assembly - 修改 6502 中断返回

Question

我正在尝试在中断返回时切换正常程序流:

START
    SEI
    LDX #<IRQ
    LDY #>IRQ
    STX $FFFE
    STY $FFFF
    CLI

LOOP1
    INC $D020
    JMP LOOP1

LOOP2
    INC $D021
    JMP LOOP2

IRQ
    STA SAVEA+1
    STX SAVEX+1
    STY SAVEY+1

    // Some Routines

    LDA #$00
    PHA
    LDA #<LOOP2
    PHA
    LDA #>LOOP2
    PHA

SAVEA   
    LDA #$00
SAVEX   
    LDX #$00
SAVEY   
    LDY #$00
    RTI

我根据该来源编写了这段代码: http://6502.org/tutorials/interrupts.html#1.3

但是PHA导致崩溃，如何在中断中将正常流程LOOP1切换到LOOP2？

Answer 1

最简单的事情可能是有两个堆栈区域——每个任务一个。例如，$100-$17f 和 $180-$1ff。然后，您将拥有如下中断任务切换代码:

  pha
  txa
  pha
  tya
  pha ;saving task's registers on its stack,
      ;where flags and PC are already saved
      ;by entering the interrupt

  tsx
  stx ... ;save task's stack position

  ... ;select new task to run/etc.

  ldx ...
  txs ;load other task's stack position

  pla
  tay
  pla
  tax
  pla ;restore other task's registers

  rti ;and finally continue other task