Can $() always replace backticks for command substitution? - bash

Question

The following code fails to generate binary numbers if backticks are replaced by dollar-parenthesis syntax:
#!/bin/bash
rm test.bin 2>/dev/null
for character in {0..255}
do
char=`printf '\\\\x'"%02x" $character`
printf "$char" >> test.bin
done
hexdump -C test.bin
Result:
00000000 00 01 02 03 04 05 06 07 08 09 0a 0b 0c 0d 0e 0f |................|
00000010 10 11 12 13 14 15 16 17 18 19 1a 1b 1c 1d 1e 1f |................|
00000020 20 21 22 23 24 25 26 27 28 29 2a 2b 2c 2d 2e 2f | !"#$%&'()*+,-./|
00000030 30 31 32 33 34 35 36 37 38 39 3a 3b 3c 3d 3e 3f |0123456789:;<=>?|
00000040 40 41 42 43 44 45 46 47 48 49 4a 4b 4c 4d 4e 4f |#ABCDEFGHIJKLMNO|
00000050 50 51 52 53 54 55 56 57 58 59 5a 5b 5c 5d 5e 5f |PQRSTUVWXYZ[\]^_|
00000060 60 61 62 63 64 65 66 67 68 69 6a 6b 6c 6d 6e 6f |`abcdefghijklmno|
00000070 70 71 72 73 74 75 76 77 78 79 7a 7b 7c 7d 7e 7f |pqrstuvwxyz{|}~.|
00000080 80 81 82 83 84 85 86 87 88 89 8a 8b 8c 8d 8e 8f |................|
00000090 90 91 92 93 94 95 96 97 98 99 9a 9b 9c 9d 9e 9f |................|
000000a0 a0 a1 a2 a3 a4 a5 a6 a7 a8 a9 aa ab ac ad ae af |................|
000000b0 b0 b1 b2 b3 b4 b5 b6 b7 b8 b9 ba bb bc bd be bf |................|
000000c0 c0 c1 c2 c3 c4 c5 c6 c7 c8 c9 ca cb cc cd ce cf |................|
000000d0 d0 d1 d2 d3 d4 d5 d6 d7 d8 d9 da db dc dd de df |................|
000000e0 e0 e1 e2 e3 e4 e5 e6 e7 e8 e9 ea eb ec ed ee ef |................|
000000f0 f0 f1 f2 f3 f4 f5 f6 f7 f8 f9 fa fb fc fd fe ff |................|
That's ok so far. Let's replace backticks and see what we get:
#!/bin/bash
rm test.bin 2>/dev/null
for character in {0..255}
do
char=$(printf '\\\\x'"%02x" $character)
printf "$char" >> test.bin
done
hexdump -C test.bin
Result:
00000000 5c 78 30 30 5c 78 30 31 5c 78 30 32 5c 78 30 33 |\x00\x01\x02\x03|
00000010 5c 78 30 34 5c 78 30 35 5c 78 30 36 5c 78 30 37 |\x04\x05\x06\x07|
00000020 5c 78 30 38 5c 78 30 39 5c 78 30 61 5c 78 30 62 |\x08\x09\x0a\x0b|
00000030 5c 78 30 63 5c 78 30 64 5c 78 30 65 5c 78 30 66 |\x0c\x0d\x0e\x0f|
00000040 5c 78 31 30 5c 78 31 31 5c 78 31 32 5c 78 31 33 |\x10\x11\x12\x13|
00000050 5c 78 31 34 5c 78 31 35 5c 78 31 36 5c 78 31 37 |\x14\x15\x16\x17|
00000060 5c 78 31 38 5c 78 31 39 5c 78 31 61 5c 78 31 62 |\x18\x19\x1a\x1b|
00000070 5c 78 31 63 5c 78 31 64 5c 78 31 65 5c 78 31 66 |\x1c\x1d\x1e\x1f|
.
.
While I prefer dollar-parenthesis syntax it appears to fail in this case but why ?
Credits for the code snippet:
http://code.activestate.com/recipes/578441-a-building-block-bash-binary-file-manipulation

Answer 1

You are running into one of the reasons that $() is preferred to the backtick notation. The shell parsing of $() is more consistent (as it introduces a new parsing context as I understand it).
So your escaping, while correct for the backtick code, is excessive for the $() code.
Try this:
$ : > test.bin; for character in {0..255}
do
char=$(printf '\\x'"%02x" $character)
printf "$char" >> test.bin
done; hexdump -C test.bin
00000000 00 01 02 03 04 05 06 07 08 09 0a 0b 0c 0d 0e 0f |................|
00000010 10 11 12 13 14 15 16 17 18 19 1a 1b 1c 1d 1e 1f |................|
00000020 20 21 22 23 24 25 26 27 28 29 2a 2b 2c 2d 2e 2f | !"#$%&'()*+,-./|
00000030 30 31 32 33 34 35 36 37 38 39 3a 3b 3c 3d 3e 3f |0123456789:;<=>?|
00000040 40 41 42 43 44 45 46 47 48 49 4a 4b 4c 4d 4e 4f |#ABCDEFGHIJKLMNO|
00000050 50 51 52 53 54 55 56 57 58 59 5a 5b 5c 5d 5e 5f |PQRSTUVWXYZ[\]^_|
00000060 60 61 62 63 64 65 66 67 68 69 6a 6b 6c 6d 6e 6f |`abcdefghijklmno|
00000070 70 71 72 73 74 75 76 77 78 79 7a 7b 7c 7d 7e 7f |pqrstuvwxyz{|}~.|
00000080 80 81 82 83 84 85 86 87 88 89 8a 8b 8c 8d 8e 8f |................|
00000090 90 91 92 93 94 95 96 97 98 99 9a 9b 9c 9d 9e 9f |................|
000000a0 a0 a1 a2 a3 a4 a5 a6 a7 a8 a9 aa ab ac ad ae af |................|
000000b0 b0 b1 b2 b3 b4 b5 b6 b7 b8 b9 ba bb bc bd be bf |................|
000000c0 c0 c1 c2 c3 c4 c5 c6 c7 c8 c9 ca cb cc cd ce cf |................|
000000d0 d0 d1 d2 d3 d4 d5 d6 d7 d8 d9 da db dc dd de df |................|
000000e0 e0 e1 e2 e3 e4 e5 e6 e7 e8 e9 ea eb ec ed ee ef |................|
000000f0 f0 f1 f2 f3 f4 f5 f6 f7 f8 f9 fa fb fc fd fe ff |................|
00000100
A little more clearly compare this
$ printf %s\\n `printf %s "\\\\ff"`
\ff
$ printf %s\\n `printf %s '\\\\ff'`
\\ff
to this
$ printf %s\\n $(printf %s "\\\\ff")
\\ff
$ printf %s\\n $(printf %s '\\\\ff')
\\\\ff

Answer 2

This is the difference:
echo `echo '\\'`
\
echo $(echo '\\')
\\
From the manual, Command substitution section:
When the old-style backquoted form of substitution is used, backslash retains its literal meaning except when followed by "$", "`", or "\".
When using the "$(COMMAND)" form, all characters between the parentheses make up the command; none are treated specially.
$() doesn't need extra escaping for \. Use:
char=$(printf '\\x'"%02x" $character)