shell, assigning value to variable from subscript

0

(using zsh with the following)

# i=for loop ((i;i<l,i++))

typeset -A _opts

AILABLE_ARG_OPTS=('v|verbose' 0 'd|destination' 1)

currentOpt=("${(@s/=/)${@[$i]##*-}}")



for k v in ${(kv)AVAILABLE_ARG_OPTS}; do

  if (( $v != 0 )); then

    _opts[${k#*|}]=$([[ -v currentOpt[2] ]] &&  #<---------- subscript

      echo "${currentOpt[2]}" ||

      echo "${@[$((++i))]}"   # <---------------------------- HERE ?

    )

  fi

done

At 'HERE', $i using correct value but doesn't set increment for the loop.

- Did I miss something with the substitution ?

- Does the variable became readonly inner sub execution ?

---

edit add same style question..

_opts[${k#*|}]=${currentOpt[2]:-"${@[$((++i))]}"}

is it "normal" that the second statement is evaluate in all case ?
`$currentOpt[2]``setted or not

---

edit: (as comment answer)

_opts[${k#*|}]=$(

  [[ -v currentOpt[2] ]] &&  #<---------- subscript 

    echo "${currentOpt[2]}" ||

    echo "${@[$((++i))]}"   # <--------------------- $i is not incremented

)

"~~equivalent~~" to (exept that increment..)

if [[ -v currentOpt[2] ]]; then

  _opts[${k#*|}]="${currentOpt[2]}"

else

  _opts[${k#*|}]="${@[$((++i))]}" # <--------  $i is incremented

fi

that the point I don't understand, in both case the right value is setted to _opts[$k].

But in the first exemple, $((++i)) don't assign it's value to$i
.. and i would know why

thanks for help

shell zsh

share|improve this question

edited Nov 19 '18 at 22:38

miken32

23.5k84672

asked Nov 19 '18 at 12:36

dim

11

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You have a || on the preceding line, which means unless there is an error with the echo command, the "HERE" line never executes.
  – chepner
  Nov 19 '18 at 13:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks, but that ok with that, assume we talk of an case where $currentOpt[2] is not setted, hilight for code is wrong at the line above the first echo, regex parse as an comment but i assign =$(. I added more explain
  – dim
  Nov 19 '18 at 13:56
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You're incrementing i in a subshell created by the command substitution. i in the current shell isn't affected.
  – chepner
  Nov 19 '18 at 14:03
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so parent shell variable are readonly ? does have a way to "reference" them ?
  – dim
  Nov 19 '18 at 14:26
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I guess that's one way of looking at it. More precisely, the child gets its own variable, initialized to the value of the parent's variable.
  – chepner
  Nov 19 '18 at 14:30
  

  
  
  
  

 | 
show 1 more comment

0

(using zsh with the following)

# i=for loop ((i;i<l,i++))

typeset -A _opts

AILABLE_ARG_OPTS=('v|verbose' 0 'd|destination' 1)

currentOpt=("${(@s/=/)${@[$i]##*-}}")



for k v in ${(kv)AVAILABLE_ARG_OPTS}; do

  if (( $v != 0 )); then

    _opts[${k#*|}]=$([[ -v currentOpt[2] ]] &&  #<---------- subscript

      echo "${currentOpt[2]}" ||

      echo "${@[$((++i))]}"   # <---------------------------- HERE ?

    )

  fi

done

At 'HERE', $i using correct value but doesn't set increment for the loop.

- Did I miss something with the substitution ?

- Does the variable became readonly inner sub execution ?

---

edit add same style question..

_opts[${k#*|}]=${currentOpt[2]:-"${@[$((++i))]}"}

is it "normal" that the second statement is evaluate in all case ?
`$currentOpt[2]``setted or not

---

edit: (as comment answer)

_opts[${k#*|}]=$(

  [[ -v currentOpt[2] ]] &&  #<---------- subscript 

    echo "${currentOpt[2]}" ||

    echo "${@[$((++i))]}"   # <--------------------- $i is not incremented

)

"~~equivalent~~" to (exept that increment..)

if [[ -v currentOpt[2] ]]; then

  _opts[${k#*|}]="${currentOpt[2]}"

else

  _opts[${k#*|}]="${@[$((++i))]}" # <--------  $i is incremented

fi

that the point I don't understand, in both case the right value is setted to _opts[$k].

But in the first exemple, $((++i)) don't assign it's value to$i
.. and i would know why

thanks for help

shell zsh

share|improve this question

edited Nov 19 '18 at 22:38

miken32

23.5k84672

asked Nov 19 '18 at 12:36

dim

11

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You have a || on the preceding line, which means unless there is an error with the echo command, the "HERE" line never executes.
  – chepner
  Nov 19 '18 at 13:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks, but that ok with that, assume we talk of an case where $currentOpt[2] is not setted, hilight for code is wrong at the line above the first echo, regex parse as an comment but i assign =$(. I added more explain
  – dim
  Nov 19 '18 at 13:56
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You're incrementing i in a subshell created by the command substitution. i in the current shell isn't affected.
  – chepner
  Nov 19 '18 at 14:03
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so parent shell variable are readonly ? does have a way to "reference" them ?
  – dim
  Nov 19 '18 at 14:26
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I guess that's one way of looking at it. More precisely, the child gets its own variable, initialized to the value of the parent's variable.
  – chepner
  Nov 19 '18 at 14:30
  

  
  
  
  

 | 
show 1 more comment

0

0

0

(using zsh with the following)

# i=for loop ((i;i<l,i++))

typeset -A _opts

AILABLE_ARG_OPTS=('v|verbose' 0 'd|destination' 1)

currentOpt=("${(@s/=/)${@[$i]##*-}}")



for k v in ${(kv)AVAILABLE_ARG_OPTS}; do

  if (( $v != 0 )); then

    _opts[${k#*|}]=$([[ -v currentOpt[2] ]] &&  #<---------- subscript

      echo "${currentOpt[2]}" ||

      echo "${@[$((++i))]}"   # <---------------------------- HERE ?

    )

  fi

done

At 'HERE', $i using correct value but doesn't set increment for the loop.

- Did I miss something with the substitution ?

- Does the variable became readonly inner sub execution ?

---

edit add same style question..

_opts[${k#*|}]=${currentOpt[2]:-"${@[$((++i))]}"}

is it "normal" that the second statement is evaluate in all case ?
`$currentOpt[2]``setted or not

---

edit: (as comment answer)

_opts[${k#*|}]=$(

  [[ -v currentOpt[2] ]] &&  #<---------- subscript 

    echo "${currentOpt[2]}" ||

    echo "${@[$((++i))]}"   # <--------------------- $i is not incremented

)

"~~equivalent~~" to (exept that increment..)

if [[ -v currentOpt[2] ]]; then

  _opts[${k#*|}]="${currentOpt[2]}"

else

  _opts[${k#*|}]="${@[$((++i))]}" # <--------  $i is incremented

fi

that the point I don't understand, in both case the right value is setted to _opts[$k].

But in the first exemple, $((++i)) don't assign it's value to$i
.. and i would know why

thanks for help

shell zsh

share|improve this question

edited Nov 19 '18 at 22:38

miken32

23.5k84672

asked Nov 19 '18 at 12:36

dim

11

(using zsh with the following)

# i=for loop ((i;i<l,i++))

typeset -A _opts

AILABLE_ARG_OPTS=('v|verbose' 0 'd|destination' 1)

currentOpt=("${(@s/=/)${@[$i]##*-}}")



for k v in ${(kv)AVAILABLE_ARG_OPTS}; do

  if (( $v != 0 )); then

    _opts[${k#*|}]=$([[ -v currentOpt[2] ]] &&  #<---------- subscript

      echo "${currentOpt[2]}" ||

      echo "${@[$((++i))]}"   # <---------------------------- HERE ?

    )

  fi

done

At 'HERE', $i using correct value but doesn't set increment for the loop.

- Did I miss something with the substitution ?

- Does the variable became readonly inner sub execution ?

---

edit add same style question..

_opts[${k#*|}]=${currentOpt[2]:-"${@[$((++i))]}"}

is it "normal" that the second statement is evaluate in all case ?
`$currentOpt[2]``setted or not

---

edit: (as comment answer)

_opts[${k#*|}]=$(

  [[ -v currentOpt[2] ]] &&  #<---------- subscript 

    echo "${currentOpt[2]}" ||

    echo "${@[$((++i))]}"   # <--------------------- $i is not incremented

)

"~~equivalent~~" to (exept that increment..)

if [[ -v currentOpt[2] ]]; then

  _opts[${k#*|}]="${currentOpt[2]}"

else

  _opts[${k#*|}]="${@[$((++i))]}" # <--------  $i is incremented

fi

that the point I don't understand, in both case the right value is setted to _opts[$k].

But in the first exemple, $((++i)) don't assign it's value to$i
.. and i would know why

thanks for help

shell zsh

shell zsh

share|improve this question

edited Nov 19 '18 at 22:38

miken32

23.5k84672

asked Nov 19 '18 at 12:36

dim

11

share|improve this question

edited Nov 19 '18 at 22:38

miken32

23.5k84672

asked Nov 19 '18 at 12:36

dim

11

share|improve this question

share|improve this question

edited Nov 19 '18 at 22:38

miken32

23.5k84672

edited Nov 19 '18 at 22:38

miken32

23.5k84672

edited Nov 19 '18 at 22:38

miken32

23.5k84672

23.5k84672

asked Nov 19 '18 at 12:36

dim

11

asked Nov 19 '18 at 12:36

dim

11

asked Nov 19 '18 at 12:36

dim

11

11

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You have a || on the preceding line, which means unless there is an error with the echo command, the "HERE" line never executes.
  – chepner
  Nov 19 '18 at 13:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks, but that ok with that, assume we talk of an case where $currentOpt[2] is not setted, hilight for code is wrong at the line above the first echo, regex parse as an comment but i assign =$(. I added more explain
  – dim
  Nov 19 '18 at 13:56
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You're incrementing i in a subshell created by the command substitution. i in the current shell isn't affected.
  – chepner
  Nov 19 '18 at 14:03
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so parent shell variable are readonly ? does have a way to "reference" them ?
  – dim
  Nov 19 '18 at 14:26
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I guess that's one way of looking at it. More precisely, the child gets its own variable, initialized to the value of the parent's variable.
  – chepner
  Nov 19 '18 at 14:30
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You have a || on the preceding line, which means unless there is an error with the echo command, the "HERE" line never executes.
  – chepner
  Nov 19 '18 at 13:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks, but that ok with that, assume we talk of an case where $currentOpt[2] is not setted, hilight for code is wrong at the line above the first echo, regex parse as an comment but i assign =$(. I added more explain
  – dim
  Nov 19 '18 at 13:56
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You're incrementing i in a subshell created by the command substitution. i in the current shell isn't affected.
  – chepner
  Nov 19 '18 at 14:03
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so parent shell variable are readonly ? does have a way to "reference" them ?
  – dim
  Nov 19 '18 at 14:26
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I guess that's one way of looking at it. More precisely, the child gets its own variable, initialized to the value of the parent's variable.
  – chepner
  Nov 19 '18 at 14:30
  

  
  
  
  

1

1

You have a || on the preceding line, which means unless there is an error with the echo command, the "HERE" line never executes.
– chepner
Nov 19 '18 at 13:13

You have a || on the preceding line, which means unless there is an error with the echo command, the "HERE" line never executes.
– chepner
Nov 19 '18 at 13:13

thanks, but that ok with that, assume we talk of an case where $currentOpt[2] is not setted, hilight for code is wrong at the line above the first echo, regex parse as an comment but i assign =$(. I added more explain
– dim
Nov 19 '18 at 13:56

thanks, but that ok with that, assume we talk of an case where $currentOpt[2] is not setted, hilight for code is wrong at the line above the first echo, regex parse as an comment but i assign =$(. I added more explain
– dim
Nov 19 '18 at 13:56

You're incrementing i in a subshell created by the command substitution. i in the current shell isn't affected.
– chepner
Nov 19 '18 at 14:03

You're incrementing i in a subshell created by the command substitution. i in the current shell isn't affected.
– chepner
Nov 19 '18 at 14:03

so parent shell variable are readonly ? does have a way to "reference" them ?
– dim
Nov 19 '18 at 14:26

so parent shell variable are readonly ? does have a way to "reference" them ?
– dim
Nov 19 '18 at 14:26

I guess that's one way of looking at it. More precisely, the child gets its own variable, initialized to the value of the parent's variable.
– chepner
Nov 19 '18 at 14:30

I guess that's one way of looking at it. More precisely, the child gets its own variable, initialized to the value of the parent's variable.
– chepner
Nov 19 '18 at 14:30

 | 
show 1 more comment

active

oldest

votes

Your Answer

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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
});


}
});

Thanks for contributing an answer to Stack Overflow!

• Please be sure to answer the question. Provide details and share your research!

But avoid …

• Asking for help, clarification, or responding to other answers.


• Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

Some of your past answers have not been well-received, and you're in danger of being blocked from answering.

Please pay close attention to the following guidance:

• Please be sure to answer the question. Provide details and share your research!

But avoid …

• Asking for help, clarification, or responding to other answers.


• Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

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%2fstackoverflow.com%2fquestions%2f53374807%2fshell-assigning-value-to-variable-from-subscript%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

active

oldest

votes

active

oldest

votes

active

oldest

votes

active

oldest

votes

Thanks for contributing an answer to Stack Overflow!

• Please be sure to answer the question. Provide details and share your research!

But avoid …

• Asking for help, clarification, or responding to other answers.


• Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

Some of your past answers have not been well-received, and you're in danger of being blocked from answering.

Please pay close attention to the following guidance:

• Please be sure to answer the question. Provide details and share your research!

But avoid …

• Asking for help, clarification, or responding to other answers.


• Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Thanks for contributing an answer to Stack Overflow!

• Please be sure to answer the question. Provide details and share your research!

But avoid …

• Asking for help, clarification, or responding to other answers.


• Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

Some of your past answers have not been well-received, and you're in danger of being blocked from answering.

Please pay close attention to the following guidance:

• Please be sure to answer the question. Provide details and share your research!

But avoid …

• Asking for help, clarification, or responding to other answers.


• Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

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%2fstackoverflow.com%2fquestions%2f53374807%2fshell-assigning-value-to-variable-from-subscript%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

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

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

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

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

Name

Name

Email

Required, but never shown

Email

Required, but never shown

Email

Required, but never shown

Email

Required, but never shown

Name

Name

Email

Required, but never shown

Email

Required, but never shown

Email

Required, but never shown

Email

Required, but never shown

This page is only for reference, If you need detailed information, please check here