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Answer:

0000 1111 1010 0101           0FA5 
0011 0110 1000 1111           368F 
-------------------           ----
0011 1111 1010 1111           3FAF

Running the Program

register display in SPIM

Doing the problem by hand yields the result, 0x3faf. Of course, when the program runs the result will be 32 bits long: 0x00003faf.

The picture shows the result of running the program. The result in $10 is what was expected.

To run the program:


  1. Create a source file (copy and paste from the previous page).
  2. Start QtSpim.
  3. Simulator → Settings → MIPS:
    • ON — Bare Machine (click on the button)
    • OFF — Load Exception Handler (clear the check box)
  4. Display registers in Hex (Registers → Hex)
  5. Load the source file (File → Reinitialize and Load File).
    • This assembles the source and loads the resulting machine code at address 0x400000
  6. Push F10 once per instruction.
    • Twice for this program
    • With the third push, the simulator will show an error message

QUESTION 7:

Here is the the first instruction from the source window: 

[0x00400000]    0x34080fa5    ori  $8, $0, 4005       ori  $8,$0,0x0FA5

Look at the 32-bit machine instruction at address 0x00400000. Do you see the immediate operand in it?

(Recall that the SPIM window shows the dis-assembly of the machine code, which in the above is ori $8, $0, 4005. The dis-assembly uses base 10 to show the bit pattern as if it were representing an integer.)


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