Example:

Assume the following tables depict memory and register values:

Memory

Address	Value
0x100	0xFF
0x104	0xAB
0x108	0x13
0x10C	0x11

Registers

Register	Value
%rax	0x100
%rcx	0x1
%rdx	0x3

---

Determine the value accessed from the following operands:

1. %rax

   Since there are now parenthesis around %rax, we are simply accessing the value stored at rax, which is 0x100.

   Solution: 0x100

2. 0x104

   Since this is a hex value, and there is not a $ symbol in front of it, then this represents a memory address. We access the value stored at the memory address 0x104 which is 0xAB.

   Solution: 0xAB

3. $0x108

   Since this value has a $ in front of it, this represents an immediate value. That is, it represents itself, 0x108.

   Solution: 0x108

4. (%rax)

   rax holds the value 0x100. The parentheses tell us that we are going to need to visit memory and get the value at 0x100. The value stored at 0x100 is 0xFF.

   Solution: 0xFF

5. 4(%rax)

   rax holds the value 0x100. The parentheses tell us that we need to visit memory. Specifically, since there is a 4 out front, we need to visit the memory addresss that is 4 over from the address stored in rax. We then get the value from there.

   Thus, %rax holds 0x100, and 4 memory locations over from 0x100 is memory location 0x104.

   The value at 0x104 is 0xAB

   Solution: 0xAB

6. 9(%rax, %rdx)

   %rax holds 0x100.%rdxholds0x3`.

   We need to add these two values together to get a resulting address. Note that in the following calculation we padded the hex value with 0s to match in digit length

     0x100
   + 0x003
   -------
     0x103

   Now, we move 9 memory locations over from 0x103.

     0x103
   + 0x009    <- 9 in hex
   -------
     0x10C    <- 3 + 9 = 12, 12 is C in hex

   Now we get the value at 0x10C which is 0x11.