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Print out argument value in for loop

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}

-1

This is the script

user@linux:~$ cat script.sh 

for i in `seq $#`

do

  echo $i

done

user@linux:~$ 

Output

user@linux:~$ ./script.sh a b c

1

2

3

user@linux:~$ 

Desired Output

I would like to get argument value like this .... and not just the number

user@linux:~$ ./script.sh a b c

1 - a

2 - b

3 - c

user@linux:~$ 

bash arguments

share|improve this question

asked 28 mins ago

SabrinaSabrina

38411 silver badge1111 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You included the bash tag, but gave your script a .sh extension, and the syntax of the file looks more like zsh (because of the unquoted variable expansions). Which is it?
  
  – Stéphane Chazelas
  2 mins ago
  

  
  
  
  

add a comment | 

-1

This is the script

user@linux:~$ cat script.sh 

for i in `seq $#`

do

  echo $i

done

user@linux:~$ 

Output

user@linux:~$ ./script.sh a b c

1

2

3

user@linux:~$ 

Desired Output

I would like to get argument value like this .... and not just the number

user@linux:~$ ./script.sh a b c

1 - a

2 - b

3 - c

user@linux:~$ 

bash arguments

share|improve this question

asked 28 mins ago

SabrinaSabrina

38411 silver badge1111 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You included the bash tag, but gave your script a .sh extension, and the syntax of the file looks more like zsh (because of the unquoted variable expansions). Which is it?
  
  – Stéphane Chazelas
  2 mins ago
  

  
  
  
  

add a comment | 

This is the script

user@linux:~$ cat script.sh 

for i in `seq $#`

do

  echo $i

done

user@linux:~$ 

Output

user@linux:~$ ./script.sh a b c

1

2

3

user@linux:~$ 

Desired Output

I would like to get argument value like this .... and not just the number

user@linux:~$ ./script.sh a b c

1 - a

2 - b

3 - c

user@linux:~$ 

bash arguments

share|improve this question

asked 28 mins ago

SabrinaSabrina

38411 silver badge1111 bronze badges

user@linux:~$ cat script.sh 

for i in `seq $#`

do

  echo $i

done

user@linux:~$ 

user@linux:~$ ./script.sh a b c

1

2

3

user@linux:~$ 

user@linux:~$ ./script.sh a b c

1 - a

2 - b

3 - c

user@linux:~$ 

share|improve this question

asked 28 mins ago

SabrinaSabrina

38411 silver badge1111 bronze badges

share|improve this question

asked 28 mins ago

SabrinaSabrina

38411 silver badge1111 bronze badges

asked 28 mins ago

SabrinaSabrina

38411 silver badge1111 bronze badges

asked 28 mins ago

SabrinaSabrina

38411 silver badge1111 bronze badges

2 Answers
2

active

oldest

votes

1

#! /bin/sh -

i=1

for arg do

  printf '%2d: %sn' "$i" "$arg"

  i=$((i + 1))

done

That is, instead of looping over the indexes, loop over the arguments and increment an index separately.

A few notes on your approach:

• leaving a parameter expansion unquoted in list context has a very special meaning, you almost never want to do that.


• 
  echo can't be used to output arbitrary data.


• The `...` form of command substitution should be avoided these days. Use $(...) instead.


• Because unquoted command substitution also invokes split+glob, it has to be used with care. Here, you're using the split part which means you have a dependency on the current value of $IFS. The output of seq won't include wildcards, so you won't have a problem with the glob part.


• 
  seq is not a standard command and not found on all systems. To loop on numbers, it's not the best approach anyway, as that means storing the whole output in memory and split it (resulting in extra copies in memory), and that means running a separate utility in a separate process even though the shell has builtin support for that.

share|improve this answer

edited 5 mins ago

answered 15 mins ago

Stéphane ChazelasStéphane Chazelas

324k5757 gold badges627627 silver badges996996 bronze badges

add a comment | 

0

#!/bin/bash

for ((i=1; i<=$#; i++))

do

  echo "$i - ${!i}"

done

share

answered 5 mins ago

FreddyFreddy

5,51611 gold badge66 silver badges2323 bronze badges

add a comment | 

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1

#! /bin/sh -

i=1

for arg do

  printf '%2d: %sn' "$i" "$arg"

  i=$((i + 1))

done

That is, instead of looping over the indexes, loop over the arguments and increment an index separately.

A few notes on your approach:

• leaving a parameter expansion unquoted in list context has a very special meaning, you almost never want to do that.


• 
  echo can't be used to output arbitrary data.


• The `...` form of command substitution should be avoided these days. Use $(...) instead.


• Because unquoted command substitution also invokes split+glob, it has to be used with care. Here, you're using the split part which means you have a dependency on the current value of $IFS. The output of seq won't include wildcards, so you won't have a problem with the glob part.


• 
  seq is not a standard command and not found on all systems. To loop on numbers, it's not the best approach anyway, as that means storing the whole output in memory and split it (resulting in extra copies in memory), and that means running a separate utility in a separate process even though the shell has builtin support for that.

share|improve this answer

edited 5 mins ago

answered 15 mins ago

Stéphane ChazelasStéphane Chazelas

324k5757 gold badges627627 silver badges996996 bronze badges

add a comment | 

1

#! /bin/sh -

i=1

for arg do

  printf '%2d: %sn' "$i" "$arg"

  i=$((i + 1))

done

That is, instead of looping over the indexes, loop over the arguments and increment an index separately.

A few notes on your approach:

• leaving a parameter expansion unquoted in list context has a very special meaning, you almost never want to do that.


• 
  echo can't be used to output arbitrary data.


• The `...` form of command substitution should be avoided these days. Use $(...) instead.


• Because unquoted command substitution also invokes split+glob, it has to be used with care. Here, you're using the split part which means you have a dependency on the current value of $IFS. The output of seq won't include wildcards, so you won't have a problem with the glob part.


• 
  seq is not a standard command and not found on all systems. To loop on numbers, it's not the best approach anyway, as that means storing the whole output in memory and split it (resulting in extra copies in memory), and that means running a separate utility in a separate process even though the shell has builtin support for that.

share|improve this answer

edited 5 mins ago

answered 15 mins ago

Stéphane ChazelasStéphane Chazelas

324k5757 gold badges627627 silver badges996996 bronze badges

add a comment | 

#! /bin/sh -

i=1

for arg do

  printf '%2d: %sn' "$i" "$arg"

  i=$((i + 1))

done

That is, instead of looping over the indexes, loop over the arguments and increment an index separately.

A few notes on your approach:

• leaving a parameter expansion unquoted in list context has a very special meaning, you almost never want to do that.


• 
  echo can't be used to output arbitrary data.


• The `...` form of command substitution should be avoided these days. Use $(...) instead.


• Because unquoted command substitution also invokes split+glob, it has to be used with care. Here, you're using the split part which means you have a dependency on the current value of $IFS. The output of seq won't include wildcards, so you won't have a problem with the glob part.


• 
  seq is not a standard command and not found on all systems. To loop on numbers, it's not the best approach anyway, as that means storing the whole output in memory and split it (resulting in extra copies in memory), and that means running a separate utility in a separate process even though the shell has builtin support for that.

share|improve this answer

edited 5 mins ago

answered 15 mins ago

Stéphane ChazelasStéphane Chazelas

324k5757 gold badges627627 silver badges996996 bronze badges

#! /bin/sh -

i=1

for arg do

  printf '%2d: %sn' "$i" "$arg"

  i=$((i + 1))

done

share|improve this answer

edited 5 mins ago

answered 15 mins ago

Stéphane ChazelasStéphane Chazelas

324k5757 gold badges627627 silver badges996996 bronze badges

answered 15 mins ago

Stéphane ChazelasStéphane Chazelas

324k5757 gold badges627627 silver badges996996 bronze badges

answered 15 mins ago

Stéphane ChazelasStéphane Chazelas

324k5757 gold badges627627 silver badges996996 bronze badges

0

#!/bin/bash

for ((i=1; i<=$#; i++))

do

  echo "$i - ${!i}"

done

share

answered 5 mins ago

FreddyFreddy

5,51611 gold badge66 silver badges2323 bronze badges

add a comment | 

0

#!/bin/bash

for ((i=1; i<=$#; i++))

do

  echo "$i - ${!i}"

done

share

answered 5 mins ago

FreddyFreddy

5,51611 gold badge66 silver badges2323 bronze badges

add a comment | 

#!/bin/bash

for ((i=1; i<=$#; i++))

do

  echo "$i - ${!i}"

done

share

answered 5 mins ago

FreddyFreddy

5,51611 gold badge66 silver badges2323 bronze badges

#!/bin/bash

for ((i=1; i<=$#; i++))

do

  echo "$i - ${!i}"

done

share

answered 5 mins ago

FreddyFreddy

5,51611 gold badge66 silver badges2323 bronze badges

answered 5 mins ago

FreddyFreddy

5,51611 gold badge66 silver badges2323 bronze badges

answered 5 mins ago

FreddyFreddy

5,51611 gold badge66 silver badges2323 bronze badges