Is the displacement signed or unsigned? So is the displacement 0xFFF8 in memory +FFF8 or -8? E.g. SS:BP+FFF8, does this wrap due to overflow or just substract? Which one is the correct one in disassembly? So does SS:[BP-4] exist as assembly, or is this simply SS:[BP+FFFC]?
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That is the same? You add the offset, and it will just wrap around 16bit causing it sometimes to look like a subtraction of the complemented value.
Base : 0x20
Offset: 0xFFF8
Base + Offset = 0x18
Base - 8 = 0x18
So yes in assembly you can have SS:[BP-4] but the assembler will translate that to SS:[BP+FFFC] (which is the 2's complement).
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it wraps.
fff8 = -8
same thing.
use debug, spend 2 seconds writing test case. look at the code generated. its not hard.
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What about 8-bit displacements? Are they ALWAYS signed? So a 8-bit displacmeents 0x80 substracts 128 from [(E)SI] etc.? Or is it always added(So e.g. SI(0x8000)+0x80 results in offset 0x8080)?
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The question you are asking is if the 8 bit displacement is sign extended. I assume so in the same manner the 8 bit jump offsets are sign extended for Jxx instructions.
Should be easy to find out with debug.exe.
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Maybe reading actual documentation helps?
http://css.csail.mit.edu/6.858/2014/readings/i386.pdf
Yes, 8 bit displacements in effective addresses are sign-extended (not zero-extended) to 16 bits. This is useful for compiled code because usually positive offsets from BP are arguments to the current function, and negative offsets from BP are local variables. With the displacement being signed, you can have ~128 bytes of each using single-byte displacements (which is nearly always enough). If the displacements were unsigned, then accesses to most local variables not in registers would be a byte longer.
