Krdytkyu

After reading Stephen Kitt's reply, xargs waits for receiving the stdin input before processing any of the input, such as splitting it into arguments.

How does xargs know when a stdin input ends, so that it can start processing it?

Is -E used for specifying the end of a stdin input?

Without it, how does xargs knows when it ends? Is there some timeout?

To read its input, xargs uses the read function (see also the corresponding Linux manpage).

read reads the next available data from a file descriptor into an area of memory specified by the calling program. As used by xargs, read waits until either data is available, or an error occurs, or it determines that no more data will ever be available. It returns respectively a positive integer, -1, or 0 in each of these cases.

To determine when it’s finished reading its input (its standard input, or the input specified by the -a option), xargs looks for the specified end-of-file marker (if the -E option was used), or a 0 return value from read.

You can see this in action by running

printf '%s ' {1..1024} | strace -e read xargs -s 2048 -x

How does xargs know when a stdin input ends, so that it can start processing it?

xargs does not wait for the end of stdin before starting to process it:

$ while date +%H:%S; do sleep 1; done | xargs -n2 echo

16:51 16:52

16:53 16:54

16:55 16:56

16:57 16:58

^C

xargs knows that the stdin has ended just like any other program, eg. cat(1) or tee(1), by checking the return value of read(2), calling feof(3), etc. xargs will also treat a read error on stdin just as an end-of-file.