ModR/M Historical Notation /0.../7

Question

While learning about x64, I struggled to understand some notations in the intel manual.

Let's look at 0xC7 MOV:

opcode	instruction	Op/Enc	Description
C7 /0 iw	MOV r/m16, imm16	MI	Move imm16 to r/m16.
C7 /0 id	MOV r/m32, imm32	MI	Move imm32 to r/m32.
REX.W + C7 /0 id	MOV r/m64, imm32	MI	Move imm32 sign extended to 64-bits to r/m64.

About the /0../7 it says:

Historically this document only specified the ModR/M.reg field restrictions with the notation /0 ... /7 and did not specify restrictions on the ModR/M.mod and ModR/M.r/m fields in the encoding boxes.

My questions are:

• That number after /0 is a restriction code, but which restriction? What does it mean while interpreting an instruction?
• If I'm reading bytes like c7c701020304... How to know if I need to ready imm16(0102 or imm32(01020304)? By testing, I know that in this imm32, but I don't understand why. I deduce that it is related to rex+modrm.

{ xxd --ps -r | ndisasm -b64 -; } <<<c7c701020304 
00000000  C7C701020304      mov edi,0x4030201

• In other words: Why ndisasm parses that as imm32 instead imm16?

Answer 1

/digit still only restricts the ModRM.reg field. It means, put whatever digit is after the slash, in the reg field of the ModRM byte.

That note is about new Intel AMX instructions, some of which have a different ModRM specification. For example TDPBSSD/TDPBSUD/TDPBUSD/TDPBUU require mod=11 (in other words, they cannot have a memory operand)

And TILELOADD/TILELOADDT1 require mod!=11 and rm=100 (they must have a memory operand, and it must be encoded with SIB)