How to pad a file to a desired size?

Question

I have a file that I want to pad until it reaches 16 MiB (16777216 bytes). Currently it is 16515072 bytes. The difference is 262144 bytes.

How do I pad it?

This doesn't seem to be working:

cp smallfile.img largerfile.img
dd if=/dev/zero of=largerfile.img bs=1 count=262144

Answer 1

Besides the answers to get a physical padding you may also leave most of the padding space in the file just empty ("holes"), by seeking to the new end-position of the file and writing a single character:

dd if=/dev/zero of=largerfile.txt bs=1 count=1 seek=16777215

(which has the advantage to be much more performant, specifically with bs=1, and does not occupy large amounts of additional disk space).

That method seems to work even without adding any character, by using if=/dev/null and the final desired file size:

dd if=/dev/null of=largerfile.txt bs=1 count=1 seek=16777216

A performant variant of a physical padding solution that uses larger block-sizes is:

padding=262144 bs=32768 nblocks=$((padding/bs)) rest=$((padding%bs))
{
  dd if=/dev/zero bs=$bs count=$nblocks
  dd if=/dev/zero bs=$rest count=1
} 2>/dev/null >>largerfile.txt

Answer 2

Drop the of=largerfile.txt and append stdout to the file:

dd if=/dev/zero bs=1 count=262144 >> largerfile.txt

Answer 3

Simple answer, courtesy of @frostschutz, posted as an answer

truncate -s 16M thefile

Answer 4

The best anwser here is Janis's (above) because it lets you forget about the current file size and pad directly to the desired size with no calculation.

It also takes advantage of sparse files, which appending /dev/zero doesn't.

The answer could be tidier though, because 'count' is allowed to be 0 and you still get the padding:

dd if=/dev/null of=largerfile.txt bs=1 count=0 seek=16777216

(Edit: this is correct for GNU dd, but the behaviour of count=0 is platform-specific, see comments)

Answer 5

Do you have to use dd? If you want a file to have a particular (logical) length, just write a zero to the position you want. The bytes between the previous end and the written byte will be displayed as having null bytes. Here's an example using perl.

$ echo Hello > file
$ ls -l file
-rw-r--r-- 1 user group 6 Apr 16 22:59 file
$ perl -le 'open(my $f,"+<","file"); seek($f, 16777216 - 2, 0); print $f "\0"'
$ ls -ln file
-rw-r--r-- 1 user group 16777216 Apr 16 22:59 file

Why the "- 2" in the line? The script will write a byte, so we subtract 1 to seek to the position before that byte. We take off the other because the seek position is zero-indexed.

Answer 6

Here is a solution that specifically does not rely on math.

echo 'The secret meetup location is at the seventh stairway at seven.' > message.txt
dd if=/dev/zero bs=1 count=300 > square.rgb
dd if=message.txt of=square.rgb conv=notrunc

I am using this as steganography to encode a message into an image for a Su Square.