How to determine if ModR/M is needed through Opcod

问题:

I am reading the ia-32 instruction format and found that ModR/M is one byte if required, but how to determine if it is required, someone says it is determined by Opcode, but how? I want to know the details, and is there some useful and authoritative documents which explain the details?

回答1:

Intel's vol.2 manual has details on the encoding of operands for each form of each instruction. e.g. taking just the 8-bit operand size versions of the well-known add instruction, which has 2 reg,rm forms ; a rm,immediate form ; and a no-ModRM 2-byte short for for add al, imm8

Opcode    Instruction    | Op/En |  64-bit Mode | Compat/Leg Mode |  Description
04 ib     ADD AL, imm8   |  I    |   Valid           Valid         Add imm8 to AL.
80 /0 ib  ADD r/m8, imm8 |  MI   |   Valid           Valid         Add imm8 to r/m8.
00 /r     ADD r/m8, r8   |  MR   |   Valid           Valid         Add r8 to r/m8.
02 /r     ADD r8, r/m8   |  RM   |   Valid           Valid         Add r/m8 to r8.

And below that, the Instruction Operand Encoding ¶ table details what those I / MI / MR / RM codes from the Op/En (operand encoding) column above mean:

Op/En   | Operand 1        | Operand 2     | Operand 3  Operand 4
RM      | ModRM:reg (r, w) | ModRM:r/m (r) |  NA        NA
MR      | ModRM:r/m (r, w) | ModRM:reg (r) |  NA        NA
MI      | ModRM:r/m (r, w) | imm8/16/32    |  NA        NA
I       | AL/AX/EAX/RAX    | imm8/16/32    |  NA        NA

Notice that the "I" operand form doesn't mention a ModRM, so there isn't one. But MI does have one. (With the /r field being filled in with the /0 from the 80 /0 in the opcode table.)

Notice that RM and MR differ only in whether the r/m operand (that can be memory) is the destination or source.