I'm checking the assembly code for a MOV instruction on Intel 64 (long) compatibility sub-mode.
For a mov edx,408ACB77 the debugger (x32dbg for Windows) shows
BA 408ACB77
From Intel 64 SDM this MOV instruction in compatibility mode should use a one-byte opcode plus a 32bit immediate value (77CB8A40). In Intel SDM the opcode for MOV should be B8 +rd id. +rd means look at the lower three bit of one-byte opcode for the target dword register (i.e. 32 bit register).
From Table 3.1 the EDX register has index 0x2 then we get 0x8+0x2 = 0xA, hence the opcode byte is actually BA.
Is the above correct ? Thanks in advance.