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 <unistd.h>
#include <sys/types.h>
#include <stdio.h>
#include <fcntl.h>
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 <filename>
", 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.
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