Does deleting sed pattern space with 'd' erase hold space as well?

Question

Can someone please explain why this is happening?

This is expected:

$ echo -e "foo\nbar" | sed -n 'h; x; p'
foo
bar

I put every line in the hold space, then swap hold space and pattern space, then print the pattern space, so every line is printed. Now, why is the following different?

$ echo -e "foo\nbar" | sed -n 'h; d; x; p'

I thought that wouldn't be, because I delete the pattern space before swapping, so the stored line should be put back to the pattern space anyway. It's the hold space that should be empty after x;, right? I delete the pattern space, then swap. Where does the line I've saved go?

Answer 1

When you use d, the pattern space is cleared, the next line is read, and processing starts over from the beginning of the script. Thus, you never actually reach the x and p steps, instead just copying to the hold space and deleting.

Answer 2

I guess it's related to the following line in man sed:

d Delete pattern space. Start next cycle.

The following works as expected:

$ echo -e "foo\nbar" | sed -n 'h; s/.*//; g; p'
foo
bar

Sorry for bothering you guys.