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

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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

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1 Answer
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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

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