Let's write the shortest code to perform a simplified variant of the DoD 5220.22-M Wipe Method with only two writing passes.

Any programming language accepted, but the use of disk-wiping-oriented libraries is prohibited.

Here's how we are to implement it in pseudocode:

Set x to 0

'Write Pass
For each sector in disk write the content of x
'Verification Pass
For each sector in disk {
    If sector does not contain the content of x then goto [Start]
'Check whether we already did the pass with 1
If x is not 1 then {
    Set x to 1
    GoTo [Start]
Else end

In other words, this code will run twice, with a write pass and verification pass for 0, and a write pass and verification pass for 1.

Anyone ballsy enough to implement it code-golf style? ;)

  • 6
    \$\begingroup\$ I doubt this will get any answers because of how hard it will be to test. \$\endgroup\$ – James Apr 3 '16 at 15:16
  • 12
    \$\begingroup\$ Seriously, what did your SSD do to you to deserve this kind of treatment? Did it kill your entire teddy bear collection or something? \$\endgroup\$ – R. Kap Apr 4 '16 at 0:41
  • 2
    \$\begingroup\$ Really want to try and tackle this...also really don't want to sacrifice my HDD to test it. \$\endgroup\$ – Michelfrancis Bustillos Apr 15 '16 at 12:16
  • 6
    \$\begingroup\$ I'm voting to close this question as off-topic because this challenge asks for malicious code. \$\endgroup\$ – AdmBorkBork Aug 4 '16 at 13:45
  • 2
    \$\begingroup\$ I would argue otherwise. This challenge asks for code that promotes privacy and information security. \$\endgroup\$ – MathuSum Mut Aug 4 '16 at 13:48

x86 machine code (Linux), 116 bytes

00000000: 89cb 6a02 5931 d289 4c24 f4b0 05cd 8050  ..j.Y1..L$.....P
00000010: 5b53 6a02 5a31 c9b0 13cd 8089 c65b 5331  [Sj.Z1.......[S1
00000020: d231 c9b0 13cd 8089 f75b 53b2 018d 4c24  .1.......[S...L$
00000030: f8b0 04cd 804e 09f6 75ef 5b53 31d2 31c9  .....N..u.[S1.1.
00000040: b013 cd80 89fe 5b53 8d4c 24f4 b201 b003  ......[S.L$.....
00000050: cd80 4e09 f674 0c8a 4424 f838 4424 f474  ..N..t..D$.8D$.t
00000060: e5eb ad89 fe8a 4424 f8c6 4424 f801 3c01  ......D$..D$..<.
00000070: 759e 58c3                                u.X.

Takes filename as argument

Assembly (NASM):

section .text
	global func
func:			;this function uses fastcall conventions
	;seems like no enter instr needed

	;open file
	mov ebx, ecx	;first argument to func (filename)
	push 0x2	;flag O_RDWR
	pop ecx		;pushing a constant is shorter than mov
	xor edx, edx	;mode=0
	mov [esp-12], ecx ;set first byte (msg) to write to 0. msg later in esp-8, buf in esp-12
	mov al, 5	;using 8 bit regs is smaller
	int 0x80	;syscall open
	push eax	;save file descriptor


	;get file size
	pop ebx		;get fd
	push ebx
	push 0x2	;flag SEEK_END
	pop edx
	xor ecx, ecx 	;offset 0
	mov al, 0x13
	int 0x80	;syscall lseek
	mov esi, eax	;save file byte count in esi

	;reset index of file to 0
	pop ebx		;get fd
	push ebx
	xor edx, edx	;flag SEEK_SET=0
	xor ecx, ecx	;ecx=0
	mov al, 0x13
	int 0x80	;syscall lseek

	;***write pass***
	mov edi, esi
	write_loop:	;for each byte in byte count, write [msg]
		;write to file
		pop ebx		;file descriptor
		push ebx
		mov dl, 1	;bytes to write
		lea ecx, [esp-8] ;buffer to write from
		mov al, 4
		int 0x80	;syscall write
		dec esi	;decrement esi (byte count) to 0
		or esi, esi	;cmp esi to 0
		jne write_loop	;while esi!=0, keep looping

	;reset index of file to 0
	pop ebx		;get fd
	push ebx
	xor edx, edx	;flag SEEK_SET=0
	xor ecx, ecx	;ecx=0
	mov al, 0x13
	int 0x80	;syscall lseek

	;***verification pass***
	mov esi, edi
	verify_loop:	;for each byte in byte count, verify written byte
		pop ebx		;get fd
		push ebx
		lea ecx, [esp-12] ;buffer to store read byte
		mov dl, 1	;read 1 byte
		mov al, 3
		int 0x80	;syscall read
		dec esi
		or esi, esi	;cmp esi to 0 
		je end_verify	;at final byte, end verification
		mov al, [esp-8]
		cmp byte [esp-12],al
		je verify_loop	 ;keep looping if expected value found
		jmp write_verify ;if byte!=expected value, restart

	mov esi, edi
	mov al, [esp-8]
	mov byte [esp-8],0x1	;set new byte to write to 1
	cmp al, 0x1
	jne write_verify	;if old byte to write!=1, goto start
	pop eax			;reset stack

Try it online! (Uses a temporary file)

-11 bytes by optimizing moving registers and using the stack as a buffer instead of a constant location in memory.

| improve this answer | |

On a linux system there's no need for special handling of devices. Just use the device file interface.

Python 3 (byte strings) - 141 bytes

while i<2:
 if f.read()==x:i+=1

It's fairly straightforward, and not really optimised heavily, but it works. Here's a basic rundown.

  • Get input (device file path)
  • Open device file
  • Seek to the end, get filesize (block devices are always their real size)
  • enter write-and-check loop
  • construct 0-bit and 1-bit strings (x)
  • write bitstring
  • flush output (I could have set buffering=0 but this is shorter)
  • test file against x, and increment step of loop if it passes

exit loop when the increment is high enough

As a bonus, you could modify this for any set and number of byte-modification patterns, like 0x55/0xaa for stronger overwriting effects.

I did actually test this on a device file, using loopback. However, I'm not 100% sure the checking actually works. It might be necessary to close and reopen the file each pass, due to buffering behaviors. I would hope flush prevents this.

*edited to incorporate some suggestions in the comments

| improve this answer | |
  • 1
    \$\begingroup\$ Welcome to the site. You certainly don't need whitespace around an = in python. You can also reduce your byte count by using ; to reduce indentation. \$\endgroup\$ – Wheat Wizard Oct 28 '18 at 17:26
  • \$\begingroup\$ Normally, we count submissions in bytes, not characters. You can also alias some of your functions, e.g. f.seek(0);f.seek(0) (19 bytes) can be s=f.seek;s(0);s(0) (18 bytes). Furthermore, if f.read()==x:i+=1 can be i+=f.read()==x. \$\endgroup\$ – Jonathan Frech Oct 28 '18 at 17:53
  • \$\begingroup\$ You also shouldn’t need the empty string as an argument to input. \$\endgroup\$ – Quintec Oct 28 '18 at 18:18
  • \$\begingroup\$ I think b'\0' instead of b'\x00' should work. \$\endgroup\$ – Jason Oct 29 '18 at 8:29
  • \$\begingroup\$ Just realized an important feature. This program will consume RAM equal to the size of the device wiped. \$\endgroup\$ – William Shipley Oct 30 '18 at 0:26

C (clang), -DZ=lseek(d,0 + 139 = 152 bytes


Try it online!

Takes the filename as an argument


#include <unistd.h>
int g,j,x,d,s,i,c;
void f(char*n){
	x=0; /*Byte to write*/
	s=lseek(d,0,SEEK_END); /*Get size of file*/
		/*Write Pass*/
		lseek(d,0,SEEK_SET); /*Start writing from start of file*/
		/*Verification Pass*/
			if(c!=x)x--,j--; /*If verification fails, repeat loop with the same value*/
| improve this answer | |
  • \$\begingroup\$ Hmm ... will the TiO wipe the internet? ;-D \$\endgroup\$ – Titus Oct 30 '18 at 0:21
  • \$\begingroup\$ @Titus It creates a temporary file and wipes it. \$\endgroup\$ – Logern Oct 30 '18 at 0:21

Tcl, 286 bytes

proc f {o i l} {seek $o 0
set s [string repeat $i $l]
puts -nonewline $o $s
flush $o
seek $o 0
return [string equal [read $o $l] $s]}
set n [open $argv r+]
fconfigure $n -translation binary
seek $n 0 end
set m [tell $n]
while {![f $n "\0" $m]} {}
while {![f $n "\xff" $m]} {}

Not really optimized that well. I tried what I could, but I don't know that much about Tcl.

Save as "f.tcl" and run on Unix with tclsh f.tcl "your filename". Make sure there's exactly one argument! I tested this on a plain file, but it should work on a device file as well.

Setting variables and indexing is more involved in Tcl, so I decided to put the common code between the passes into a function. Then I call it first with "\0", and repeat while it fails to verify. I do the same thing with "\xff".

I flushed after writes; it might not be necessary. fconfigure -translation binary -buffering none is longer.

-2 bytes by removing quotes around r+.

| improve this answer | |

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