Cfrtjryk

Is it possible to use xargs to invoke a command so that the last argument of the command is fixed?

My attempt:

printf '%sn' a b c d | xargs -I{} echo {} LAST

ends up doing

echo a LAST  

echo b LAST  

echo c LAST  

echo d LAST  

I want for xargs to invoke

echo a b c d LAST

#fit as many as you can but always finish wiht LAST

Is this possible to do, preferably in a portable way?

tl;dr; this is how you could do it portably, without -I and other broken fancy options:

$ echo a b c d f g | xargs -n 2 sh -c 'echo "$@" LAST' sh

a b LAST

c d LAST

f g LAST



$ seq 1 100000 | xargs sh -c 'echo "$#" LAST' sh

23692 LAST

21841 LAST

21841 LAST

21841 LAST

10785 LAST

The problem with the -I option is that it's broken by design, and there is no way around it:

$ echo a b c d f g | xargs -I {} -n 1 echo {} LAST

a b c d f g LAST

$ echo a b c d f g | xargs -I {} -n 2 echo {} LAST

{} LAST a b

{} LAST c d

{} LAST f g

But they're probably covered, because that's what the standard says:

-I replstr
^[XSI] [Option Start] Insert mode: utility is executed for each
line from standard input, taking the entire line as a single
argument, inserting it in arguments for each occurrence of
replstr.

And it doesn't say anything about the interaction with the -n and -d options, so they're free to do whatever they please.

This is how it is on an (older) FreeBSD, less unexpected but non-standard:

fzu$ echo a b c d f g | xargs -I {} -n 2 echo {} LAST

a b LAST

c d LAST

f g LAST

fzu$ echo a b c d f g | xargs -I {} -n 1 echo {} LAST

a LAST

b LAST

c LAST

d LAST

f LAST

g LAST

Not with xargs (alone). If you have an item list of unpredictable length, how should xargs know from the beginning (= first element) which element would be the last one?
You'll need some additional logics around it to separate the desired element from the others.