why $($var) gives an error?

up vote
1
down vote

favorite

I am trying to execute the following code in shell but I am not getting the output as expected. The command is as follows:

i=1
echo $($i)

I am getting an error:

Command not found error

I think it should show the value of the first command line argument, rather it is giving error.

edited Aug 21 at 10:12

perror

1,83141833

asked Aug 21 at 7:52

jeffery3015

1

@steve answers your question nicely, but if you're trying to get access to arguments dynamically, you can look at $@, shift, getopts and bash arrays.
â€“Â mikst
Aug 21 at 8:37

Accessing a positional parameter through a variable
â€“Â Mark Plotnick
Aug 21 at 10:42

add a commentÂ |Â

up vote
1
down vote

favorite

I am trying to execute the following code in shell but I am not getting the output as expected. The command is as follows:

i=1
echo $($i)

I am getting an error:

Command not found error

I think it should show the value of the first command line argument, rather it is giving error.

edited Aug 21 at 10:12

perror

1,83141833

asked Aug 21 at 7:52

jeffery3015

1

@steve answers your question nicely, but if you're trying to get access to arguments dynamically, you can look at $@, shift, getopts and bash arrays.
â€“Â mikst
Aug 21 at 8:37

Accessing a positional parameter through a variable
â€“Â Mark Plotnick
Aug 21 at 10:42

add a commentÂ |Â

up vote
1
down vote

favorite

I am trying to execute the following code in shell but I am not getting the output as expected. The command is as follows:

i=1
echo $($i)

I am getting an error:

Command not found error

I think it should show the value of the first command line argument, rather it is giving error.

edited Aug 21 at 10:12

perror

1,83141833

asked Aug 21 at 7:52

jeffery3015

I am trying to execute the following code in shell but I am not getting the output as expected. The command is as follows:

i=1
echo $($i)

I am getting an error:

Command not found error

I think it should show the value of the first command line argument, rather it is giving error.

linux shell-script shell command-line programming

edited Aug 21 at 10:12

perror

1,83141833

asked Aug 21 at 7:52

jeffery3015

edited Aug 21 at 10:12

perror

1,83141833

asked Aug 21 at 7:52

jeffery3015

edited Aug 21 at 10:12

perror

1,83141833

edited Aug 21 at 10:12

perror

1,83141833

edited Aug 21 at 10:12

perror

1,83141833

asked Aug 21 at 7:52

jeffery3015

asked Aug 21 at 7:52

jeffery3015

asked Aug 21 at 7:52

jeffery3015

1

@steve answers your question nicely, but if you're trying to get access to arguments dynamically, you can look at $@, shift, getopts and bash arrays.
â€“Â mikst
Aug 21 at 8:37

Accessing a positional parameter through a variable
â€“Â Mark Plotnick
Aug 21 at 10:42

add a commentÂ |Â

1

@steve answers your question nicely, but if you're trying to get access to arguments dynamically, you can look at $@, shift, getopts and bash arrays.
â€“Â mikst
Aug 21 at 8:37

Accessing a positional parameter through a variable
â€“Â Mark Plotnick
Aug 21 at 10:42

@steve answers your question nicely, but if you're trying to get access to arguments dynamically, you can look at $@, shift, getopts and bash arrays.
â€“Â mikst
Aug 21 at 8:37

Accessing a positional parameter through a variable
â€“Â Mark Plotnick
Aug 21 at 10:42

add a commentÂ |Â

1 Answer
1

active

oldest

votes

up vote
2
down vote

Because the construct $(...) is a command substitution, so $($i) means to run the contents of variable $i. In your case, it would try to execute 1, which the shell cannot find.

See below how it fails, but if we create a script named '1', it works.

Example:

$ i=1
$ echo "$($i)"
-bash: 1: command not found 
$ cat >1
#!/bin/bash
echo Hello World
$ chmod 755 1
$ PATH=$PATH:`pwd`
$ echo "$($i)"
Hello World
$

To display the value of the first command line argument, you need $1 or $1

$ cat >mytest
#!/bin/bash
echo "arg 1 is $1"
echo "arg 1 is $1"
$ chmod 755 mytest
$ ./mytest foo
arg 1 is foo
arg 1 is foo
$

edited Aug 21 at 10:33

ilkkachu

51.2k678141

answered Aug 21 at 8:01

steve

12.9k22149

add a commentÂ |Â

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',
convertImagesToLinks: false,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);

);

draft saved

draft discarded

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2funix.stackexchange.com%2fquestions%2f463798%2fwhy-var-gives-an-error%23new-answer', 'question_page');

);

Post as a guest

Name

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
2
down vote

Because the construct $(...) is a command substitution, so $($i) means to run the contents of variable $i. In your case, it would try to execute 1, which the shell cannot find.

See below how it fails, but if we create a script named '1', it works.

Example:

$ i=1
$ echo "$($i)"
-bash: 1: command not found 
$ cat >1
#!/bin/bash
echo Hello World
$ chmod 755 1
$ PATH=$PATH:`pwd`
$ echo "$($i)"
Hello World
$

To display the value of the first command line argument, you need $1 or $1

$ cat >mytest
#!/bin/bash
echo "arg 1 is $1"
echo "arg 1 is $1"
$ chmod 755 mytest
$ ./mytest foo
arg 1 is foo
arg 1 is foo
$

edited Aug 21 at 10:33

ilkkachu

51.2k678141

answered Aug 21 at 8:01

steve

12.9k22149

add a commentÂ |Â

up vote
2
down vote

Because the construct $(...) is a command substitution, so $($i) means to run the contents of variable $i. In your case, it would try to execute 1, which the shell cannot find.

See below how it fails, but if we create a script named '1', it works.

Example:

$ i=1
$ echo "$($i)"
-bash: 1: command not found 
$ cat >1
#!/bin/bash
echo Hello World
$ chmod 755 1
$ PATH=$PATH:`pwd`
$ echo "$($i)"
Hello World
$

To display the value of the first command line argument, you need $1 or $1

$ cat >mytest
#!/bin/bash
echo "arg 1 is $1"
echo "arg 1 is $1"
$ chmod 755 mytest
$ ./mytest foo
arg 1 is foo
arg 1 is foo
$

edited Aug 21 at 10:33

ilkkachu

51.2k678141

answered Aug 21 at 8:01

steve

12.9k22149

add a commentÂ |Â

up vote
2
down vote

Because the construct $(...) is a command substitution, so $($i) means to run the contents of variable $i. In your case, it would try to execute 1, which the shell cannot find.

See below how it fails, but if we create a script named '1', it works.

Example:

$ i=1
$ echo "$($i)"
-bash: 1: command not found 
$ cat >1
#!/bin/bash
echo Hello World
$ chmod 755 1
$ PATH=$PATH:`pwd`
$ echo "$($i)"
Hello World
$

To display the value of the first command line argument, you need $1 or $1

$ cat >mytest
#!/bin/bash
echo "arg 1 is $1"
echo "arg 1 is $1"
$ chmod 755 mytest
$ ./mytest foo
arg 1 is foo
arg 1 is foo
$

edited Aug 21 at 10:33

ilkkachu

51.2k678141

answered Aug 21 at 8:01

steve

12.9k22149

Because the construct $(...) is a command substitution, so $($i) means to run the contents of variable $i. In your case, it would try to execute 1, which the shell cannot find.

See below how it fails, but if we create a script named '1', it works.

Example:

$ i=1
$ echo "$($i)"
-bash: 1: command not found 
$ cat >1
#!/bin/bash
echo Hello World
$ chmod 755 1
$ PATH=$PATH:`pwd`
$ echo "$($i)"
Hello World
$

To display the value of the first command line argument, you need $1 or $1

$ cat >mytest
#!/bin/bash
echo "arg 1 is $1"
echo "arg 1 is $1"
$ chmod 755 mytest
$ ./mytest foo
arg 1 is foo
arg 1 is foo
$

edited Aug 21 at 10:33

ilkkachu

51.2k678141

answered Aug 21 at 8:01

steve

12.9k22149

edited Aug 21 at 10:33

ilkkachu

51.2k678141

edited Aug 21 at 10:33

ilkkachu

51.2k678141

edited Aug 21 at 10:33

ilkkachu

51.2k678141

answered Aug 21 at 8:01

steve

12.9k22149

answered Aug 21 at 8:01

steve

12.9k22149

answered Aug 21 at 8:01

steve

12.9k22149

add a commentÂ |Â

draft saved

draft discarded

draft saved

draft discarded

Post as a guest

Name

ZzllwK1RtfEy68W39d14f lWZP,fy,10DgZZZcoVYFb1lyRT2tCUR1ze1RBivx5,D,DqmEMoPil5kVQu6 2d QsQ2ErBkjdmFMM 12Be1z

搜尋此網誌

mjhjmtu