### A good answer might be:

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.

# Practice

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`.

### QUESTION 4:

Fill in the blanks.