Question

程序将接受来自用户的两个数字，并显示这两个数字的总和，产品和功率（a ^ b）。这是捕获，但是......

程序必须：使用AddNumbers函数在MultiplyNumbers函数中使用AddNumbers函数在CalculatePower函数中使用MultiplyNumbers函数

我无法通过实现乘法加法功能找出如何找到两个数的乘积。我无法理解它背后的逻辑。我显然做错了什么，但我不确定是什么。（请忽略代码的低效率，我只想知道在找到两者的产品时我做错了什么）

我的代码到目前为止..

INCLUDE Irvine32.inc
.data
str1 BYTE "Enter a positive integer: ",0
str2 BYTE "The sum is: ",0
str3 BYTE "The product is: ",0
str4 BYTE "The power result is: ",0

.code
main PROC

call GetInteger
call Crlf
mov eax, ebx
call AddNumbers

mov edx, OFFSET str2
call WriteString
call WriteInt
call Crlf
mov eax, 0
mov ecx, edi
call MultiplyNumber
mov eax, edx
mov edx, OFFSET str3
call WriteString
call WriteInt
call Crlf
call CalculatePower
mov eax, esi
mov edx, OFFSET str4
call WriteString
call WriteInt
call Crlf

exit
main ENDP

GetInteger PROC 
mov edx, OFFSET str1
call WriteString
call ReadInt
mov ebx, eax
call WriteString
call ReadInt
mov edi, eax
ret
GetInteger ENDP

CalculatePower PROC USES edi ebx 

mov ecx, edi
mov esi, 0
L2:

    call MultiplyNumber
    add esi, edx
    loop L2
    ret
CalculatePower ENDP

MultiplyNumber PROC USES ebx edi ecx

mov edx, 0
L1:
    mov edi, ebx
    mov eax, 0
    call AddNumbers
    add edx, eax
    loop L1
    ret
MultiplyNumber ENDP

AddNumbers PROC 

    add eax, edi

ret
AddNumbers ENDP

END main

Answer 1

试试这个。您的代码分析很复杂，因为您在不同的过程中有不同的循环而无需再次初始化ecx等。

OPTION CASEMAP:NONE

INCLUDE Irvine32.inc

ExitProcess proto, dwExitCode:dword

.data
    str1 BYTE "Enter a positive integer: ",0
    str2 BYTE "The sum is: ",0
    str3 BYTE "The product is: ",0
    str4 BYTE "The power result is: ",0
    num1 DWORD 0
    num2 DWORD 0
    sum  DWORD 0
    prod DWORD 0
    pow  DWORD 0
    tmp  DWORD 0

.code
    main PROC

        mov edx, OFFSET str1    ;// Input
        call WriteString
        call ReadInt
        mov [num1], eax
        mov edx, OFFSET str1
        call WriteString
        call ReadInt
        mov [num2], eax

        call doSum              ;// Calculations
        call doMul
        call doPow

        mov edx, OFFSET str2    ;// Output
        call WriteString
        mov eax, [sum]
        call WriteInt
        call Crlf

        mov edx, OFFSET str3
        call WriteString
        mov eax, [prod]
        call WriteInt
        call Crlf

        mov edx, OFFSET str4
        call WriteString
        mov eax, [pow]
        call WriteInt
        call Crlf

        invoke ExitProcess, 0

    main ENDP

    doSum PROC              ;// Sum
        mov eax, [num1]
        add eax, [num2]
        mov [sum], eax
        ret
    doSum ENDP

    doMul PROC              ;// Multiply: Add num1 x num2 times
        xor eax, eax
        mov ecx, [num2]
        ADD_LOOP:
            add eax, [num1]
        loop ADD_LOOP
        mov [prod], eax
        ret
    doMul ENDP

    doPow PROC              ;// Power: Add num1 x num2 times and 
        mov eax, [num1]     ;// add result x num2 times till ebx=0
        mov [tmp], eax
        mov ebx, [num2]
        dec ebx
        POW_LOOP:
            xor eax, eax
            mov ecx, [num1]
            ADDPOW_LOOP:
                add eax, [tmp]
            loop ADDPOW_LOOP
            mov [tmp], eax
            dec ebx
        jnz POW_LOOP
        mov [pow], eax
        ret

    doPow ENDP

END main

通过乘法和幂处理调用添加proc的修改版本：

OPTION CASEMAP:NONE

INCLUDE Irvine32.inc

ExitProcess proto, dwExitCode:dword

.data
    str1 BYTE "Enter a positive integer: ",0
    str2 BYTE "The sum is: ",0
    str3 BYTE "The product is: ",0
    str4 BYTE "The power result is: ",0
    num1 DWORD 0
    num2 DWORD 0
    sum  DWORD 0
    prod DWORD 0
    pow  DWORD 0
    tmp  DWORD 0

.code
    main PROC

        mov edx, OFFSET str1    ;// Input
        call WriteString
        call ReadInt
        mov [num1], eax
        mov edx, OFFSET str1
        call WriteString
        call ReadInt
        mov [num2], eax

        mov eax, [num1]         ;// Calculations
        mov ebx, [num2]
        mov ecx, 1
        call doSum          
        mov [sum], eax
        call doMul
        call doPow

        mov edx, OFFSET str2    ;// Output
        call WriteString
        mov eax, [sum]
        call WriteInt
        call Crlf

        mov edx, OFFSET str3
        call WriteString
        mov eax, [prod]
        call WriteInt
        call Crlf

        mov edx, OFFSET str4
        call WriteString
        mov eax, [pow]
        call WriteInt
        call Crlf

        invoke ExitProcess, 0

    main ENDP

    doSum PROC              ;// Sum: (add ebx to eax) * ecx times
        SUM_LOOP:           ;// when ecx = 1 => simple addition a+b
        add eax, ebx
        loop SUM_LOOP
        ret
    doSum ENDP

    doMul PROC              ;// Multiply: Add num1 x num2 times
        xor eax, eax
        mov ebx, [num1]
        mov ecx, [num2]
        call doSum          ;// call Sum proc
        mov [prod], eax
        ret
    doMul ENDP

    doPow PROC              ;// Power: Add num1 x num2 times and 
        mov eax, [num1]     ;// add result x num2 times till end
        mov [tmp], eax
        mov esi, [num2]
        dec esi
        POW_LOOP:
            xor eax, eax
            mov ebx, [tmp]
            mov ecx, [num1]
            call doSum      ;// Call Sum proc
            mov [tmp], eax
            dec esi
        jnz POW_LOOP
        mov [pow], eax
        ret

    doPow ENDP

END main

Answer 2

这一点很容易相信我。我只是不知道它是如何起作用的。呵呵。

 %include "asm_io.inc"
segment .data
    prompt db "Enter first num ",0
    prompt1 db "Enter second num ",0
    prompt2 db "Equals ",0
segment .bss
    input1 resd 1
    input2 resd 1
segment .text
    global _asm_main
_asm_main:
    enter 0,0
    pusha

    mov eax, prompt
    call print_string

    call read_int
    mov [input1], eax

    mov eax, prompt1
    call print_string

    call read_int
    mov [input2], eax

    mov eax, prompt2
    call print_string

    mov eax, [input1]
    mov ebx, [input2]
    mov ebx, eax
    mul ebx
    call print_int

popa
mov eax,0
mov ebx,0
leave
ret

x86装配 - 使用添加来查找功率

2 个答案: