Self-Printing Machine Code
The following 12-byte program composed of pure x86 machine code writes itself to standard output when executed in a DOS environment:
fc b1 0c ac 92 b4 02 cd 21 e2 f8 c3
We can write these bytes to a file with the .COM extension and execute it in DOS. It runs successfully in MS-DOS 6.22, Windows 98, as well as in DOSBox and writes a copy of itself to standard output.
- Quine Conundrums
- Proper Quines
- A Note on DOS Services
- Writing to Video Memory Directly
- Boot Program
On a Unix or Linux system, the following commands demonstrate this program with the help of DOSBox:
echo fc b1 0c ac 92 b4 02 cd 21 e2 f8 c3 | xxd -r -p > foo.com dosbox -c 'MOUNT C .' -c 'C:\FOO > C:\OUT.COM' -c 'EXIT' diff foo.com OUT.COM
diff command should produce no output confirming
that the output of the program is identical to the program itself.
On an actual MS-DOS 6.22 system or a Windows 98 system, we can
demonstrate this program in the following manner:
C:\>DEBUG -E 100 fc b1 0c ac 92 b4 02 cd 21 e2 f8 c3 -N FOO.COM -R CX CX 0000 :C -W Writing 0000C bytes -Q C:\>FOO > OUT.COM C:\>FC FOO.COM OUT.COM Comparing files FOO.COM and OUT.COM FC: no differences encountered
DEBUG session shown above, we use the debugger
E to enter the machine code at offset 0x100 of
the code segment. Then we use the
N command to name the
file we want to write this machine code to. The command
CX is used to specify that we want to write 0xC (decimal 12)
bytes to this file. The
W command writes the 12 bytes
entered at offset 0x100. The
Q command quits the
debugger. Then we run the new
FOO.COM program while
redirecting its output to
OUT.COM. Finally, we use
FC command to compare the two files and confirm
that they are exactly the same.
Let us disasssemble this program now and see what it does. The output below is generated using the Netwide Disassembler (NDISASM), a tool that comes with Netwide Assembler (NASM):
$ ndisasm -o 0x100 foo.com 00000100 FC cld 00000101 B10C mov cl,0xc 00000103 AC lodsb 00000104 92 xchg ax,dx 00000105 B402 mov ah,0x2 00000107 CD21 int 0x21 00000109 E2F8 loop 0x103 0000010B C3 ret
When DOS executes a program in .COM file, it loads the machine code
in the file at offset 0x100 of the code segment chosen by DOS. That
is why we ask the disassembler to assume a load address of 0x100
-o command line option. The first instruction
clears the direction flag. The purpose of this instruction is
explained later. The next instruction sets the register CL to 0xc
(decimal 12). The register CH is already set to 0 by default when a
.COM program starts. Thus setting the register CL to 0 effectively
sets the entire register CX to 0xc. The register CX is used as a
loop counter for the
loop 0x103 instruction that comes
later. Everytime this loop instruction executes, it decrements CX
and makes a near jump to offset 0x103 if CX is not 0. This results
in 12 iterations of the loop.
In each iteration of the loop, the instructions from offset 0x103 to
offset 0x109 are executed. The
lodsb instruction loads
a byte from address DS:SI into AL. When DOS starts executing this
program, DS and SI are set to CS and 0x100 by default, so at the
beginning DS:SI points to the first byte of the program.
xchg instruction exchanges the values in AX and DX.
Thus the byte we just loaded into AL ends up in DL. Then we set AH
to 2 and generate the software interrupt 0x21 (decimal 33) to write
the byte in DL to standard output. This is how each iteration reads
a byte of this program and writes it to standard output.
lodsb instruction increments or decrements SI
depending on the state of the direction flag (DF). When DF is
cleared, it increments SI. If DF is set, it decrements SI. We use
cld instruction at the beginning to clear DF, so
that in each iteration of the loop, SI moves forward to point to the
next byte of the program. This is how the 12 iterations of the loop
write 12 bytes of the program to standard output. In many DOS
environments, the DF flag is already in cleared state when a .COM
program starts, so the CLD instruction could be omitted in such
environments. However, there are some environments where DF may not
be in cleared state when our program starts, so it is a best
practice to clear DF before relying on it.
Finally, when the loop terminates, we execute the
instruction to terminate the program.
While reading the description of the self-printing program presented earlier, one might wonder if it is a quine. While there is no standardized definition of the term quine, it is generally accepted that a quine is a computer program that takes no input and produces an exact copy of its own source code as its output. Since a quine cannot take any input, tricks involving reading its own source code or evaluating itself are ruled out.
For example, this shell script is a valid quine:
s='s=\47%s\47;printf "$s" "$s"\n';printf "$s" "$s"
However, the following shell script is not considered a proper quine:
The shell script above reads its own source code which is considered cheating. Improper quines like this are often called cheating quines.
Is our 12-byte x86 program a quine? It turns out that we have a conundrum. There is no notion of source code for our program. There would have been one if we had written out the source code of this program in assembly language. In such a case we would first need to choose an assembler and a proper quine would need to produce an exact copy of the assembly language source code (not the machine code bytes) for the chosen assembler. But we are not doing that here. We want the machine code to produce an exact copy of itself. There is no source code involved. We only have machine code. So we could argue that the whole notion of machine code quine is nonsense. No machine code quine can exist because there is no source code to produce as output.
However, we could also argue that the machine code is the input for the CPU that the CPU fetches, decodes, and converts to a sequence of state changes in the CPU. If we define a machine code quine to be a machine code program that writes its own bytes, then we could say that we have a machine code quine here.
Let us now entertain the thought that our 12-byte program is indeed a machine code quine. Now we have a new conundrum. Is it a proper quine? This program reads its own bytes from memory and writes them. Does that make it a cheating quine? What would a proper quine written in pure machine code even look like? If we look at the shell script quine above, we see that it contains parts of the executable part of the script code embedded in a string as data. Then we format the string cleverly to produce a new string that looks exactly like the entire shell script. It is a common pattern followed in many quines. The quine does not read its own code but it reads some data defined by the code and formats that data to look like its own code. However, in a pure machine code like this the lines between data and code are blurred. Even if we try to keep the bytes we want to read at a separate place in the memory and treat it like data, they would look exactly like machine instructions, so one might wonder if there is any point in trying to make a machine quine that does not read its own bytes. Nevertheless the next section shows how to accomplish this.
If the thought of a machine code quine program reading its own bytes from the memory makes you uncomfortable, here is an adapation of the previous program that keeps the machine instructions to be executed separate from the data bytes to be read by the program.
fc b3 02 b1 14 be 14 01 ac 92 b4 02 cd 21 e2 f8 4b 75 f0 c3 fc b3 02 b1 14 be 14 01 ac 92 b4 02 cd 21 e2 f8 4b 75 f0 c3
Here is how we can demonstrate this 40-byte program:
echo fc b3 02 b1 14 be 14 01 ac 92 b4 02 cd 21 e2 f8 4b 75 f0 c3 | xxd -r -p > foo.com echo fc b3 02 b1 14 be 14 01 ac 92 b4 02 cd 21 e2 f8 4b 75 f0 c3 | xxd -r -p >> foo.com dosbox -c 'MOUNT C .' -c 'C:\FOO > C:\OUT.COM' -c 'EXIT' diff foo.com OUT.COM
Here is the disassembly:
$ ndisasm -o 0x100 foo.com 00000100 FC cld 00000101 B302 mov bl,0x2 00000103 B114 mov cl,0x14 00000105 BE1401 mov si,0x114 00000108 AC lodsb 00000109 92 xchg ax,dx 0000010A B402 mov ah,0x2 0000010C CD21 int 0x21 0000010E E2F8 loop 0x108 00000110 4B dec bx 00000111 75F0 jnz 0x103 00000113 C3 ret 00000114 FC cld 00000115 B302 mov bl,0x2 00000117 B114 mov cl,0x14 00000119 BE1401 mov si,0x114 0000011C AC lodsb 0000011D 92 xchg ax,dx 0000011E B402 mov ah,0x2 00000120 CD21 int 0x21 00000122 E2F8 loop 0x11c 00000124 4B dec bx 00000125 75F0 jnz 0x117 00000127 C3 ret
The first 20 bytes is the executable part of the program. The next 20 bytes is the data read by the program. The executable bytes are identical to the data bytes. The executable part of the program has an outer loop that iterates twice. In each iteration, it reads the data bytes and writes them to standard output. Therefore, in two iterations of the outer loop, it writes the data bytes twice. In this manner, the output is identical to the program itself.
Here is another simpler 32-byte quine based on this approach:
b8 23 09 fe c0 a2 20 01 ba 10 01 cd 21 cd 21 c3 b8 23 09 fe c0 a2 20 01 ba 10 01 cd 21 cd 21 c3
Here are the commands to demostrate this quine:
echo b8 23 09 fe c0 a2 20 01 ba 10 01 cd 21 cd 21 c3 | xxd -r -p > foo.com echo b8 23 09 fe c0 a2 20 01 ba 10 01 cd 21 cd 21 c3 | xxd -r -p >> foo.com dosbox -c 'MOUNT C .' -c 'C:\FOO > C:\OUT.COM' -c 'EXIT' diff foo.com OUT.COM
Here is the disassembly:
$ ndisasm -o 0x100 foo.com 00000100 B82309 mov ax,0x923 00000103 FEC0 inc al 00000105 A22001 mov [0x120],al 00000108 BA1001 mov dx,0x110 0000010B CD21 int 0x21 0000010D CD21 int 0x21 0000010F C3 ret 00000110 B82309 mov ax,0x923 00000113 FEC0 inc al 00000115 A22001 mov [0x120],al 00000118 BA1001 mov dx,0x110 0000011B CD21 int 0x21 0000011D CD21 int 0x21 0000011F C3 ret
This example too has two parts. The first half has the executable
bytes and the second half has the data bytes. Both parts are
identical. This example sets AH to 9 in the first instruction and
then later uses
int 0x21 to invoke the DOS service that
prints a dollar-terminated string beginning at the address specifed
in DS:DX. When a .COM program starts, DS already points to the
current code segment, so we don't have to set it explicitly. The
dollar symbol has an ASCII code of 0x24 (decimal 36). We need to be
careful about not having this value anywhere within the the data
bytes or this DOS function would prematurely stop printing our data
bytes as soon as it encounters this value. That is why we set AL to
0x23 in the first instruction, then increment it to 0x24 in the
second instruction, and then copy this value to the end of the data
bytes in the third instruction. Finally, we execute
0x21 twice to write the data bytes twice to standard output,
so that the output matches the program itself.
While both these programs take care not to read the same memory region that is being executed by the CPU, the data bytes they read look exactly like the executable bytes. This is what I meant when I mentioned earlier that the lines between code and data are blurred in an exercise like this. This is why I don't really see a point in keeping the executable bytes separate from the data bytes while writing machine code quines.
A Note on DOS Services
The self-printing programs presented above use
which offers DOS services that support various input/output
functions. In the first two programs, we selected the function to
write a character to standard output by setting AH to 2 before
invoking this software interrupt. In the next program, we selected
the function to write a dollar-terminated string to standard output
by setting AH to 9.
ret instruction in the end too relies on DOS
services. When a .COM program starts, the register SP contains
0xfffe. The stack memory locations at offset 0xfffe and 0xffff
contain 0x00 and 0x00, respectively. Further, the memory address at
offset 0x0000 contains the instruction
int 0x20 which
is a DOS service that terminates the program. As a result, executing
ret instruction pops 0x0000 off the stack at 0xfffe
and loads it into IP. This results in the instruction
0x20 at offset 0x0000 getting executed. This instruction
terminates the program and returns to DOS.
Relying on DOS services gives us a comfortable environment to work
with. In particular, DOS implements the notion of standard
output which lets us redirect standard output to a file. This
lets us conveniently compare the original program file and the
output file with the
FC command and confirm that they
But one might wonder if we could avoid relying on DOS services completely and still write a program that prints its own bytes to screen. We definitely can. We could write directly to video memory at address 0xb800:0x0000 and show the bytes of the program on screen. We could also forgo DOS completely and let BIOS load our program from the boot sector and execute it. The next two sections discuss these things.
Writing to Video Memory Directly
Here is an example of an 18-byte self-printing program that writes directly to the video memory at address 0xb800:0x0000.
fc b4 b8 8e c0 31 ff b1 12 b4 0a ac ab e2 fc f4 eb fd
Here are the commands to create and run this program:
echo fc b4 b8 8e c0 31 ff b1 12 b4 0a ac ab e2 fc f4 eb fd | xxd -r -p > foo.com dosbox foo.com
With the default code page active, i.e., with code page 437 active, the program should display an output that looks approximately like the following and halt:
Now of course this type of output looks gibberish but there is a
quick and dirty way to confirm that this output indeed represents
the bytes of our program. We can use the
of DOS to print the program and check if the symbols that appear in
its output seem consistent with the output above. Here is an
C:\>TYPE FOO.COM ⁿ┤╕Ä└1 ▒↕┤ ¼½Γⁿ⌠δ² C:\>
This output looks very similar to the previous one except that the byte value 0x0a is rendered as a line break in this output whereas in the previous output this byte value is represented as a circle in a box. This method would not have worked if there were any control characters such as backspace or carriage return that result in characters being erased in the displayed output.
A proper way to verify that the output of the program represents the bytes of the program would be to find each symbol in the output in a chart for code page 437 and confirm that the byte value of each symbol matches each byte value in the program. Here is one such chart that approximates the symbols in code page 437 with Unicode symbols: cp437.html.
Here is the disassembly of the above program:
$ ndisasm -o 0x100 foo.com 00000100 FC cld 00000101 B4B8 mov ah,0xb8 00000103 8EC0 mov es,ax 00000105 31FF xor di,di 00000107 B112 mov cl,0x12 00000109 B40A mov ah,0xa 0000010B AC lodsb 0000010C AB stosw 0000010D E2FC loop 0x10b 0000010F F4 hlt 00000110 EBFD jmp short 0x10f
This program sets ES to 0xb800 and DI to 0. Thus ES:DI points to the
video memory at address 0xb800:0x0000. DS:SI points to the first
instruction of this program by default. Further AH is set to 0xa.
This is used to specify the colour attribute of the text to be
displayed on screen. Each iteration of the loop in this program
loads a byte of the program and writes it along with the colour
attribute to video memory. The
lodsb instruction loads
a byte of the program from the memory address specified by DS:SI
into AL and increments SI by 1. AH is already set to 0xa. The value
0xa (binary 00001010) here specifies black as the background colour
and bright green as the foreground colour. The
instruction stores a word from AX to the memory address specified by
ES:DI and increments DI by 2. In this manner, the byte in AL and its
colour attribute in AH gets copied to the video memory.
Once again, if you are not happy about the program reading its own executable bytes, we can keep the bytes we read separate from the bytes the CPU executes. Here is a 54-byte program that does this:
fc b3 02 b4 b8 8e c0 31 ff be 1b 01 b9 1b 00 b4 0a ac ab e2 fc 4b 75 f1 f4 eb fd fc b3 02 b4 b8 8e c0 31 ff be 1b 01 b9 1b 00 b4 0a ac ab e2 fc 4b 75 f1 f4 eb fd
Here is how we can create and run this program:
echo fc b3 02 b4 b8 8e c0 31 ff be 1b 01 b9 1b 00 b4 | xxd -r -p > foo.com echo 0a ac ab e2 fc 4b 75 f1 f4 eb fd fc b3 02 b4 b8 | xxd -r -p >> foo.com echo 8e c0 31 ff be 1b 01 b9 1b 00 b4 0a ac ab e2 fc | xxd -r -p >> foo.com echo 4b 75 f1 f4 eb fd | xxd -r -p >> foo.com dosbox foo.com
With code page 437 active, the output should look approximately like this:
ⁿ│☻┤╕Ä└1 ╛←☺╣← ┤◙¼½ΓⁿKu±⌠δ²ⁿ│☻┤╕Ä└1 ╛←☺╣← ┤◙¼½ΓⁿKu±⌠δ²
We can clearly see in this output that the first 27 bytes of output are identical to the next 27 bytes of the output. Like the proper quines discussed earlier, this one too has two halves that are identical to each other. The executable code in the first half reads the data bytes from the second half and prints the data bytes twice so that the output bytes is an exact copy of all 54 bytes in the program. Here is the disassembly:
$ ndisasm -o 0x100 foo.com 00000100 FC cld 00000101 B302 mov bl,0x2 00000103 B4B8 mov ah,0xb8 00000105 8EC0 mov es,ax 00000107 31FF xor di,di 00000109 BE1B01 mov si,0x11b 0000010C B91B00 mov cx,0x1b 0000010F B40A mov ah,0xa 00000111 AC lodsb 00000112 AB stosw 00000113 E2FC loop 0x111 00000115 4B dec bx 00000116 75F1 jnz 0x109 00000118 F4 hlt 00000119 EBFD jmp short 0x118 0000011B FC cld 0000011C B302 mov bl,0x2 0000011E B4B8 mov ah,0xb8 00000120 8EC0 mov es,ax 00000122 31FF xor di,di 00000124 BE1B01 mov si,0x11b 00000127 B91B00 mov cx,0x1b 0000012A B40A mov ah,0xa 0000012C AC lodsb 0000012D AB stosw 0000012E E2FC loop 0x12c 00000130 4B dec bx 00000131 75F1 jnz 0x124 00000133 F4 hlt 00000134 EBFD jmp short 0x133
This disassembly is rather long but we can clearly see that the bytes from offset 0x100 to offset 0x11a are identical to the bytes from offset 0x11b to 0x135. These are the bytes we see in the output of the program too.
The 32-byte program below writes itself to video memory when executed from the boot sector:
ea 05 7c 00 00 fc b8 00 b8 8e c0 8c c8 8e d8 31 ff be 00 7c b9 20 00 b4 0a ac ab e2 fc f4 eb fd
We can create a boot image that contains these bytes, write it to the boot sector of a drive and boot an IBM PC compatible computer with it. On booting, this program prints its own bytes on the screen.
On a Unix or Linux system, the following commands can be used to create a boot image with the above program:
echo ea 05 7c 00 00 fc b8 00 b8 8e c0 8c c8 8e d8 31 | xxd -r -p > boot.img echo ff be 00 7c b9 20 00 b4 0a ac ab e2 fc f4 eb fd | xxd -r -p >> boot.img echo 55 aa | xxd -r -p | dd seek=510 bs=1 of=boot.img
Now we can test this boot image using DOSBox with the following command:
dosbox -c cls -c 'boot boot.img'
We can also test this image using QEMU x86 system emulator as follows:
qemu-system-i386 -fda boot.img
We could also write this image to the boot sector of an actual
physical storage device, such as a USB flash drive, and then boot
the computer with it. Here is an example command that writes the
boot image to the drive represented by the device
cp a.img /dev/sdx
CAUTION: You need to be absolutely sure of the device path of the
device being written to. The device path
only an example here. If the boot image is written to the wrong
device, access to the data on that would be lost.
On testing this boot image with an emulator or a real computer, the output should look approximately like this:
Ω♣| ⁿ╕ ╕Ä└î╚Ä╪1 ╛ |╣ ┤◙¼½Γⁿ⌠δ²
This looks like gibberish, however every symbol in the above output corresponds to a byte of the program mentioned earlier. For example, the first symbol (omega) represents the byte value 0xea, the second symbol (club) represents the byte value 0x05, and so on. The chart at cp437.html can be used to confirm that every symbol in the output indeed represents every byte of the program.
Here is the disassembly of the program:
$ ndisasm -o 0x7c00 boot.img 00007C00 EA057C0000 jmp 0x0:0x7c05 00007C05 FC cld 00007C06 B800B8 mov ax,0xb800 00007C09 8EC0 mov es,ax 00007C0B 8CC8 mov ax,cs 00007C0D 8ED8 mov ds,ax 00007C0F 31FF xor di,di 00007C11 BE007C mov si,0x7c00 00007C14 B92000 mov cx,0x20 00007C17 B40A mov ah,0xa 00007C19 AC lodsb 00007C1A AB stosw 00007C1B E2FC loop 0x7c19 00007C1D F4 hlt 00007C1E EBFD jmp short 0x7c1d 00007C20 0000 add [bx+si],al 00007C22 0000 add [bx+si],al ...
The ellipsis in the end represents the remainder of the bytes that contains zeroes and the boot sector magic bytes 0x55 and 0xaa in the end. They have been omitted here for the sake of brevity.
When a computer boots, the BIOS reads the boot sector code from the first sector of the boot device into the memory at physical address 0x7c00 and jumps to this address. Most BIOS implementations jump to 0x0000:0x7c00 but there are some implementations that jump to 0x07c0:0x0000 instead. Both these jumps are jumps to the same physical address 0x7c00 but this difference poses a problem for us because the offsets in our program depend on which jump the BIOS executed. In order to ensure that our program can run with both types of BIOS implementations, we use a popular trick of having the first instruction of our program execute a jump to address 0x0000:0x7c05 in order to reach the second instruction. This sets the register CS to 0 and IP to 0x7c05 and we don't have to worry about the differences between BIOS implementations anymore. We can now pretend as if a BIOS implementation that jumps to 0x0000:0x7c00 is going to load our program.
The remainder of the program is similar to the one in the previous
section. However, there are some small but important differences.
While the DOS environment guarantees that AH and CH are initialized
to 0 when a .COM program starts, the BIOS offers no such guarantee
while loading and executing a boot program. This is why we use the
registers AX and CX (as opposed to only AH and CL) in
mov instructions to initialize them. Similarly,
while DOS initializes SI to 0x100 when a .COM program starts, for a
boot program, we set the register SI ourselves.
If you feel uncomfortable about calling the above program a quine because it reads its own bytes from the memory, we could have the program read the bytes it needs to print from a separate place in memory. We do not execute these bytes. We only read them and copy them to video memory. The following 76-byte program does this:
ea 05 7c 00 00 fc bb 02 00 b8 00 b8 8e c0 8c c8 8e d8 31 ff be 26 7c b9 26 00 b4 0a ac ab e2 fc 4b 75 f1 f4 eb fd ea 05 7c 00 00 fc bb 02 00 b8 00 b8 8e c0 8c c8 8e d8 31 ff be 26 7c b9 26 00 b4 0a ac ab e2 fc 4b 75 f1 f4 eb fd
Here is how we can create a boot image with this:
echo ea 05 7c 00 00 fc bb 02 00 b8 00 b8 8e c0 8c c8 | xxd -r -p > boot.img echo 8e d8 31 ff be 26 7c b9 26 00 b4 0a ac ab e2 fc | xxd -r -p >> boot.img echo 4b 75 f1 f4 eb fd ea 05 7c 00 00 fc bb 02 00 b8 | xxd -r -p >> boot.img echo 00 b8 8e c0 8c c8 8e d8 31 ff be 26 7c b9 26 00 | xxd -r -p >> boot.img echo b4 0a ac ab e2 fc 4b 75 f1 f4 eb fd | xxd -r -p >> boot.img echo 55 aa | xxd -r -p | dd seek=510 bs=1 of=boot.img
Here are the commands to test this boot image:
dosbox -c cls -c 'boot boot.img' qemu-system-i386 -fda boot.img
The output should look like this:
Ω♣| ⁿ╗☻ ╕ ╕Ä└î╚Ä╪1 ╛&|╣& ┤◙¼½ΓⁿKu±⌠δ²Ω♣| ⁿ╗☻ ╕ ╕Ä└î╚Ä╪1 ╛&|╣& ┤◙¼½ΓⁿKu±⌠δ²
Here is the disassembly of this program:
$ ndisasm -o 0x7c00 boot.img 00007C00 EA057C0000 jmp 0x0:0x7c05 00007C05 FC cld 00007C06 BB0200 mov bx,0x2 00007C09 B800B8 mov ax,0xb800 00007C0C 8EC0 mov es,ax 00007C0E 8CC8 mov ax,cs 00007C10 8ED8 mov ds,ax 00007C12 31FF xor di,di 00007C14 BE267C mov si,0x7c26 00007C17 B92600 mov cx,0x26 00007C1A B40A mov ah,0xa 00007C1C AC lodsb 00007C1D AB stosw 00007C1E E2FC loop 0x7c1c 00007C20 4B dec bx 00007C21 75F1 jnz 0x7c14 00007C23 F4 hlt 00007C24 EBFD jmp short 0x7c23 00007C26 EA057C0000 jmp 0x0:0x7c05 00007C2B FC cld 00007C2C BB0200 mov bx,0x2 00007C2F B800B8 mov ax,0xb800 00007C32 8EC0 mov es,ax 00007C34 8CC8 mov ax,cs 00007C36 8ED8 mov ds,ax 00007C38 31FF xor di,di 00007C3A BE267C mov si,0x7c26 00007C3D B92600 mov cx,0x26 00007C40 B40A mov ah,0xa 00007C42 AC lodsb 00007C43 AB stosw 00007C44 E2FC loop 0x7c42 00007C46 4B dec bx 00007C47 75F1 jnz 0x7c3a 00007C49 F4 hlt 00007C4A EBFD jmp short 0x7c49 00007C4C 0000 add [bx+si],al 00007C4E 0000 add [bx+si],al ...
This program has two identical halves. The first half from offset 0x7c00 to offset 0x7c25 are executable bytes. The second half from offset 0x7c26 to 0x7c4b are the data bytes read by the executable bytes. The executable part of the code has an outer loop that uses the register BX as the counter variable. It sets BX to 2 so that the outer loop iterates twice. In each iteration, it reads data bytes from the second half of the program and prints them. The code to read bytes and print them is very similar to our earlier program. Since the data bytes in the second half are identical to the executable bytes in the first half, printing the data bytes twice amounts to printing all bytes of the program.
While this program does avoid reading the bytes that the CPU executes, the data bytes look exactly like the executable bytes. Although I do not see any point in trying to avoid reading executable bytes in an exercise like, this program serves as an example of a self-printing boot program that does not execute the bytes it reads.