03/03/2014 · Show activity on this post. Technically x86 simply refers to a family of processors and the instruction set they all use. It doesn't actually say anything specific about data sizes. x86 started out as a 16-bit instruction set for 16-bit processors (the 8086 and 8088 processors), then was extended to a 32-bit instruction set for 32-bit processors ...

Show activity on this post. x86 is the name of the architecture that it's built to run on (the name comes from a series of old Intel processors, the names of which all ended in 86, The first of which was the 8086). Although x86 was originally a 16-bit architecture, the version in use today is the 32-bit extension.

31/05/2019 · I have come upon x86 assembly code I don't understand and can't find anything about it. The bit of code i don't understand is parentheses around register ecx: movb (%ecx),%al. What does this code mean? I am familiar with registers and …

02/11/2013 · it means compute the address like this way :DS * size_of_segment + OFFSET. normally, for x86 the size of segment is 16byte. For example: DS: 07C0H 0000 0111 1100 0000 + OFFSET: 0000H 0000 0000 0000 0000 = 07C00H 0000 0111 1100 0000 0000

24/04/2017 · With x86 masm the "$" will be decoded as an ASCII character, which is byte value 36. So it is equivalent to INSTR1 DB 30 DUP(36) , but the character $ in source gives hint to the reader of source code, that purpose of that line is to pre-fill buffer with DOS-string terminator value.

25/10/2021 · The $ means immediate. So it's a number. It's not a property of 0x400 itself. If you wrote mov 0x400, %eax (without the $) it would be a read from memory at address 0x400. –

01/11/2011 · ds:si and es:di mean the segment:offset referred to by the registers in question. This is primarily important when you're working in real mode (where offsets are a maximum of 64K apiece). In real mode, the segment are offset are combined as segment * 16 + offset. In protected mode, a segment register holds a "selector".

10/12/2015 · $0x0 is an immediate constant, zero.0 XOR y = y, just like adding zero to anything is a no-op. (xor is addition without the carry from bit to bit).. 0x6100 is an absolute memory address. Was there a segement override on that? Where was this instruction? It seems like an unlikely thing to do. The insn will still fault if that address isn't read/writeable, because xor immediate-0 isn't …

23/03/2013 · This code copies the address of the variable to be incremented off the stack into the ecx register, then it does lock incl (%ecx) to atomically increment that variable by 1. The next two instructions set the eax register (which holds the return value from the function) to 0 if the new value of the variable is 0, and 1 otherwise.

10/01/2017 · 1 Answer1. Show activity on this post. The point of this hint is to tell the size of the operand. You're writing to a point in memory. As you're only giving a 0, it doesn't know if it should write a byte, or word, or doubleword. The dword ptr means "write a doubleword". Yes, it stands for pointer, because you put a memory address as destination.