Answer:
No. A jump instruction causes the processor to fetch whatever bit pattern is stored in memory at the new address. This might not be part of the program.
No. A jump instruction causes the processor to fetch whatever bit pattern is stored in memory at the new address. This might not be part of the program.
Here is another schematic program. The instructions are just sketched in as place holders. Don't pay much attention to them, but look at the jump instruction and its target.
| Address | Instruction (details omitted) | PC just after this
instruction has executed (at the bottom of the cycle) |
|---|---|---|
| ............... | ........... | 00450000 |
| 00450000 | load | 00450004 |
| 00450004 | load | 00450008 |
| 00450008 | add | 0045000C |
| 0045000C | store | 00450010 |
| 00450010 | jump 0x00450000 | 004500__ |
| 00450014 | no-op | 004500__ |
The target of the jump instruction is the address
0x00450000.
Fill in the blanks.