File holes are the empty spaces in file, which, however, doesn\'t take up any disk space and contains null bytes. Therefore, the file size is larger than its actual size on
Use the dd command with a seek parameter.
dd if=/dev/urandom bs=4096 count=2 of=file_with_holes
dd if=/dev/urandom bs=4096 seek=7 count=2 of=file_with_holes
That creates for you a file with a nice hole from byte 8192 to byte 28671.
Here's an example, demonstrating that indeed the file has holes in it (the ls -s command tells you how many disk blocks are being used by a file):
$ dd if=/dev/urandom bs=4096 count=2 of=fwh # fwh = file with holes
2+0 records in
2+0 records out
8192 bytes (8.2 kB) copied, 0.00195565 s, 4.2 MB/s
$ dd if=/dev/urandom seek=7 bs=4096 count=2 of=fwh
2+0 records in
2+0 records out
8192 bytes (8.2 kB) copied, 0.00152742 s, 5.4 MB/s
$ dd if=/dev/zero bs=4096 count=9 of=fwnh # fwnh = file with no holes
9+0 records in
9+0 records out
36864 bytes (37 kB) copied, 0.000510568 s, 72.2 MB/s
$ ls -ls fw*
16 -rw-rw-r-- 1 hopper hopper 36864 Mar 15 10:25 fwh
36 -rw-rw-r-- 1 hopper hopper 36864 Mar 15 10:29 fwnh
As you can see, the file with holes takes up fewer disk blocks, despite being the same size.
If you want a program that does it, here it is:
#include
#include
#include
#include
int main(int argc, const char *argv[])
{
char random_garbage[8192]; /* Don't even bother to initialize */
int fd = -1;
if (argc < 2) {
fprintf(stderr, "Usage: %s \n", argv[0]);
return 1;
}
fd = open(argv[1], O_WRONLY | O_CREAT | O_TRUNC, 0666);
if (fd < 0) {
perror("Can't open file: ");
return 2;
}
write(fd, random_garbage, 8192);
lseek(fd, 5 * 4096, SEEK_CUR);
write(fd, random_garbage, 8192);
close(fd);
return 0;
}
The above should work on any Unix. Someone else replied with a nice alternative method that is very Linux specific. I highlight it here because it's a method distinct from the two I gave, and can be used to put holes in existing files.