TI-83 + z80十六进制程序出现问题

Question

我做了一个运行这个十六进制代码的AsmProgram:

9D95:00
9D96:00
9D97:00
9D98:00
9D99:0E
9D9A:09
9D9B:3E
9D9C:00
9D9D:EF
9D9E:5D
9D9F:49
9DA0:06
9DA1:80
9DA2:A8
9DA3:CB
9DA4:47
9DA5:CB
9DA6:0F
9DA7:CA
9DA8:95
9DA9:9D
9DAA:10
9DAB:F9
9DAC:C9

它应该等到按下+按钮然后结束,但它似乎不起作用.

Answer 1

这是你的原始代码,据我所知:

        .org 9D95h
        nop
        nop
        nop
        nop
        ld c, 09h       ; ???
        ld a, 0         ; ???
        rst 28h         ; \
        .db 5Dh         ;  > DispTail, destroys AF, BC, DE, HL, aka
        .db 49H         ; /  undefined behaviour in this case (we don't
                        ;    know what A contains)
        ld b, 80h       ; B <- 0x80
        xor b           ; A <- A XOR B
        bit 0, a        ; A[0] == 0, Z is set, if so
loop:   rrc a           ; rotate right A, C <- A[0]
        jp z, 9D95h     ; Jump to 9D95, if Z set
        djnz loop       ; Decrease B, jump if not zero to loop
        ret             ; After 128 jumps, returns

总而言之,他们的密钥代码检查部分从get go开始是错误的(调用错误的ROM调用并且过度复杂).这似乎有用:

        .org 9D95h
        rst 28h         ; \
        .db 72h         ;  > Call GetKey, A <- key code
        .db 49h         ; /
        cp 80h          ; Compare A with immediate value 0x80 (subtract),
                        ; key code for [+] is 0x80
        jp nz, 9D95h    ; Jump if Z is not set (was not 0x80)
        ret             ; return

或以十六进制:

EF 72 49 FE 80 C2 95 9D C9