor

JShorthouse 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 | 

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 8 hours ago

answered yesterday

JShorthouseJShorthouse

50728

New contributor

JShorthouse is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

function exitIfEmpty()

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


exitIfEmpty "$someKey" someKey

share|improve this answer

edited 8 hours ago

answered yesterday

JShorthouseJShorthouse

50728

New contributor

JShorthouse is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered yesterday

JShorthouseJShorthouse

50728

New contributor

JShorthouse is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered yesterday

JShorthouseJShorthouse

50728

2

echo "Exiting because $1 is empty"

should do the trick.

share|improve this answer

answered yesterday

PankiPanki

838412

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Wouldn't this always print Exiting because is empty, if the test is against $1 also? That doesn't seem very useful.
  
  – ilkkachu
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  yesterday
  

  
  
  
  

add a comment | 

2

echo "Exiting because $1 is empty"

should do the trick.

share|improve this answer

answered yesterday

PankiPanki

838412

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Wouldn't this always print Exiting because is empty, if the test is against $1 also? That doesn't seem very useful.
  
  – ilkkachu
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  yesterday
  

  
  
  
  

add a comment | 

echo "Exiting because $1 is empty"

should do the trick.

share|improve this answer

answered yesterday

PankiPanki

838412

echo "Exiting because $1 is empty"

share|improve this answer

answered yesterday

PankiPanki

838412

answered yesterday

PankiPanki

838412

answered yesterday

PankiPanki

838412