How to pass list of arguments into a function
I want to pass arguments from a loop into a function func. For simplicity let's say we are working with the loop
for x in {1..5}
do
for y in {a..c}
do
echo $x$y
done
done
I'm just echoing because I don't know what to do. I'd like to run the equivalent of func 1a 1b 1c 2a 2b 2c 3a 3b 3c 4a 4b 4c 5a 5b 5c.
How do I do this?
shell-script function arguments
asked 4 hours ago
CharacterClassCharacterClass
103
New contributor
CharacterClass is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I want to pass arguments from a loop into a function func. For simplicity let's say we are working with the loop
for x in {1..5}
do
for y in {a..c}
do
echo $x$y
done
done
I'm just echoing because I don't know what to do. I'd like to run the equivalent of func 1a 1b 1c 2a 2b 2c 3a 3b 3c 4a 4b 4c 5a 5b 5c.
How do I do this?
shell-script function arguments
asked 4 hours ago
CharacterClassCharacterClass
103
New contributor
CharacterClass is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In the loop:y="$y $x"; outside the loop:func $y. This only covers the most simple case. If you have special characters in your list, you need to do something more complex.
– Weijun Zhou
3 hours ago
@WeijunZhou Thanks, this worked!
– CharacterClass
3 hours ago
add a comment |
I want to pass arguments from a loop into a function func. For simplicity let's say we are working with the loop
for x in {1..5}
do
for y in {a..c}
do
echo $x$y
done
done
I'm just echoing because I don't know what to do. I'd like to run the equivalent of func 1a 1b 1c 2a 2b 2c 3a 3b 3c 4a 4b 4c 5a 5b 5c.
How do I do this?
shell-script function arguments
asked 4 hours ago
CharacterClassCharacterClass
103
New contributor
CharacterClass is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I want to pass arguments from a loop into a function func. For simplicity let's say we are working with the loop
for x in {1..5}
do
for y in {a..c}
do
echo $x$y
done
done
I'm just echoing because I don't know what to do. I'd like to run the equivalent of func 1a 1b 1c 2a 2b 2c 3a 3b 3c 4a 4b 4c 5a 5b 5c.
How do I do this?
shell-script function arguments
shell-script function arguments
asked 4 hours ago
CharacterClassCharacterClass
103
New contributor
CharacterClass is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
CharacterClassCharacterClass
103
New contributor
CharacterClass is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
CharacterClass
asked 4 hours ago
CharacterClassCharacterClass
103
New contributor
CharacterClass is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
CharacterClassCharacterClass
103
asked 4 hours ago
CharacterClassCharacterClass
103
103
New contributor
CharacterClass is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
CharacterClass is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
CharacterClass is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In the loop:y="$y $x"; outside the loop:func $y. This only covers the most simple case. If you have special characters in your list, you need to do something more complex.
– Weijun Zhou
3 hours ago
@WeijunZhou Thanks, this worked!
– CharacterClass
3 hours ago
add a comment |
In the loop:y="$y $x"; outside the loop:func $y. This only covers the most simple case. If you have special characters in your list, you need to do something more complex.
– Weijun Zhou
3 hours ago
@WeijunZhou Thanks, this worked!
– CharacterClass
3 hours ago
In the loop:
y="$y $x"; outside the loop: func $y. This only covers the most simple case. If you have special characters in your list, you need to do something more complex.– Weijun Zhou
3 hours ago
In the loop:
y="$y $x"; outside the loop: func $y. This only covers the most simple case. If you have special characters in your list, you need to do something more complex.– Weijun Zhou
3 hours ago
@WeijunZhou Thanks, this worked!
– CharacterClass
3 hours ago
@WeijunZhou Thanks, this worked!
– CharacterClass
3 hours ago
add a comment |
1 Answer
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func {1..5} would be equivalent to func 1 2 3 4 5.
In general, the list of words in a for statement is just like any list of words in a command, so you can just replace the loop with a single invocation of the command, with whatever list you used there moved to the command arguments.
Also, you can use multiple brace expansions together: {1..5}{a..c} would create the list 1a 1b 1c 2a 2b 2c 3a 3b 3c 4a 4b 4c 5a 5b 5c (as distinct words), so in the case you show, func {1..5}{a..c} should work.
If your loop does something more complex to create the arguments to the final command, you can use an array to collect them (in Bash/ksh/zsh). Assuming we have generate_arg that has to be run to produce the arguments to func:
args=()
for i in {1..5}; do
args+=( "$(generate_arg "$i")" )
done
func "${args[@]}"
(Using an array is better than concatenating the values to a string in that it keeps values with whitespace intact.)
answered 3 hours ago
ilkkachuilkkachu
60.9k1098174
There are actually two loops, they're nested. Is there a version of the first instance that works?
– CharacterClass
3 hours ago
@CharacterClass, well, now I'm not exactly sure what your situation is then. Can you edit your post to show those nested loops in a bit more detail?
– ilkkachu
3 hours ago
Thank you. I've updated the question. The loop will serve only to concatenate the variables.
– CharacterClass
3 hours ago
@CharacterClass, right, that looks simple since you can use multiple brace expansions. Edited, the section in the middle.
– ilkkachu
3 hours ago
add a comment |
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func {1..5} would be equivalent to func 1 2 3 4 5.
In general, the list of words in a for statement is just like any list of words in a command, so you can just replace the loop with a single invocation of the command, with whatever list you used there moved to the command arguments.
Also, you can use multiple brace expansions together: {1..5}{a..c} would create the list 1a 1b 1c 2a 2b 2c 3a 3b 3c 4a 4b 4c 5a 5b 5c (as distinct words), so in the case you show, func {1..5}{a..c} should work.
If your loop does something more complex to create the arguments to the final command, you can use an array to collect them (in Bash/ksh/zsh). Assuming we have generate_arg that has to be run to produce the arguments to func:
args=()
for i in {1..5}; do
args+=( "$(generate_arg "$i")" )
done
func "${args[@]}"
(Using an array is better than concatenating the values to a string in that it keeps values with whitespace intact.)
answered 3 hours ago
ilkkachuilkkachu
60.9k1098174
There are actually two loops, they're nested. Is there a version of the first instance that works?
– CharacterClass
3 hours ago
@CharacterClass, well, now I'm not exactly sure what your situation is then. Can you edit your post to show those nested loops in a bit more detail?
– ilkkachu
3 hours ago
Thank you. I've updated the question. The loop will serve only to concatenate the variables.
– CharacterClass
3 hours ago
@CharacterClass, right, that looks simple since you can use multiple brace expansions. Edited, the section in the middle.
– ilkkachu
3 hours ago
add a comment |
func {1..5} would be equivalent to func 1 2 3 4 5.
In general, the list of words in a for statement is just like any list of words in a command, so you can just replace the loop with a single invocation of the command, with whatever list you used there moved to the command arguments.
Also, you can use multiple brace expansions together: {1..5}{a..c} would create the list 1a 1b 1c 2a 2b 2c 3a 3b 3c 4a 4b 4c 5a 5b 5c (as distinct words), so in the case you show, func {1..5}{a..c} should work.
If your loop does something more complex to create the arguments to the final command, you can use an array to collect them (in Bash/ksh/zsh). Assuming we have generate_arg that has to be run to produce the arguments to func:
args=()
for i in {1..5}; do
args+=( "$(generate_arg "$i")" )
done
func "${args[@]}"
(Using an array is better than concatenating the values to a string in that it keeps values with whitespace intact.)
answered 3 hours ago
ilkkachuilkkachu
60.9k1098174
There are actually two loops, they're nested. Is there a version of the first instance that works?
– CharacterClass
3 hours ago
@CharacterClass, well, now I'm not exactly sure what your situation is then. Can you edit your post to show those nested loops in a bit more detail?
– ilkkachu
3 hours ago
Thank you. I've updated the question. The loop will serve only to concatenate the variables.
– CharacterClass
3 hours ago
@CharacterClass, right, that looks simple since you can use multiple brace expansions. Edited, the section in the middle.
– ilkkachu
3 hours ago
add a comment |
func {1..5} would be equivalent to func 1 2 3 4 5.
In general, the list of words in a for statement is just like any list of words in a command, so you can just replace the loop with a single invocation of the command, with whatever list you used there moved to the command arguments.
Also, you can use multiple brace expansions together: {1..5}{a..c} would create the list 1a 1b 1c 2a 2b 2c 3a 3b 3c 4a 4b 4c 5a 5b 5c (as distinct words), so in the case you show, func {1..5}{a..c} should work.
If your loop does something more complex to create the arguments to the final command, you can use an array to collect them (in Bash/ksh/zsh). Assuming we have generate_arg that has to be run to produce the arguments to func:
args=()
for i in {1..5}; do
args+=( "$(generate_arg "$i")" )
done
func "${args[@]}"
(Using an array is better than concatenating the values to a string in that it keeps values with whitespace intact.)
answered 3 hours ago
ilkkachuilkkachu
60.9k1098174
func {1..5} would be equivalent to func 1 2 3 4 5.
In general, the list of words in a for statement is just like any list of words in a command, so you can just replace the loop with a single invocation of the command, with whatever list you used there moved to the command arguments.
Also, you can use multiple brace expansions together: {1..5}{a..c} would create the list 1a 1b 1c 2a 2b 2c 3a 3b 3c 4a 4b 4c 5a 5b 5c (as distinct words), so in the case you show, func {1..5}{a..c} should work.
If your loop does something more complex to create the arguments to the final command, you can use an array to collect them (in Bash/ksh/zsh). Assuming we have generate_arg that has to be run to produce the arguments to func:
args=()
for i in {1..5}; do
args+=( "$(generate_arg "$i")" )
done
func "${args[@]}"
(Using an array is better than concatenating the values to a string in that it keeps values with whitespace intact.)
answered 3 hours ago
ilkkachuilkkachu
60.9k1098174
edited 3 hours ago
answered 3 hours ago
ilkkachuilkkachu
60.9k1098174
answered 3 hours ago
ilkkachuilkkachu
60.9k1098174
answered 3 hours ago
ilkkachuilkkachu
60.9k1098174
60.9k1098174
There are actually two loops, they're nested. Is there a version of the first instance that works?
– CharacterClass
3 hours ago
@CharacterClass, well, now I'm not exactly sure what your situation is then. Can you edit your post to show those nested loops in a bit more detail?
– ilkkachu
3 hours ago
Thank you. I've updated the question. The loop will serve only to concatenate the variables.
– CharacterClass
3 hours ago
@CharacterClass, right, that looks simple since you can use multiple brace expansions. Edited, the section in the middle.
– ilkkachu
3 hours ago
add a comment |
There are actually two loops, they're nested. Is there a version of the first instance that works?
– CharacterClass
3 hours ago
@CharacterClass, well, now I'm not exactly sure what your situation is then. Can you edit your post to show those nested loops in a bit more detail?
– ilkkachu
3 hours ago
Thank you. I've updated the question. The loop will serve only to concatenate the variables.
– CharacterClass
3 hours ago
@CharacterClass, right, that looks simple since you can use multiple brace expansions. Edited, the section in the middle.
– ilkkachu
3 hours ago
There are actually two loops, they're nested. Is there a version of the first instance that works?
– CharacterClass
3 hours ago
There are actually two loops, they're nested. Is there a version of the first instance that works?
– CharacterClass
3 hours ago
@CharacterClass, well, now I'm not exactly sure what your situation is then. Can you edit your post to show those nested loops in a bit more detail?
– ilkkachu
3 hours ago
@CharacterClass, well, now I'm not exactly sure what your situation is then. Can you edit your post to show those nested loops in a bit more detail?
– ilkkachu
3 hours ago
Thank you. I've updated the question. The loop will serve only to concatenate the variables.
– CharacterClass
3 hours ago
Thank you. I've updated the question. The loop will serve only to concatenate the variables.
– CharacterClass
3 hours ago
@CharacterClass, right, that looks simple since you can use multiple brace expansions. Edited, the section in the middle.
– ilkkachu
3 hours ago
@CharacterClass, right, that looks simple since you can use multiple brace expansions. Edited, the section in the middle.
– ilkkachu
3 hours ago
add a comment |
CharacterClass is a new contributor. Be nice, and check out our Code of Conduct.
CharacterClass is a new contributor. Be nice, and check out our Code of Conduct.
CharacterClass is a new contributor. Be nice, and check out our Code of Conduct.
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In the loop:
y="$y $x"; outside the loop:func $y. This only covers the most simple case. If you have special characters in your list, you need to do something more complex.– Weijun Zhou
3 hours ago
@WeijunZhou Thanks, this worked!
– CharacterClass
3 hours ago