问题描述
我正在阅读ia-32指令格式，发现ModR/M是一个字节(如果需要)，但是如何确定是否需要它，有人说它是由Opcode确定的，但是如何?我想知道细节，是否有一些有用的权威性文件解释这些细节?
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?
推荐答案
Intel的vol.2手册详细介绍了每种指令的每种形式的操作数编码.例如仅采用著名的 add 指令的8位操作数大小版本有2个reg，rm表格； rm，立即表格;和add al, imm8
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.
在此之下，指令操作数编码¶表详细说明了上面Op/En(操作数编码)列中的那些I/MI/MR/RM代码的含义:
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
请注意，"I"操作数形式未提及ModRM，因此没有之一.但MI确实有一个. (使用操作码表中的80 /0中的/0填充/r字段.)
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.)
请注意，RM和MR的区别仅在于r/m操作数(可以是内存)是目标还是源.
Notice that RM and MR differ only in whether the r/m operand (that can be memory) is the destination or source.
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