### A good answer might be:

1. | Write the full 32-bit jump address: 0x00400000 | 0000 0000 0100 0000 0000 0000 0000 0000
2. | Write the 26-bit field of the jump instruction: | 00 0001 0000 0000 0000 0000 0000
3. | Shift it left two positions: | 0000 0100 0000 0000 0000 0000 0000
4. | What are the high-order four bits of the PC? | 0000
5. | Copy (4) to the left of (3): | 0000 0000 0100 0000 0000 0000 0000 0000
6. | Is (5) the same as (1)? | Yes

How do you know what the high-order four bits of the PC are? Well, since you have the address of the jump instruction, you know that its address is in the PC. So the high order four bits come from that address.

# Symbolic Address

With some trickery, a 26-bit field can specify a 32-bit address. But it is a nuisance to figure out! If you were doing machine language programming, that is what you would have to do. But the assembler does the work for you. Here is a tiny program:

    ## jump.asm
    ##

          .text
          .globl  main

    main:
          sll    $0,$0,0
          sll    $0,$0,0
          sll    $0,$0,0
          sll    $0,$0,0
          j      main
          addiu $8,$8,1

    ## End of file

It is similar to the previous example. The symbolic address `main` stands for the address of the first instruction. The instruction `j main` tells the assembler to assemble a machine instruction with the proper 26-bit field so that control is transfered to `main`.

The branch delay slot is filled with an instruction that increments register $8.

### QUESTION 8:

After the loop has executed five times, what value will be in register $8? SPIM initializes all registers to zero when a program starts.