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Editorial Team
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Asked: 2026-06-04T16:29:42+00:00

I disassembled the code resulting from compiling the very simple source file test.c, which

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I disassembled the code resulting from compiling the very simple source file test.c, which looked like this:

void main() {}

I ran these commands to link the main function into a static executable (editor’s note: with no CRT start code so it would just crash), then extract to a flat binary with a couple sections removed, so I could feed that to ndisasm (editor’s note: which doesn’t understand ELF metadata like objdump -drwC -Mintel does)

 gcc -c test.c 
 ld -o test -Ttext 0x0 -e main test.o 
 objcopy -R .note -R .comment -S -O binary test test.bin
 ndisasm -b 32 test.bin

And this is what I got:

00000000  55                push ebp
00000001  89E5              mov ebp,esp
00000003  5D                pop ebp
00000004  C3                ret
00000005  0000              add [eax],al
00000007  001400            add [eax+eax],dl
0000000A  0000              add [eax],al
0000000C  0000              add [eax],al
0000000E  0000              add [eax],al
00000010  017A52            add [edx+0x52],edi
00000013  0001              add [ecx],al
00000015  7C08              jl 0x1f
00000017  011B              add [ebx],ebx
00000019  0C04              or al,0x4
0000001B  0488              add al,0x88
0000001D  0100              add [eax],eax
0000001F  001C00            add [eax+eax],bl
00000022  0000              add [eax],al
00000024  1C00              sbb al,0x0
00000026  0000              add [eax],al
00000028  D8FF              fdivr st7
0000002A  FF                db 0xff
0000002B  FF05    00000000      inc dword [dword 0x0]
00000031  41                inc ecx
00000032  0E                push cs
00000033  088502420D05      or [ebp+0x50d4202],al
00000039  41                inc ecx
0000003A  0C04              or al,0x4
0000003C  04C5              add al,0xc5
0000003E  0000              add [eax],al

What is the purpose of everything past the first four lines? Why is it adding to the memory locations pointed to by eax, 2*eax, edx+0x52, comparing, and so on? Is it all about checking that the program executed correctly or something else?

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1 Answer

  1. Editorial Team

    I believe you have disassembled bits that are not code, which is why it doesn’t make a lot of sense.

    To get an idea of what the file might contain, I would recommend running objdump on the full binary (ELF), to see if you can recognize the above byte sequences in any of the sections.

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