CODEWITHSUNDEEP
objdumpdisassembly
j1nma
j1nma

Reputation: 467

What does this dissasembly mean?

I am a beginner in assembler. I objdumped a file that validates a password entered through command line. One of its sections was the one below. I don't understand a thing of what the movs are copying into eax register. I am also using evans debugger. A general understanding of what is happening would help.

<fillpassword>:
804851d:    55                      push   ebp
804851e:    89 e5                   mov    ebp,esp
8048520:    8b 45 08                mov    eax,DWORD PTR [ebp+0x8]
8048523:    c6 00 53                mov    BYTE PTR [eax],0x53
8048526:    8b 45 08                mov    eax,DWORD PTR [ebp+0x8]
8048529:    83 c0 01                add    eax,0x1
804852c:    c6 00 30                mov    BYTE PTR [eax],0x30
804852f:    8b 45 08                mov    eax,DWORD PTR [ebp+0x8]
8048532:    83 c0 02                add    eax,0x2
8048535:    c6 00 52                mov    BYTE PTR [eax],0x52
8048538:    8b 45 08                mov    eax,DWORD PTR [ebp+0x8]
804853b:    83 c0 03                add    eax,0x3
804853e:    c6 00 50                mov    BYTE PTR [eax],0x50
8048541:    8b 45 08                mov    eax,DWORD PTR [ebp+0x8]
8048544:    83 c0 04                add    eax,0x4
8048547:    c6 00 52                mov    BYTE PTR [eax],0x52
804854a:    8b 45 08                mov    eax,DWORD PTR [ebp+0x8]
804854d:    83 c0 05                add    eax,0x5
8048550:    c6 00 33                mov    BYTE PTR [eax],0x33
8048553:    8b 45 08                mov    eax,DWORD PTR [ebp+0x8]
8048556:    83 c0 06                add    eax,0x6
8048559:    c6 00 53                mov    BYTE PTR [eax],0x53
804855c:    8b 45 08                mov    eax,DWORD PTR [ebp+0x8]
804855f:    83 c0 07                add    eax,0x7
8048562:    c6 00 34                mov    BYTE PTR [eax],0x34
8048565:    8b 45 08                mov    eax,DWORD PTR [ebp+0x8]
8048568:    83 c0 08                add    eax,0x8
804856b:    c6 00 00                mov    BYTE PTR [eax],0x0
804856e:    5d                      pop    ebp
804856f:    c3                      ret

Upvotes: 1

Views: 384

Answers (1)

William McBrine
William McBrine

Reputation: 2266

EBP+0x8 looks like a parameter to a function call -- the address of a string area to write to. The code repeatedly loads this address to AX and increments it, then writes a hard-wired character to the location pointed to by AX, so that the net result is to store the string "S0RPR3S4" (0-delimited) to the address passed in EBP+0x8.

In C, it would look something like this:

void fillpassword(char *p)
{
    p[0] = 'S';
    p[1] = '0';
    p[2] = 'R';
    p[3] = 'P';
    p[4] = 'R';
    p[5] = '3';
    p[6] = 'S';
    p[7] = '4';
    p[8] = '\0';
}

It's kind of an inefficient way to accomplish its task, and it looks to have been compiled without optimization (assuming it did start in C, or another higher-level language).

Upvotes: 1

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