Bash％s格式说明符消除了打印输出时字符串中的空格

Question

I am starting to use the printf instead of echo . My first forray into printf %s is this:

#!/bin/bash
danny=$(tail -1 /come/and/play/with/US.log| ~/walt/convert_gm_est)
printf "%s" $danny

08:02.020ZINFO<-casper/casperbox001wowSYSSTATS[sz=21,tag=0,aux=0]process_log2221397800

It eliminates all the spaces in the string. So I added a space after the format specifier. the string looks nicer with the spaces plus I could rip out the time with awk very easily. I don't see anything about this on the interwebs. I tried a "%s\\s" and that did not work. Is this standard procedure for using format specifiers - use a space after %s ? Is this the way to do it or am I missing something?

#!/bin/bash
danny=$(tail -1 /come/and/play/with/US.log| ~/walt/convert_gm_est)
printf "%s " $danny

08:02.020Z INFO <- casper/casperbox001wow SYSSTATS[sz=21, tag=0, aux=0] process_log 2221397 80 0 casper@casperbox001wow:~$

Answer 1

When the shell evaluates:

printf "%s" $danny

the shell will expand the value of the variable danny and then split it into words. It will also expand globs in those words. Once that is done, the expression will look something like this (quotes added for clarification):

printf '%s' '08:02.020Z' 'INFO' '<-' 'casper/casperbox001' 'wow0' 'SYSSTATS'...

printf repeats its format string until all of the arguments are consumed. So using the format string %s causes the arguments to be concatenated without intervening spaces.

You probably meant to quote $danny so that it would be presented as a single argument to printf:

printf "%s" "$danny"