Linux xargs command

Question

A quick question. A command

cat * | wc -c

doesn't require xargs, but the command

ls | xargs echo

requires xargs command. Can please someone explain me the concept of xargs more clearly. Thanks.

Answer 1

In short, xargs converts stdin (standard input) to arguments for the command you specify. For example

$ seq 1 3
1
2
3
$ seq 1 3 | xargs echo
1 2 3

seq, as you can see, prints a sequence to stdout. We pipe (|) that output to xargs on stdin. xargs calls echo with stdin as arguments, so we get echo 1 2 3.

As el.pescado said, wc accepts input on stdin (you can also give it a file argument). Because cat prints a file to stdout, you can pipe it directly to wc.

$ cat text
This is only
a test
$ cat text | wc
      2       5      20
$ wc text
 2  5 20 text

echo does not accept anything from stdin. That would be weird, since echo's job is to print to stdout - you could already print anything you'd pipe it. So, you use xargs to convert the stream to arguments.

echo might be too trivial a command to see what's going on, so here's a more real-world example. Say we've got a directory with some files in it:

$ ls
bar1 foo1 foo2 foo3 foo4 foo5 foo6

We've had it up to here with foo, and we want to delete all of them, but we can't be bothered to type rm foo1 foo2 .... After all, we're programmers, and we're lazy. What we can do is...

$ ls foo* | xargs rm
$ ls
bar1

rm expects arguments, ls foo* prints every file we want to remove, and xargs does the translation.

---

As a side note, sometimes you want to split up stdin into smaller pieces. xargs -n is highly useful for that, and passes N arguments at a time to your final command.

$ ls foo* | xargs -n2 echo
foo1 foo2
foo3 foo4
foo5 foo6

Answer 2

wc reads data from its standard input, whereas echo prints its command line arguments.