1

所以我有这个程序：


[BITS 16]
[ORG 7C00h]

    JMP 0:start

start:
    MOV AX, 0
    MOV DS, AX
    MOV ES, AX
    MOV SS, AX
    MOV SP, 7C00h

    MOV SI, Hello
    CALL _puts16
    MOV AL, LogLvl1
    INT 21h
    CALL .getinput
.getinput:
    MOV AH, 2h
    MOV AL, Success ;Echo was successful
    INT 21h
    MOV AH, 1h ;Set AH to 1h, which means character input
    INT 21h
    MOV INPUT, AL ;Store the character in INPUT
    MOV AH, 2h ;Change AH back to 2h, which means character output
    MOV DL, Message ;Move Message to DL, which will be echoed
    INT 21h
    MOV DL, INPUT ;Move the inputed key to DL to be echoed
    INT 21h
    RET
;input= es:si -> string
_puts16:
    PUSHA

.BEGIN:
    LODSB ;Load a single byte
    CMP AL, 0 ;Calculate if that byte is 0, or null
    JE .END ;Jump to end if it is not 0

    MOV AH, 0Eh ;Otherwise, set 24 bit graphics modes
    INT 10h
    JMP .BEGIN

.END:
    POPA
    RET
.define:
    Hello DB "Enter a string: ",0
    Message DB "        Grats! You said",0
    INPUT DB " ",0
    LogLvl1 DB "Calling .getinput...\n",0
    Success DB "Successful.\n",0
times 510 - ($ - $$) db 0

DW 0xAA55

我对汇编非常陌生，我需要将键入的密钥移动到 .define 中定义的字符串变量“INPUT”中，但是网络范围的汇编器给了我一个错误：

1.asm:24：错误：操作码和操作数的组合无效

我认为这意味着由于某种原因我无法将 AL 移动到 INPUT，但我不知道为什么。正如我所说，我是组装新手，所以这可能是一些明显的错误。我怎样才能解决这个问题？我需要将键入的字符或键存储在内存中，以便稍后在程序中回显它。

4

1 回答 1

1

因为您实际上是在尝试将AL常量存储到空格字符的地址中。这就像要求汇编器（或编译器）将 2 分配给 1。

Change the instruction to mov [INPUT], al.

Looks like you have several instances of this same problem, confusing memory contents with memory addresses.

Also, DOS int 21h isn't available at the time when your PC boots a boot sector from a disk. Use BIOS functions (e.g. int 16h and int 10h) instead.

于 2013-03-29T15:53:28.157 回答