Using xargs to run multiple commands

Question

Using this post as a starting point I am running the following in bash:

seq 1 5 | xargs -d $'\n' sh -c 'for arg do echo $arg; done'

Expected output

1
2
3
4
5

Actual output

2
3
4
5

i.e. is missing the first of the intended arguments.

Am probably being a tool, but wondering why this is.

Answer 1

You have to pass some dummy value at position 0 to sh script like this:

seq 1 5 | xargs -d $'\n' sh -c 'for arg do echo $arg; done' _
1
2
3
4
5

Without passing _ to sh script 1 is passed as $0 whereas for arg loops through positional arguments starting with position 1 only.