Why can't sed replace 0x24 byte in binary file?

Question

I'm want to replace some bytes in binary file to another. Created sample (6 bytes long) file via

echo -ne '\x8f\x15\x42\x02\x24\xc2' > test

Then tried to replace bytes \x15\x42\x02 to \x12\x12\x02 via sed:

sed 's \x15\x42\x02 \x12\x12\x02 g' test > test1

sed replaced bytes:

cat test test1 | xxd -c 6
0000000: 8f15 4202 24c2  ..B.$.
0000006: 8f12 1202 24c2  ....$.
           ^^ ^^^^

Tried then replace bytes \x42\x02\x24 to \x12\x02\x24:

sed 's \x42\x02\x24 \x12\x02\x24 g' test > test2

sed NOT replaced bytes:

cat test test2 | xxd -c 6
0000000: 8f15 4202 24c2  ..B.$.
0000006: 8f15 4202 24c2  ..B.$.
              ^^^^ ^^

What's wrong? I have sed (GNU sed) 4.2.2 (Kubuntu 13.10)

Thank You.

Answer 1

It is because 0x24 is the hex code for the $ character. Note that ^ and $ have special meaning in regex (which sed uses for matching), and literally means "at the beginning of the line", and "at the end of the line", respectively. As such, those characters are ignored unless escaped. Therefore,

echo "this and that" | sed 's/this$/cat/'

will leave the string unchanged because it you're telling sed to look for 'this' at the end of the line -- since 'that' is at the end, it doesn't match. However,

echo "It costs $50.00." | sed 's/\$50\.00/47.00 Swiss Francs/'

Will change the line as expected because the $ was escaped. For binary data, just include the hex code for the \ character (\x5c) just prior to \x24 in the regex section, and it should work just fine.

Cheers.

Answer 2

You can try this,

hexdump -ve '1/1 "%.2X"' file1 | sed 's/420224/121224/g' | xxd -r -p > new_updated

Test:

sat:~# xxd -c 6  file1
0000000: 8f15 4202 24c2  ..B.$.
sat:~# hexdump -ve '1/1 "%.2X"' file1 | sed 's/420224/121224/g' | xxd -r -p > new_updated
sat:~# xxd -c 6 new_updated
0000000: 8f15 1212 24c2  ....$.