Question

有人可以帮我弄清楚为什么这段代码不起作用？我在比较指令的3行收到错误。这是为了上课，我必须自学，所以任何帮助将不胜感激。这个赋值用于传递值和分析堆栈的章节。

; procedure to compute the minimum of 3 DWORD values
; The MIN is calculated, returned in eax, and displayed.
; Other registers are unchanged.

.386            ; assembler use 80386 instructions
.MODEL  FLAT    ; use modern standard memory model 

ExitProcess PROTO NEAR32 stdcall, dwExitCode:DWORD

INCLUDE io.h    ; header file for input/output

cr      EQU     0dh     ; carriage return
Lf      EQU     0ah     ; line feed

.STACK  4096    ; reserve 4096-byte stack

.DATA   ; reserve storage for data

num1    DWORD   0
num2    DWORD   0
num3    DWORD   0

directions  BYTE     cr, Lf, 'Please enter 3 numbers.', cr, Lf
            BYTE     'This program will then report the minimum',cr,Lf, 0
numlabel    BYTE        cr,Lf,Lf, 'The MIN is:  '
minValue    BYTE     16 DUP (?), cr,Lf,0    

.CODE   ; program code

MIN3  PROC   NEAR32

    push    ebp         ; save base pointer
    mov     ebp,esp     ; copy stack pointer
    push    ebx         ; save registers
    push    ecx
    push    edx

    cmp     num1, num2
    jg      second
    cmp     num1, num3
    jg      second
    mov     eax, num1
    jmp     retpop

second:
    cmp     num2, num3
    jg      third
    mov     eax, num2
    jmp     retpop

third:
    mov     eax, num3
    jmp     retpop

retpop: 
    pop     edx         ; restore registers
    pop     ecx 
    pop     ebx          
    pop     ebp         ; restore base pointer
    ret                 ; return

MIN3  ENDP

_start:     ; program entry point

    output  directions      ; display directions
    input   minValue,16     ; get number
    atod    minValue        ; convert to integer
    mov     num1, eax
    input   minValue,16     ; get number
    atod    minValue        ; convert to integer
    mov     num2, eax
    input   minValue,16     ; get number
    atod    minValue        ; convert to integer
    mov     num3, eax

    call   MIN3             ; find minimum value, ret eax

    dtoa    minValue, eax
    output  numlabel


INVOKE ExitProcess, 0   ; exit with return code 0
PUBLIC _start           
END

Answer 1

x86没有支持2个内存操作数的cmp指令版本。因此，cmp num1, num2无效，您应将其中一个加载到寄存器中并将其用于比较，例如：

mov edx, num1
cmp edx, num2

你应该注意汇编程序试图通过错误消息告诉你什么，并且还有方便的指令集参考。

在3 - MASM数组中查找MIN值

1 个答案: