On Wed, 29 Mar 2006 20:18:07 -0600, Sumo Wrestler (or just ate too much) wrote:
>> the command printf seems to be ignoring the ending '\n' that passes to >> it: >> >> [...] > > It's an interaction between the echo command and the shell that's causing > the trailing \n to be removed. Printf is working just fine: $ printf > "hello\n" | od -t x1 - > 0000000 68 65 6c 6c 6f 0a > 0000006 > This is the correct output. But what happens when you use echo: $ echo > -n `printf "hello\n\n\n\n\n"` | od -t x1 - 0000000 68 65 6c 6c 6f > 0000005 oh, thanks for the input. It really made me thinking, and I found that it is not echo's fault, because the previous \n was printed ok. here is more test: $ echo -n "`printf "hello\n\n"`" | od -t x1 - 0000000 68 65 6c 6c 6f 0000005 but: $ echo -n "`printf "hello\n\n "`" | od -t x1 - 0000000 68 65 6c 6c 6f 0a 0a 20 0000010 So is it bash that is eating the trailing \n's? > Avoid using "echo" when you need trailing whitespace; instead, write the Actually, I use echo as an example, the assignment variable is then used in awk. :-) I need to get the result into a variable. thanks tong ps, my bash $ bash --version GNU bash, version 2.05b.0(1)-release (i386-pc-linux-gnu) -- To UNSUBSCRIBE, email to [EMAIL PROTECTED] with a subject of "unsubscribe". Trouble? Contact [EMAIL PROTECTED]
