No. Addresses are 32-bit patterns, machine instructions are 32-bit
patterns, and many data are 32-bit patterns.
There is no way to tell them apart, except by context.
Here is an example. Say that this bit pattern got fetched
as an instruction: 0x00000000.
Is this the address of the first byte of memory,
or the sll $0,$0,0 instruction?
The following illustrates how the jump instruction is constructed. For simplicity, all instructions other than the jump instruction are no-ops. The jump instruction jumps to the first instruction of the program. The very last instruction fills the delay slot.
The left-most six bits of the j
machine instruction are the opcode.
You need to decide if the next 26 bits are correct.
Do this by filling in the blanks of the question.
(Copy-and-paste works well for this.)
| Address | Machine Instruction | Assembly Instruction |
|---|---|---|
| 00400000 | 0000 0000 0000 0000 0000 0000 0000 0000 | sll $0,$0,0 |
| 00400004 | 0000 0000 0000 0000 0000 0000 0000 0000 | sll $0,$0,0 |
| 00400008 | 0000 0000 0000 0000 0000 0000 0000 0000 | sll $0,$0,0 |
| 0040000C | 0000 0000 0000 0000 0000 0000 0000 0000 | sll $0,$0,0 |
| 00400010 | 000010 00 0001 0000 0000 0000 0000 0000 | j firstInstruction |
| 00400014 | 0000 0000 0000 0000 0000 0000 0000 0000 | sll $0,$0,0 |
| 1. | Write the jump address 0x00400000 as 32 bits | |
| 2. | Write the 26-bit field of the jump instruction: | |
| 3. | Shift it left two positions: | |
| 4. | What are the high-order four bits of the PC? | |
| 5. | Copy (4) to the left of (3): | |
| 6. | Is (5) the same as (1)? |