Fdtxjr

Why didn't this character "real die" when they blew their stack out in Altered Carbon?

If a contract sometimes uses the wrong name, is it still valid?

Is it ethical to give a final exam after the professor has quit before teaching the remaining chapters of the course?

When were vectors invented?

What does this icon in iOS Stardew Valley mean?

Can an alien society believe that their star system is the universe?

Should I discuss the type of campaign with my players?

How widely used is the term Treppenwitz? Is it something that most Germans know?

Fundamental Solution of the Pell Equation

Generate an RGB colour grid

English words in a non-english sci-fi novel

How to bypass password on Windows XP account?

Why aren't air breathing engines used as small first stages

How do pianists reach extremely loud dynamics?

Is it true that "carbohydrates are of no use for the basal metabolic need"?

How discoverable are IPv6 addresses and AAAA names by potential attackers?

How to react to hostile behavior from a senior developer?

What does an IRS interview request entail when called in to verify expenses for a sole proprietor small business?

What is the logic behind the Maharil's explanation of why we don't say שעשה ניסים on Pesach?

How to find out what spells would be useless to a blind NPC spellcaster?

Why did the rest of the Eastern Bloc not invade Yugoslavia?

Why am I getting the error "non-boolean type specified in a context where a condition is expected" for this request?

Naming the result of a source block

Denied boarding although I have proper visa and documentation. To whom should I make a complaint?

Print name if parameter passed to function

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

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

28.3k1958114

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

28.3k1958114

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

28.3k1958114

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

28.3k1958114

asked Mar 26 at 15:06

XerxesXerxes

1956

share|improve this question

edited Mar 26 at 17:42

GAD3R

28.3k1958114

asked Mar 26 at 15:06

XerxesXerxes

1956

edited Mar 26 at 17:42

GAD3R

28.3k1958114

edited Mar 26 at 17:42

GAD3R

28.3k1958114

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

63.5k10104181

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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 14 hours ago

answered Mar 26 at 15:07

PankiPanki

872512

• 
  
  
  

  
  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 | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "106"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2funix.stackexchange.com%2fquestions%2f508764%2fprint-name-if-parameter-passed-to-function%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

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

63.5k10104181

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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

63.5k10104181

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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

63.5k10104181

#!/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

63.5k10104181

answered Mar 26 at 15:24

ilkkachuilkkachu

63.5k10104181

answered Mar 26 at 15:24

ilkkachuilkkachu

63.5k10104181

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 14 hours ago

answered Mar 26 at 15:07

PankiPanki

872512

• 
  
  
  

  
  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 14 hours ago

answered Mar 26 at 15:07

PankiPanki

872512

• 
  
  
  

  
  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 14 hours ago

answered Mar 26 at 15:07

PankiPanki

872512

echo "Exiting because $1 is empty"

share|improve this answer

edited 14 hours ago

answered Mar 26 at 15:07

PankiPanki

872512

answered Mar 26 at 15:07

PankiPanki

872512

answered Mar 26 at 15:07

PankiPanki

872512