How to interpret the opcode manually?

Question

77f4bcbc 8945fc          mov     dword ptr [ebp-4],eax

And here's the rule:

88  /r   MOV r/m8,r8       2/2           Move byte register to r/m byte
89  /r   MOV r/m16,r16     2/2           Move word register to r/m word
89  /r   MOV r/m32,r32     2/2           Move dword register to r/m dword

How to interpret 8945fc to mov dword ptr [ebp-4],eax?

Answer 1

We have here a three-byte instruction: 89 45 fc. The first byte is the opcode byte. Looking it up in the table, we can see that it's a MOV instruction and it takes a Mod R/M byte. The Mod R/M byte has the following layout:

 7  6   5  4  3   2  1  0
+-----+---------+---------+
| Mod |   Reg   |   R/M   | 
+-----+---------+---------+

Let's look at the second byte of the instruction. 0x45 is 01.000.101 in binary. Thus, Mod is 01, Reg is 000 and R/M is 101.

Looking up in the reference, e.g. here, we can see that the combination of Mod=01 and R/M=101 corresponds to the [EBP+sbyte] operand. The "sbyte" is an 8-bit signed displacement which is encoded in the third byte: 0xFC. Since the displacement is signed, it has to be interpreted as such number, i.e. -4.

The "/r" note next to the instruction tells us that the register (second) operand is specified by the Reg field of the instruction. Reg=000 is al/ax/eax. Assuming a 32-bit mode by default, this will mean eax.

Assembling all of the above, we get

MOV [EBP-4], EAX