Shell Programming and Scripting

Code:

$ ksh
$ echo "\047"0"\047"

result:

Code:

8'

Code:

# ksh
# echo "\047"0"\047"
\0470\047

But not in cygwin or aix.

Wow. On AIX, I had to treble the number of '\' to get the intended result.

# echo "\\\047"0"\\\047"
\0470\047
# echo "\\047"0"\\047"
8'
# echo "\047"0"\047"
8'

Tried with other numbers also:
# echo "\046"0"\046"
0&
# echo "\045"0"\045"
(%
# echo "\044"0"\044"
$
# echo "\043"0"\043"
#

Who knew?...

use 'printf'

Code:

$ ksh
$ echo "\047"0"\047"
8'
$ printf "\047"0"\047"
'0'

To answer the question, the echo man page says:

Code:

 \0n   where n is the 8-bit character whose ASCII  code
                 is the 1-, 2- or 3-digit octal number represent-
                 ing that character.

So the \0 just says that 1, 2, or 3 digits may follow. That zero in \0 is not counted as of of the digits. So
echo "\047"0
is seen by the echo command exactly like
echo "\0470"
and is interrpreted as an attempt to use a byte with the octal value "470". This is too big and so leading bits are dropped. In this case the bit represented by the 4 is dropped and character becomes "\070".

Code:

# echo "\0470" | od -to1
0000000 070 012
0000002
# bc
obase=2
ibase=8
470
100111000
#

Edit:
By the way, a couple of ways that do work:

Code:

# echo "'0'"
'0'
# echo "\00470\0047"
'0'
#

That answers it. Thank you very much!