I was asked a question recently and I have not been able to deduce the answer and so am appealing to the group for further education. I could not find a match for my question in the FAQ.
$ printf "\n" | od -tx1 0000000 0a $ printf "\n" | wc -c 1 But when this is used with a command substitution the newline disappears. $ printf "$(printf "\n")" | wc -c 0 Thinking that the IFS was reason for this I tried the following, where the ' ' is a single space character. $ export IFS=' ' $ printf "\n" | wc -c 1 $ printf "$(printf "\n")" | wc -c 0 Hmm... Shouldn't this last case have printed a single newline? The IFS no longer contains a newline. Why wouldn't a literal 0x0a be embedded into the "$(...)" command substitution string and then be printed by printf? The following shows that a literal newline would be printed in that case. $ printf " " | wc -c 1 $ printf " > " | od -tx1 0000000 0a I can tell that I am missing something basic in the understanding of the behavior. Thanks Bob _______________________________________________ Bug-bash mailing list Bug-bash@gnu.org http://lists.gnu.org/mailman/listinfo/bug-bash
