CPSC 418: Solution 4

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Problem 1

      R0   I1
      F0   F1
     &R1   I2
copy &R2   AI
     &R3   AF
copy &R4   S
     &R0   R

Note: & = pass address of parameter

Problem 2

Feldman & Retter 2.7

Does Byte (R6,R7) = (R9,R10)?

Pseudocode:                  Code:
R7 && 3 -> R8                LANDx R7, 3, R8
R10 && 3 -> R11              LANDx R10, 3, R11
R8 = R11                     SUBc R8, R11, NUL 
JUMP_Z (RA)                  JUMP_Z _back        ;check if byte indices are the same 
R7 >> 2 -> R7                SHIFTux -2(NULL), R7, R7
R10 >> 2 -> R10              SHIFTux -2(NULL), R10, R10 
R6 + R7 -> R7                ADDx R6, R7, R7
R9 + R10 -> R10              ADDx R9, R10, R10   
R7 = R10                     SUBc R7, R11, NUL
Return                 _back:                    ;rest of program

Problem 3

Yes, the CALL instruction is necessary because instructions can't directly read or write to PC. However, remember that NULL and PC are both aliases for R31, and PC is a valid {i>base register for indexed and displacement addressing modes. So when using indexed addressing mode (NULL,R3) = M[PC + R3], but (R3, NULL) = M[R3 + 0].

Problem 4

Yes, the LODSET instruction is necessary because it is atomic Read-Modify-Write (i.e., read value, change it, and then store it back).

Problem 5

PC = R31	ISA	If the PC were R3, programs would break
SP = R30	ISA	F&R save R30 in hardware when an interrupt occurs (Fig 2.14), because different interrupt levels use different stacks. This is ISA type behaviour, in that programs can assume that R30 will be saved by hardware and so do not need to save it themselves.
RA = R29	Compiler	The use of R29 as the return address is just a convention, hardware does not need to treat R29 any differently than any other register. So, one compiler could always put the return address in R29 and another could put it in another register. However, programs compiled with these different compilers could not call each other. So, one might argue that the most correct definition is that the operating system defines RA = 29.

Problem 6

Time
----->

IL2                        T________
IL1                T_______|        |________   I________
IL0 _______________|                R        |__|        |_____________________
      P2          MEM              IL1       R     KBD   R   MEM        P1   

         Call                      ^C          INT      RFI        JMP   
        (100)    (150)   (200)    (250)   (300)(310)   (410)      (460)    (560)

T: Trap
R: Return from Trap
I: Interrupt
JMP(RA) can also simply be a RETURN

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Last modified: 1 Feb 96