LX0-103 · Question #13
Which of the following commands overwrites the bootloader located on /dev/sda without overwriting the partition table or any data following it?
The correct answer is C. dd if=/dev/zero of=/dev/sda bs=440 count=1. The MBR bootloader code occupies only the first 440 bytes of a 512-byte MBR; bytes 446-511 hold the partition table, so writing exactly 440 bytes zeros only the bootloader.
Question
Which of the following commands overwrites the bootloader located on /dev/sda without overwriting the partition table or any data following it?
Options
- Add if=/dev/zero of=/dev/sda bs=512
- Bdd if=/dev/zero of=/dev/sda bs=512 count=1
- Cdd if=/dev/zero of=/dev/sda bs=440 count=1
- Ddd if=/dev/zero of=/dev/sda bs=440
How the community answered
(37 responses)- A5% (2)
- B30% (11)
- C54% (20)
- D11% (4)
Why each option
The MBR bootloader code occupies only the first 440 bytes of a 512-byte MBR; bytes 446-511 hold the partition table, so writing exactly 440 bytes zeros only the bootloader.
Without a count limit, dd will write 512-byte zero blocks continuously until the entire disk is overwritten, destroying all partitions and data.
A block size of 512 with count=1 writes a full 512 bytes, which includes bytes 446-511 containing the partition table entries and boot signature, corrupting the partition layout.
The x86 MBR layout reserves bytes 0-439 for bootstrap code (the bootloader), bytes 440-445 for disk signature and reserved space, and bytes 446-511 for partition table entries and the boot signature. Specifying bs=440 count=1 writes exactly 440 zero bytes, overwriting only the bootloader code area while leaving the partition table entries at byte offset 446 and beyond completely untouched.
Without count=1, dd will write 440-byte zero blocks indefinitely across the entire disk, destroying all filesystem data and partition structures beyond the first block.
Concept tested: dd command targeting MBR bootloader byte range
Source: https://www.gnu.org/software/coreutils/manual/html_node/dd-invocation.html
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