问题
By looking at intel volume of instructions, I found this:
1) 88 /r MOV r/m8,r8
2) 8A /r MOV r8,r/m8
When I write a line like this in NASM, and assemble it with the listing option:
mov al, bl
I get this in the listing:
88D8 mov al, bl
So obviously NASM chosed the first instruction of the two above, but isn't the second instruction an option two? if so, on what basis did NASM chosed the first?
回答1:
These two encodings exist because a modr/m byte can only encode one memory operand. So to allow both mov r8,m8 and mov m8,r8, two encodings are needed. Of course, this way we can encode uses of mov with both operands being registers using either encoding and nasm just picks one at random. There is no special reason for the choice and I have seen assemblers make different choices.
I have also heard about an assembler that watermarked binaries it assembles by choosing an instruction encoding in a specific way. This way the author of the assembler could track down and sue people who used his assembler without paying.
回答2:
Interesting side effect of having different encodings: From the A86 manual.
A86 takes advantage of situations in which more than one set of opcodes can be generated for the same instruction. (For example, MOV AX,BX can be generated using either an 89 or 8B opcode, by reversing fields in the following effective address byte. Both forms are absolutely identical in functionality and execution speed.) A86 adopts an unusual mix of choices in such situations. This creates a code-generation "footprint" that occupies no space in your program file, but will enable me to tell, and to demonstrate in a court of law, if a non-trivial object file has been produced by A86. The specification for this "footprint" is sufficiently obscure and complicated that it would be impossible to duplicate by accident. I claim exclusive rights to the particular "footprint" I have chosen, and prohibit anyone from duplicating it. This has at least two specific implications:
a. Any assembler that duplicates the "footprint" is mine. If it is not identified as mine and issued under these terms, then those who sell or distribute the assembler will be subject to prosecution.
b. Any program marked with the "footprint" has been produced by my assembler. It is subject to condition 5 above.
回答3:
As Harold pointed out: there is no reason.
Maybe the authors found the use of a forward move more appealing than a reverse move or maybe they just picked up the first opcode.
I took a look at the NASM source code and found out that the encoding is done essentially with a big lookup table so it is really a matter of taste.
Using the other opcode (8AC3) would have simplified the code (I guess) if the parsing had not used a lookup table: instructions like addps are asymmetric and by using 8A /r for mov al, bl the code could be reused to compute the ModR/M byte for addps and similar instructions too. addps xmm0, xmm3 use the same ModR/M byte (C3) as mov al, bl when the 8A /r is used.
Note that the registers A (B) and xmm0 (xmm0) are encoded with the same numbers.
It is still fun, however, to figure out why there are two encodings.
As Mark Hopkins (re)discovered, the earlier x86 instructions encoding make a lot more sense in octal.
A byte in octal has three digits that I'll call G P F (Group, oPeration, Flags).
G is the octal group, instructions in the same group tend to perform similar tasks (e.g. arithmetic vs moves).
However, this is not a strict division.
P is the operation; for example, in the arithmetic group an operation is the subtraction and another is the addition.
F is a bitset used to control the behaviour of the operation. Each group and operation use the digit F as they please, it may not even be a bit set (for example G=2, P=7 is mov r16, imm16 and F is used to select r16).
For the mov instructions that move from memory/register into a register or the other way around the G is 2 and P is 1.
The F is a 3-bit field with semantic:
2 1 0 bit
+---+---+---+
| s | d | b |
+---+---+---+
s = 1 if moving to/from a segment register
0 if moving to/from a gp register
d = 1 if moving mem -> reg
0 if moving mem <- reg
b = 1 if moving a WORD
0 if moving a BYTE
We can begin to form opcodes but we still miss a way to select the operands.
G=2, P=1, F={s=0, d=0, b=0} 210 (88) mov r/m8, r8
G=2, P=1, F={s=0, d=0, b=1} 211 (89) mov r/m16, r16
G=2, P=1, F={s=0, d=1, b=0} 212 (8A) mov r8, r/m8
G=2, P=1, F={s=0, d=1, b=1} 213 (8B) mov r16, r/m16
G=2, P=1, F={s=1, d=0, b=0} 214 (8C) mov r/m16, Sreg
G=2, P=1, F={s=1, d=0, b=1} 215 (8D) Not a move, segment registers are 16-bit
G=2, P=1, F={s=1, d=1, b=0} 216 (8E) mov Sreg, r/m16
G=2, P=1, F={s=1, d=1, b=1} 217 (8F) Not a move, segment registers are 16-bit
After the opcode it must come the ModR/M byte, it is used to select the addressing mode and the register.
The ModR/M byte can be regarded, in octal, as three fields: X R M.
X and M are combined together to form the addressing mode.
R selects the register (e.g. 0 = A, 3 = B).
One of the addressing mode (X=3, M=any) lets us address the registers (through M) and not the memory.
For example, X=3, R=0, M=3 (C3) sets the register B as the "memory" operand and the register A as the register operand.
While X=3, R=3, M=0 (D8) sets the register A as the "memory" operand and the register B as the register operand.
Here we can see where the ambiguity lies: the ModR/M byte lets us encode a source register and a destination register. Meanwhile, the opcode let us encode a move from the source to the destination or from the destination to the source - this gives us the freedom to choose which register is what.
For example, suppose we want to move B into A.
If we settle on A as the register operand (source) and B as the memory operand (destination) then the ModR/M byte is X=3, R=0, M=3 (C3).
To move from B to A, as in your example, using the lower 8 bits only, we encode the move as
G=2, P=1, F={s=0,d=1,b=0} (8A) because we move mem->reg (B->A).
Thus the final instruction is 8AC3.
If we choose A as the memory operand (destination) and B as the register operand (source) the ModR/M byte is X=3, R=3, M=0 (D8).
The move is G=2, P=1, F={s=0,d=0,b=0} (88) because we move reg->mem (B->A).
The final instruction is 88D8.
If we want to move the whole 16-bit register (we ignore operand size prefixes here) we just set the b bit of F:
G=2, P=1, F={s=0,d=1,b=1} for the first case, leading to 8BC3.
G=2, P=1, F={s=0,d=0,b=1} for the second case, leading to 89D8.
You can check this out with ndisasm
00000000 8AC3 mov al,bl
00000002 88D8 mov al,bl
00000004 8BC3 mov ax,bx
00000006 89D8 mov ax,bx
来源：https://stackoverflow.com/questions/44335265/difference-between-mov-r-m8-r8-and-mov-r8-r-m8