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通过查看我的 c64 参考书的屏幕显示部分，我在下面输入的字符可以正确打印到屏幕上，但是当稍后从记忆中再次调用时，它们不是。:(

基本上，如果输入“A”，我会得到黑色铁锹符号，“S”作为心形符号返回，“X”作为梅花符号返回。我必须声明当我以用户身份输入时不会发生这种情况，并看到我的文本打印到屏幕上——效果很好。只有当我以用户身份按 RETURN 时才会出错，并且我认为存储在 $1000,x 的字符（每次按键后都会发生 inx）在分配区域的下方显示为符号，然后是 @ 符号组成我能看到的其余部分是其余字节中的空格。

请原谅凌乱/过度新设计的代码，我学习汇编不到一周左右 - 只有 BASIC 作为基础。

我运行了调试器，当在程序中输入“ASX”时，通过调试器的内存位置看起来像这样......

41 53 58 00 00 00 00 00 00 00 00 00 00 00 00 00 ♠♥♣@@@@@@@@@@@@@

;USER INPUTS NAME OF CORPORATION

getnamea
        ldx #00
        ldy #00
        jmp getnameb

getnameb
       ; ldx $0900   ; transfer the x value here to a safe address

        jsr $FF9F   ;SCNKEY, place ASCII character into keyboard queue
        jsr $FFE4   ;GETIN, this places the ASCII value into the Accumulator 

        BEQ getnameb ;loop until keys are pressed. Branch if zero
        
        JSR $FFD2    ;CHROUT, print it to the screen as it is being typed in.
        CMP #13      ; CMP looks for the carrige return
        BEQ Name2    ; if we find it we branch using BEQ to name2 for msg2 
        ldx $0900
        STA $1000,x  ; also store what is being typed in consecutively? 
        INX
        INY
        stx $0900
        JMP getnameb ; if we don't we loop!        
                    
                    ; * store exactly what has been typed by the user into a place 
                    ;   in memory and give it the varible name: corpname
                    ; * display some more program defined text that includes the 
                    ; users inputted string

msg1    text 'What is the name of your corporation?'
        byte 0

;------------------------------------------------------------------------------------------------------

;PRINT MSG 2

Name2   LDX #00       ; load into the x registry zero


cycle2  LDA msg2,x    ; load into A the msg2, the x sequence.
        CMP #00       ; compare memory and accumulator to the value 0?    
        BEQ reveal   ; branch/jump if the result in A is 0
        STA 1704,x    ; where on the screen does msg2 start?
        INX           ; inc x to move the print along 1 space?
        
        JMP cycle2    ; jump back to the beginning of cycle and do it all again.

; WHAT DOES THIS DO WHEN WE JUST CALL THE ADDRESS $1000?

reveal  LDX #00
reveal1 
        LDA $1000,x
        STA 1784,x
        INX
        TXA
        CMP #16
        BEQ exit
        JMP reveal1
exit    jsr *



msg2    text `New astro mining corp registration...
        byte 0
4

1 回答 1

1

问题解决了，但我很想知道是否有更好的方法来做到这一点......

        CLC            ; clear the carry
        LDA $1000,x    ; load the character
        SBC #63        ; subtract 63 from it to get the right character in the first set
于 2021-07-04T10:15:47.837 回答