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Home/ Questions/Q 9187745
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Editorial Team
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Editorial Team
Asked: June 17, 20262026-06-17T19:50:59+00:00 2026-06-17T19:50:59+00:00

I have the following from an objdump. This was C code compiled by gcc

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I have the following from an objdump. This was C code compiled by gcc for an IA32.

08048e9a <my_func>:
8048e9a:    55                      push   %ebp
8048e9b:    89 e5                   mov    %esp,%ebp
8048e9d:    83 ec 48                sub    $0x48,%esp
8048ea0:    89 5d f4                mov    %ebx,-0xc(%ebp)
8048ea3:    89 75 f8                mov    %esi,-0x8(%ebp)
8048ea6:    89 7d fc                mov    %edi,-0x4(%ebp)
8048ea9:    8d 5d d0                lea    -0x30(%ebp),%ebx
8048eac:    89 5c 24 04             mov    %ebx,0x4(%esp)
8048eb0:    8b 45 08                mov    0x8(%ebp),%eax
8048eb3:    89 04 24                mov    %eax,(%esp)
8048eb6:    e8 52 04 00 00          call   804930d <read_num>
8048ebb:    8d 7d dc                lea    -0x24(%ebp),%edi
8048ebe:    be 00 00 00 00          mov    $0x0,%esi
8048ec3:    8b 03                   mov    (%ebx),%eax
8048ec5:    3b 43 0c                cmp    0xc(%ebx),%eax
8048ec8:    74 05                   je     8048ecf <my_func+0x35>
8048eca:    e8 fc 03 00 00          call   80492cb <other_func>
8048ecf:    03 33                   add    (%ebx),%esi

I am interested in finding out the values being compared on line 8048ec5 In gdb I can step to this line and I can read %eax just fine from info registers but how can I read 0xc(%ebx)? This means 0xc offset from %ebx or 0xc + %ebx?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-17T19:51:00+00:00Added an answer on June 17, 2026 at 7:51 pm

    It refers to the 32-bit value at the address %ebx + 0xc in memory.

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