Write hexadecimal values to binary file with bash

Question

How can I write an specific number of zeros in a binary file automatically?

I write:

#!/bin/bash
for i in {0..127};
do
    echo -e -n "\x00" >> file.bin
done

But looking the file with Bless outputs: 2D 65 20 2D 6E 20 5C 78 30 30 0A which corresponds to -e -n \x00.

Answer 1

printf should be portable and supports octal character escapes:

i=0
while [ "$i" -le 127 ]; do
    printf '\000'
    i=$((i+1)) 
done >> file.bin

(printf isn't required to support hex escapes like \x00, but a number of shells support that.)

See Why is printf better than echo? for the troubles with echo.

Answer 2

If it's zeros to a file, then the obvious one is:

dd if=/dev/zero of=file.bin bs=1 count=128  

Note that this is pretty inefficient as it goes, as it does single byte writes. You could just as easily use:

dd if=/dev/zero of=file.bin bs=128 count=1  

Here, bs' is the 'block size' andcount` is how many blocks. Better to write one block than lots of little ones!

Note that the above commands do not append to file.bin, they overwrite it. One way round that is:

dd if=/dev/zero bs=128 count=1 >> file.bin  

Explanation: in the absence of of=, dd writes to standard output, which is then appended to the output file.

Answer 3

1. printf will re-use the same format for all arguments.
2. The format %.0s will print no part of its argument.

Combining these leads to the rather nice

$ printf '\0%.0s' {1..128} >128nulls

for 128 nulls.