I debug a program with IDA, it have a part of code that i don't understand

loc_8048E30:                            ; CODE XREF: phase_6+9Ej
.text:08048E30 mov     esi, [esi+8]
.text:08048E33 inc     ebx
.text:08048E34 cmp     ebx, eax
.text:08048E36 jl      short loc_8048E30   

line 1: mov esi, [esi+8] when I debug address of esi is 0x804B260 so esi+8 is 0x804B268.

The value in [esi+8] is 60h so after mov esi, [esi+8], the value in esi is 60h but it really is 0x804B260. Why it is 0x804B260?

And when esi is named .data:node2, it is linked link?

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1 Answer 1


The instruction mov esi, [esi + 8] copies 4 bytes (DWORD) of data at the location pointed to by esi + 8to register esi.

In your case esi is 0804B260 so it copies 4 bytes from 0804B268. Since x86_64 is little endian the least significant byte as per the screenshot 1 is 0x60. The remaining three bytes are located below (not in the picture).

It is named node2 as it is an exported symbol.

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