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Chapter 9

Chapter 9. Trap Routines TRAP number (go to service routine) & RET (return from service routine) Subroutines (or Functions) JSR offset or JSRR rn (go to subr) & RET (return from subroutine). Traps.

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Chapter 9

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  1. Chapter 9 • Trap Routines • TRAP number(go to service routine) • & RET (return from service routine) • Subroutines (or Functions) • JSR offset or JSRR rn (go to subr) • & RET (return from subroutine)

  2. Traps • Execute TRAP “vector number” - Service Routine Address Trap Vectors are at memory locations [0000:00FF] Trap Vectors contain addresses of “Pre written (?)” Service Routines 2) [PC] is stored in R7 (save the return address) 3) Address of Trap Service Routine loaded into PC (addr of service routine) 4) Service Routine Program executed • Trap service routine program ends with an RET (to return to calling prog) [R7] loaded into PC

  3. Subroutines JSR Instruction: JSR offset(11 bit) 0100 1 xxxxxxxxxxx [PC ] R7, JMP Offset Jump to Subroutine at offset from PC JSRR Instruction: JSRR Rx 0100 0 00 xxx 000000 [PC ] R7, JMP [Reg] Jump to Subroutine at address in Rx RET Instruction: RET 1100 000 111 000000 C1C0 [R7]  PC (JMP [R7]) Return to Instruction after Jump to Subroutine (or TRAP)

  4. Subroutines 1)Execute JSR offset or JSRR rn- Call Subroutine or Method Location of Subroutine is specified in the Instruction 2) [PC] stored in R7 (store return address) 3) Address from JSR offset or JSRR rn is loaded into PC (beginning of subr) 4) Subroutine is executed R0 likely contains passed parameter (or address) R5 may be used to return error message R0 likely contains return parameter (or address) 5) Subroutine program ends with an RET ( [R7] loaded into PC) Can you pass (return) two parameters ? Can a subroutine call another subroutine ? How does this mechanism support recursion?

  5. Subroutines vs Traps How are Subroutines different from Traps ? • Traps are called using the TRAP instruction (Indirect call through the Trap Vector Table) Subroutines are called using JSR or JSRR instructions (JSR Direct call, JSRR Indirect call) • Both end with a RET ( load the return address) A Trap is a Subroutine call: -(Indirect) through a Vector Table ([x0000-x00FF]). - rather than with a PC offset (or a label)

  6. Trap Service Routine for Character Input (P 234) .ORIG x04A0 START ST R7,SaveR7 ; Save Registers JSR SaveReg LD R2,Newline ; Write “newline” JSR WriteChar ; ; LEA R1,PROMPT ; Write prompt character string Loop LDR R2,R1,#0 ; Get next prompt char BRz Input ; Go to Input when done JSR WriteChar ; Write character ADD R1,R1,#1 BRnzp Loop ; ; Input JSR ReadChar ; Read Input character ADD R2,R0,#0 ; Echo to monitor JSR WriteChar ; LD R2,Newline ; Write “newline” JSR WriteChar JSR RestoreReg ; Restore Registers & Return LD R7,SaveR7 RET ; SaveR7 .FILL x0000 Newline .FILL x000A Prompt .STRINGZ "Input a character> "

  7. Trap Service Routine for Character Input (2) ; SaveR1 .FILL x0000 SaveR2 .FILL x0000 SaveR3 .FILL x0000 SaveR4 .FILL x0000 SaveR5 .FILL x0000 SaveR6 .FILL x0000 ; .END WriteChar LDI R3,DSR ; Monitor Done? BRzp WriteChar STI R2,DDR ; Send Char RET DSR .FILL xFE04 DDR .FILL xFE06 ; ; ReadChar LDI R3,KBSR ; Keyboard Ready? BRzp ReadChar LDI R0,KBDR ; Get Char RET KBSR .FILL xFE00 KBDR .FILL xFE02 ; SaveReg ST R1,SaveR1 ; Store R1,.., R6 ST R2,SaveR2 ST R3,SaveR3 ST R4,SaveR4 ST R5,SaveR5 ST R6,SaveR6 RET ; RestoreReg LD R1,SaveR1 ; Load R1,..., R6 LD R2,SaveR2 LD R3,SaveR3 LD R4,SaveR4 LD R5,SaveR5 LD R6,SaveR6 RET Why is SaveR7 not with SaveR1 thru SaveR6 ? Why wasn’t R7 stored on each JSR ? Why wasn’t R1–R6 stored on each JSR ?

  8. Chapter 10 The Stack The Stack is a dynamic “data structure” descending from high memory to lower memory, used to store data in a First In – Last Out order

  9. Illustration of a Stack

  10. Stack Operation Implemented in memory:

  11. The STACK • The Stack is a dynamic “data structure” descending from high memory to lower memory (The stack grows down ) • The Stack Pointer (R6)always points to the top word on the stack. (The Stk Ptr is the “stack reference” that is available to the programmer) • A new stack entry (a word) is “PUSHed” onto the stack • To take a word off of the Stack, the top word is “POPed” off of the stack.

  12. The Stack Most Popular Organization Alternate Organization Initially [PC] = xFE00 Initially [PC] = begin

  13. Stack R6 is the Stack Ptr Push: Push ADD R6, R6, #-1 ; Decrement Stack Ptr STR R0, R6, #0 ; “Push” [R0] onto Stack Pop: Pop LDR R0, R6, #0 ; “Pop” Data off of Stack (into R0) ADD R6, R6, #1 ; Increment Stack Ptr Which way does the Stack grow? Where does Stack Ptr (R6) point?

  14. Stack Underflow Check (Stack Empty) ; Stack POP subroutine. SP is R6. Returns “data” in R0. ; Fails if the stack is empty (SP=x4000) and reports error (1 --> R5) POP LD R1, STK_Empty ; Compare Stk Ptr with “Empty” ADD R2, R6, R1 BRz fail_exit ; EXIT IF STACK IS EMPTY ; if ok, Pop LDR R0, R6, #0 ; Pop top of Stack into R0 and ADD R6, R6, #1 ; Increment Stk Ptr AND R5, R5, #0 ; R5 <-- 0 (POP successful) RET fail_exit AND R5, R5, #0 ; R5 <-- 1 (POP failed - underflow) ADD R5, R5, #1 RET STK_Empty .FILL xC000 ; STK_Empty  -x4000 (Stack begins at x3FFF)

  15. Stack Overflow Check (Stack too Large) ; Stack PUSH subroutine. SP is R6. Value in R0 is pushed onto the stack. ; Fails if Stack is full (SP=x3F00) and reports error (1  R5) PUSH LD R1, STK_Full ; Compare Stack Ptr with “Full” ADD R2, R6, R1 BRz fail_exit ; EXIT IF STACK IS FULL ; if ok, PUSH ADD R6, R6, #-1 ; Decrement Stk Ptr and STR R0, R6, #0 ; “Push” R0 onto top of Stack AND R5, R5, #0 ; “Report” no Error 0  R5 RET fail_exit AND R5, R5, #0 ; “Report Error” 1  R5 ADD R5, R5, #1 RET STK_Full .FILL xC100 ; STK_Full <-- -x3F00 (Stack max is x3F00)

  16. Subroutine for Push & Pop ; PUSH and POP Subroutines. STK Ptr is R6. Data in R0. Success: R0 = 0. ; Stack: x3FFF to x3F00 (256 words max). POP and PUSH are Entry Points. POP ST R2,Save2 ; R1 & R2 are used by “POP”. Save them. ST R1,Save1 LD R1,STK_Empty ; Compare Stk Ptr with “Empty” ADD R2,R6,R1 BRz fail_exit ; EXIT IF STACK IS EMPTY LDR R0,R6,#0 ; “POP” top of Stack into R0 and ADD R6,R6,#1 ; Increment stack pointer BRnzp success_exit PUSH ST R2,Save2 ; R1 & R2 are used by “PUSH”. Save them. ST R1,Save1 LD R1,STK_Full ; Compare Stk Ptr with “Full” ADD R2,R6,R1 BRz fail_exit ; EXIT IF STACK IS FULL ADD R6,R6,#-1 ; Decrement Stk Ptr and STR R0,R6,#0 ; “PUSH” R0 onto top of Stack success_exit AND R5,R5,#0 ; R5 <-- 0 (success) LD R1,Save1 ; Restore registers and Return LD R2,Save2 RET fail_exit AND R5,R5,#0 ; R5 <-- 1 (failure) ADD R5,R5,#1 LD R1,Save1 ; Restore registers and Return LD R2,Save2 RET STK_Empty .FILL xC000 ; BASE = –x4000. STK_Full .FILL xC100 ; Stack 256 words Save1 .FILL x0000 Save2 .FILL x0000 .END What is a reasonable length for the Stack ?

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