CODEWITHSUNDEEP
bash
Lukáš Velecký
Lukáš Velecký

Reputation: 17

Need to printf script's arguments

I'm trying to list script's argument by printf function. Arguments are counted by iteration of $i. What should be in printf function? I need something like

eval echo \$$i

but in printf function.

Edit: Have while cycle with iteration of $i and among other code, I have

printf "%s" $i

But, instead of $i, I need some code, that shows me value of argument. In my case, it is name of file, and I need list them. One file in one iteration.

Upvotes: 0

Views: 599

Answers (4)

smani
smani

Reputation: 1973

It is not very clear what you are asking for, but this will list the arguments passed to the script:

while [ $# -gt 0 ]; do
    printf "%s\n" "$1"
    shift
done

Upvotes: 1

Charles Duffy
Charles Duffy

Reputation: 295679

To get your arguments in such a way that they can be fed back into the shell with the exact same value, the following bash extension can be used:

printf '%q ' "$@"; printf '\n'

This works even for rather unusual cases. Let's say that one of your arguments contains a newline:

./your-script 'hello
world' 'goodbye world'

This will be represented in the printf output:

$'hello\nworld' goodbye\ world

...with something you can use again in the shell:

$ echo $'hello\nworld' goodbye\ world
hello
world goodbye world

Upvotes: 0

Diego Torres Milano
Diego Torres Milano

Reputation: 69368

I guess what you are asking for is how to make the indirect reference to positional parameter i (i containing the position):

print "%s\n" ${!i}

Upvotes: 0

Jonathan Leffler
Jonathan Leffler

Reputation: 754590

As noted in a comment, you normally do that with a loop such as:

i=1
for arg in "$@"
do
    echo "$i: [$arg]"
    ((i++))
done

(If echo isn't allowed, use printf "%s\n" … where the … is whatever would have followed echo.)

You might also use indirect expansion to avoid the use of eval:

for i in 1 2 3 4; do echo "$i: [${!i}]"; done

You can generalize that with:

for i in $(seq 1 $#); do echo "$i: [${!i}]"; done

or

for ((i = 1; i <= $#; i++)); do echo "$i: [${!i}]"; done

For example, given:

set -- a b 'c  d' '  e  f  '

all the loops produce the output:

1: [a]
2: [b]
3: [c  d]
4: [  e  f  ]

The square brackets are merely there to delimit the argument values; it allows you to see the trailing blanks on the fourth line of output.

You might also be able to use:

printf "[%s]\n" "$@"

to get:

[a]
[b]
[c  d]
[  e  f  ]

Upvotes: 2

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