Passing multiple parameters via xargs

I believe that you can’t use -I that way.  But you can get the effect / behavior you want by saying:

echo {1..8} | xargs -n2 sh -c 'echo "the number $1 comes before $2"' sh

This, essentially, creates an ad hoc one-line shell script, which xargs executes via sh -c.  The two values that xargs parses out of the input are passed to this “script”.  The shell then assigns those values to $1 and $2, which you can then reference in the “script”.

In the specific case of printf, you could always do:

echo {1..8} | xargs printf 'the number %s comes before %s\n'

because printf has an intrinsic xargs-like ability to execute multiple times if it is given more arguments than it needs for a single invocation.  Though that has little advantage over

printf 'the number %s comes before %s\n' {1..8}

And for large lists, the simple xargs command could result in xargs running several instances of printf, some of which might have odd numbers of arguments. You could pass -n 1000 to xargs to guard against that, where 1000 is an even number that should be small enough so as not to reach the arg list too long limit and large enough to avoid running so many printfs.

Note that xargs would call, not your shell's builtin printf, but the external printf, with each invocation in a separate new process.

Also note that for an empty input, except on some BSDs, it would still run printf once with no argument. GNU xargs and compatible have a -r (or --no-run-if-empty) option to avoid that.

To be clear, this simple answer is specific to your printf example, and would not work in the general case where you must pass two parameters at a time to your command (as would be the case for diff, for example).  To solve the general problem with zsh, you can use:

for i j ({1..8}) echo "the number $i comes before $j"

try this:

echo {1..8} |xargs -n 2 bash -c 'echo "the number $0 comes before $1"'