我有一个简单的计算器，如下所示。我想添加幂功能，以便计算器可以计算 2 个数字的幂，这些值取自用户输入。
这是4个主要操作的代码。我想向该程序添加电源功能。我尝试了不同的方法，但没有成功。下面的代码通过 emu8086 工作，但我想向该程序添加电源功能。有什么想法我该如何处理这个问题？
org 100h
jmp start ; jump over data declaration
msg: db "1-Add",0dh,0ah,"2-Multiply",0dh,0ah,"3-Subtract",0dh,0ah,"4-Divide", 0Dh,0Ah, '$'
msg2: db 0dh,0ah,"Enter First No: $"
msg3: db 0dh,0ah,"Enter Second No: $"
msg4: db 0dh,0ah,"Choice Error $"
msg5: db 0dh,0ah,"Result : $"
msg6: db 0dh,0ah ,'thank you for using the simple calculator! press any key to exit... ', 0Dh,0Ah, '$'
start: mov ah,9
mov dx, offset msg ;first , we initialize the program with showing a simple prompt that ask the user to choose one operation on key press
int 21h
mov ah,0
int 16h ;then we will use int 16h to read a key press, to know the operation he/she choosed
cmp al,31h ;the keypress will be stored in al so, we will comapre to 1 addition ..........
je Addition
cmp al,32h ; if the keypress wasn't equal to 1 or addition ,we check that with 2 multiply
je Multiply
cmp al,33h ; if the keypress wasn't equal to 1 or addition ,we check that with 3 subtract
je Subtract
cmp al,34h ; if the keypress wasn't equal to 1 or addition ,we check that with 3 Devide
je Divide
mov ah,09h
mov dx, offset msg4
int 21h
mov ah,0
int 16h
jmp start
Addition: mov ah,9 ;then let us handle the case of addition operation
mov dx, offset msg2 ;first we will display this message enter first no also using int 21h
int 21h
mov cx,0 ;we will call InputNo to handle our input as we will take each number seprately
call InputNo ;first we will move to cx 0 because we will increment on it later in InputNo
push dx
mov ah,9
mov dx, offset msg3
int 21h
mov cx,0
call InputNo
pop bx
add dx,bx
push dx
mov ah,9
mov dx, offset msg5
int 21h
mov cx,10000
pop dx
call View
jmp exit
InputNo: mov ah,0
int 16h ;then we will use int 16h to read a key press
mov dx,0
mov bx,1
cmp al,0dh ;the keypress will be stored in al so, we will comapre to 0d which represent the enter key, to know wheter he finished entering the number or not
je FormNo ;if it's the enter key then this mean we already have our number stored in the stack, so we will return it back using FormNo
sub ax,30h ;we will subtract 30 from the the value of ax to convert the value of key press from ascii to decimal
call ViewNo ;then call ViewNo to view the key we pressed on the screen
mov ah,0 ;we will mov 0 to ah before we push ax to the stack bec we only need the value in al
push ax ;push the contents of ax to the stack
inc cx ;we will add 1 to cx as this represent the counter for the number of digit
jmp InputNo ;then we will jump back to input number to either take another number or press enter
;we took each number separatly so we need to form our number and store in one bit for example if our number 235
FormNo: pop ax; Take the last input from the stack
push dx
mul bx;Here we are multiplying the value of ax with the value of bx
pop dx;After multiplication we will remove it from stack
add dx,ax;After removing from stack add the value of dx with the value of ax
mov ax,bx;Then set the value of bx in ax
mov bx,10
push dx;push the dx value again in stack before multiplying to resist any kind of accidental effect
mul bx;Multiply bx value by 10
pop dx;pop the dx after multiplying
mov bx,ax;Result of the multiplication is still stored in ax so we need to move it in bx
dec cx;After moving the value we will decrement the digit counter value
cmp cx,0;Check if the cx counter is 0
jne FormNo;If the cx counter is not 0 that means we have multiple digit input and we need to run format number function again
ret;If the cx counter is 0 that means all of our digits are fully formatted and stored in bx we just need to return the function
View: mov ax,dx
mov dx,0
div cx
call ViewNo
mov bx,dx
mov dx,0
mov ax,cx
mov cx,10
div cx
mov dx,bx
mov cx,ax
cmp ax,0
jne View
ret
ViewNo: push ax ;we will push ax and dx to the stack because we will change there values while viewing then we will pop them back from
push dx ;the stack we will do these so, we don't affect their contents
mov dx,ax ;we will mov the value to dx as interrupt 21h expect that the output is stored in it
add dl,30h ;add 30 to its value to convert it back to ascii
mov ah,2
int 21h
pop dx
pop ax
ret
exit: mov dx,offset msg6
mov ah, 9
int 21h
mov ah, 0
int 16h
ret
Multiply: mov ah,9
mov dx, offset msg2
int 21h
mov cx,0
call InputNo
push dx
mov ah,9
mov dx, offset msg3
int 21h
mov cx,0
call InputNo
pop bx
mov ax,dx
mul bx
mov dx,ax
push dx
mov ah,9
mov dx, offset msg5
int 21h
mov cx,10000
pop dx
call View
jmp exit
Subtract: mov ah,9
mov dx, offset msg2
int 21h
mov cx,0
call InputNo
push dx
mov ah,9
mov dx, offset msg3
int 21h
mov cx,0
call InputNo
pop bx
sub bx,dx
mov dx,bx
push dx
mov ah,9
mov dx, offset msg5
int 21h
mov cx,10000
pop dx
call View
jmp exit
Divide: mov ah,9
mov dx, offset msg2
int 21h
mov cx,0
call InputNo
push dx
mov ah,9
mov dx, offset msg3
int 21h
mov cx,0
call InputNo
pop bx
mov ax,bx
mov cx,dx
mov dx,0
mov bx,0
div cx
mov bx,dx
mov dx,ax
push bx
push dx
mov ah,9
mov dx, offset msg5
int 21h
mov cx,10000
pop dx
call View
pop bx
cmp bx,0
je exit
jmp exit
第一个输入的数字是底数，第二个输入的数字是幂。请记住，零的幂会导致结果 1！不要使用大数字，因为它可能会很快溢出。
5^0 = 1 5^1 = 5 5^2 = 25 5^3 = 125
在顶部省略号处填写您的输入代码，并在底部省略号处填写您的显示代码。
...
call InputNo ; -> DX
pop ax ; First number is the base
mov cx, dx ; Second number is the power
mov dx, 1 ; If power is 0 then by definition result is 1
sub cx, 1 ; Produces 'below' if CX was 0
; Produces 'zero' if CX was 1
jb .done ; Inputted power was 0, so we return DX=1
jz .xfer ; Inputted power was 1, so we return DX=Num1
.more:
mul ax ; Ignoring DX here...
dec cx
jnz .more
.xfer:
mov dx, ax
.done:
...
您的 FormNo 代码可以在顶部添加
jcxzpopFormNo: jcxz .None
pop ax
push dx
mul bx
pop dx
add dx,ax
mov ax,bx
mov bx,10
push dx
mul bx
pop dx
mov bx,ax
dec cx
jnz FormNo
.None: ret