How do I pass variable value as bracket expansion sequence value?

Question

The code block in particular is:

for n in {1..$PASSEDARGS}; do
      printf "%s\n" "$n"
done

The following code block doesn't work, instead it prints {1..3} where 3 is the value of the number of arguments passed to the script.

How do I make the program print the argument at $n?

Kusalananda · Accepted Answer · 2017-01-04 10:33:48Z

0

Using the C-like for-loop syntax of Bash:

for (( i = 1; i <= PASSEDARGS; ++i )); do
  printf "%s\n" "$i"
done

answered Jan 4, 2017 at 10:33

Kusalananda♦

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Ipor Sircer · Accepted Answer · 2017-01-04 10:21:25Z

-1

for n in $(seq $#); do
      printf "%s\n" "$n"
      eval echo argument at $n: \$$n
done

answered Jan 4, 2017 at 10:21

Ipor Sircer

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2 Answers 2