(XCHG AX, BX) == (XCHG BX, AX)?

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StrenoJr

Joined: 13 Mar 2014
Posts: 22
Location: Slovakia

StrenoJr 18 Aug 2014, 19:55

Hello.
I will start directly with some code.

Code:

xchg ax, bx
xchg bx, ax

Debugger (GRDB) said that both of these instructions are "0x93".
Does fasm compile these instructions as 0x93 or is it physically the same instruction?
Thanks for reading.

18 Aug 2014, 19:55

CandyMan

Joined: 04 Sep 2009
Posts: 413
Location: film "CandyMan" directed through Bernard Rose OR Candy Shop

CandyMan 18 Aug 2014, 20:00

Yes it is physically the same instruction.

Code:

Code    Mnemonic        Description
90 + rw         XCHG AX, r16    Exchange r16 with AX
90 + rw         XCHG r16, AX    Exchange AX with r16
90 + rd         XCHG EAX, r32   Exchange r32 with EAX
90 + rd         XCHG r32, EAX   Exchange EAX with r32
86 / r  XCHG r/m8, r8   Exchange r8 (byte register) with byte from r/m8
86 / r  XCHG r8, r/m8   Exchange byte from r/m8 with r8 (byte register)
87 / r  XCHG r/m16, r16         Exchange r16 with word from r/m16
87 / r  XCHG r16, r/m16         Exchange word from r/m16 with r16
87 / r  XCHG r/m32, r32         Exchange r32 with doubleword from r/m32
87 / r  XCHG r32, r/m32         Exchange doubleword from r/m32 with r32

_________________
smaller is better

18 Aug 2014, 20:00

StrenoJr

Joined: 13 Mar 2014
Posts: 22
Location: Slovakia

StrenoJr 18 Aug 2014, 20:05

Thanks for the fast reply.

18 Aug 2014, 20:05

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