第一个数字＆gt; 1
第二个数字＆gt; 5
总和将是＆gt; 1 + 2 + 3 + 4 + 5 = 15
我只做了两个数字的总和。我不知道如何做这个数字序列，我的代码如下。谢谢
MAIN:
MOV AX, SEG DATA
MOV DS,AX
mov ah, 9h ; msg for first number
lea dx, msg
int 21H
mov ah, 1h ; read char
int 21H
sub al, '0' ; strip ASCII
mov x, al ; storing first number
mov ah, 9h ; msg for second number
lea dx, msg2
int 21H
mov ah, 1h ; read char
int 21H
sub al, '0'
mov y, al ; storing second number
mov ah, 9h ; msg for sum
lea dx, msg3
int 21H
mov dl, x
add dl, y
add dl, '0' ; x + y
mov ah, 2h ; printing
int 21H
MOV AX,4C00H ;end
INT 21H
CODE ENDS
END MAIN
答案 0 :(得分：1)
以下是EMU8086的解决方案：
.stack 100h
.data
msj1 db 'Enter first number: $'
msj2 db 'Enter second number: $'
num1 dw ? ;FIRST NUMBER OF INTERVAL.
num2 dw ? ;SECOND NUMBER OF INTERVAL.
str db 6 ;MAX NUMBER OF CHARACTERS ALLOWED (4).
db ? ;NUMBER OF CHARACTERS ENTERED BY USER.
db 6 dup (?) ;CHARACTERS ENTERED BY USER.
crlf db 13,10,'$' ;LINE BREAK.
plus db '+$' ;PLUS SIGN TO DISPLAY.
back db 8,'$' ;CURSOR WILL MOVE ONE PLACE BACK (TO THE LEFT).
ekual db '=$' ;EQUAL SIGN TO DISPLAY.
suma dw 0 ;SUMATORY OF ALL NUMBERS BETWEEN NUM1 AND NUM2.
.code
;INITIALIZE DATA SEGMENT.
mov ax, @data
mov ds, ax
;DISPLAY MESSAGE FOR FIRST NUMBER.
mov ah, 9
mov dx, offset msj1
int 21h
;CAPTURE NUMBER AS STRING.
mov ah, 0Ah
mov dx, offset str
int 21h
;CONVERT CAPTURED NUMBER FROM STRING TO NUMERIC.
mov si, offset str ;PARAMETER FOR STRING2NUMBER.
call string2number
mov num1, bx ;RETURNED VALUE.
;DISPLAY TWO LINE BREAKS.
mov ah, 9
mov dx, offset crlf
int 21h
mov ah, 9
mov dx, offset crlf
int 21h
;DISPLAY MESSAGE FOR SECOND NUMBER.
mov ah, 9
mov dx, offset msj2
int 21h
;CAPTURE NUMBER AS STRING.
mov ah, 0Ah
mov dx, offset str
int 21h
;CONVERT CAPTURED NUMBER FROM STRING TO NUMERIC.
mov si, offset str ;PARAMETER FOR STRING2NUMBER.
call string2number
mov num2, bx ;RETURNED VALUE.
;DISPLAY TWO LINE BREAKS.
mov ah, 9
mov dx, offset crlf
int 21h
mov ah, 9
mov dx, offset crlf
int 21h
;ASURE FIRST NUMBER IS LESS THAN THE SECOND.
mov ax, num1
cmp ax, num2
jbe fine ;IF AX < NUM2 JUMP FINE.
xchg ax, num2 ;EXCHANGE : NUM2=AX, AX=NUM2.
mov num1, ax ;NUM1=NUM2.
fine:
;DISPLAY THE INTERVAL.
call display_interval
;WAIT UNTIL USER PRESS ANY KEY.
mov ah,7
int 21h
;FINISH PROGRAM.
mov ax, 4c00h
int 21h
;------------------------------------------
;DISPLAY ALL NUMBERS BETWEEN NUM1 AND NUM2
;SEPARATED BY '+' AND DISPLAYS THE SUM
proc display_interval
interval:
;ADD CURRENT NUMBER TO SUMA.
mov ax, num1 ;AX = CURRENT NUMBER.
add suma, ax
;CONVERT CURRENT NUMBER TO STRING TO DISPLAY IT.
call dollars ;FILL "STR" WITH '$'. NEEDED TO DISPLAY.
call number2string ;PARAMETER = AX. RETURNS IN VARIABLE "STR".
;DISPLAY NUMBER CONVERTED TO STRING.
mov ah, 9
mov dx, offset str
int 21h
;DISPLAY PLUS SIGN.
mov ah, 9
mov dx, offset plus
int 21h
;NEXT NUMBER TO DISPLAY.
inc num1
mov ax, num1
cmp ax, num2
jbe interval ;IF AX <= NUM2 THEN REPEAT.
;DISPLAY THE SUMA.
;DISPLAY BACKSPACE (TO DELETE LAST PLUS SIGN).
mov ah, 9
mov dx, offset back
int 21h
;DISPLAY EQUAL SIGN.
mov ah, 9
mov dx, offset ekual
int 21h
;CONVERT SUMA TO STRING TO DISPLAY IT.
call dollars ;FILL "STR" WITH '$'. NEEDED TO DISPLAY.
mov ax, suma
call number2string ;PARAMETER = AX. RETURNS IN VARIABLE "STR".
;DISPLAY NUMBER CONVERTED TO STRING.
mov ah, 9
mov dx, offset str
int 21h
ret
endp
;------------------------------------------
;CONVERT STRING TO NUMBER IN BX.
;SI MUST ENTER POINTING TO THE STRING.
proc string2number
;MAKE SI TO POINT TO THE LEAST SIGNIFICANT DIGIT.
inc si ;POINTS TO THE NUMBER OF CHARACTERS ENTERED.
mov cl, [ si ] ;NUMBER OF CHARACTERS ENTERED.
mov ch, 0 ;CLEAR CH, NOW CX==CL.
add si, cx ;NOW SI POINTS TO LEAST SIGNIFICANT DIGIT.
;CONVERT STRING.
mov bx, 0
mov bp, 1 ;MULTIPLE OF 10 TO MULTIPLY EVERY DIGIT.
repeat:
;CONVERT CHARACTER.
mov al, [ si ] ;CHARACTER TO PROCESS.
sub al, 48 ;CONVERT ASCII CHARACTER TO DIGIT.
mov ah, 0 ;CLEAR AH, NOW AX==AL.
mul bp ;AX*BP = DX:AX.
add bx,ax ;ADD RESULT TO BX.
;INCREASE MULTIPLE OF 10 (1, 10, 100...).
mov ax, bp
mov bp, 10
mul bp ;AX*10 = DX:AX.
mov bp, ax ;NEW MULTIPLE OF 10.
;CHECK IF WE HAVE FINISHED.
dec si ;NEXT DIGIT TO PROCESS.
loop repeat ;COUNTER CX-1, IF NOT ZERO, REPEAT.
ret
endp
;------------------------------------------
;FILLS VARIABLE STR WITH '$'.
;USED BEFORE CONVERT NUMBERS TO STRING, BECAUSE
;THE STRING WILL BE DISPLAYED.
proc dollars
mov si, offset str
mov cx, 6
six_dollars:
mov bl, '$'
mov [ si ], bl
inc si
loop six_dollars
ret
endp
;------------------------------------------
;NUMBER TO CONVERT MUST ENTER IN AX.
;ALGORITHM : EXTRACT DIGITS ONE BY ONE, STORE
;THEM IN STACK, THEN EXTRACT THEM IN REVERSE
;ORDER TO CONSTRUCT STRING (STR).
proc number2string
mov bx, 10 ;DIGITS ARE EXTRACTED DIVIDING BY 10.
mov cx, 0 ;COUNTER FOR EXTRACTED DIGITS.
cycle1:
mov dx, 0 ;NECESSARY TO DIVIDE BY BX.
div bx ;DX:AX / 10 = AX:QUOTIENT DX:REMAINDER.
push dx ;PRESERVE DIGIT EXTRACTED FOR LATER.
inc cx ;INCREASE COUNTER FOR EVERY DIGIT EXTRACTED.
cmp ax, 0 ;IF NUMBER IS
jne cycle1 ;NOT ZERO, LOOP.
;NOW RETRIEVE PUSHED DIGITS.
mov si, offset str
cycle2:
pop dx
add dl, 48 ;CONVERT DIGIT TO CHARACTER.
mov [ si ], dl
inc si
loop cycle2
ret
endp
接下来是您的代码有一些更改，现在它从堆栈中获取两个参数，并在内部将它们放在变量num1和num2中。我没有处理所有＆＃34; print＆＃34;的参数。 ：
proc display_interval
;RETRIEVE PARAMETERS.
pop ax ;RETURN ADDRESS.
pop num2 ;NUM2.
pop num1 ;NUM1.
push ax ;RETURN ADDRESS BACK (NECESSARY FOR "RET").
print msg4
interval:
;ADD CURRENT NUMBER TO SUMA.
mov ax, x ;AX = CURRENT NUMBER.
add k, ax
;CONVERT CURRENT NUMBER TO STRING TO DISPLAY IT.
call dollars ;FILL "STR" WITH '$'. NEEDED TO DISPLAY.
call number2string ;PARAMETER = AX. RETURNS IN VARIABLE "STR".
;DISPLAY NUMBER CONVERTED TO STRING.
mov ah, 9
mov dx, offset str
int 21h
;DISPLAY PLUS SIGN.
mov ah, 9
mov dx, offset plus
int 21h
;NEXT NUMBER TO DISPLAY.
inc x
mov ax, x
cmp ax, y
jbe interval ;IF AX <= NUM2 THEN REPEAT.
;DISPLAY THE SUMA.
;DISPLAY BACKSPACE (TO DELETE LAST PLUS SIGN).
print back
;DISPLAY EQUAL SIGN.
print equal
;CONVERT SUMA TO STRING TO DISPLAY IT.
call dollars ;FILL "STR" WITH '$'. NEEDED TO DISPLAY.
mov ax, k
call number2string ;PARAMETER = AX. RETURNS IN VARIABLE "STR".
;DISPLAY NUMBER CONVERTED TO STRING.
mov ah, 9
mov dx, offset str
int 21h
ret
endp
程序＆＃34; display_interval＆＃34;应该像这样调用：
mov ax, 1
push ax ;FIRST PARAMETER.
mov ax, 5
push ax ;SECOND PARAMETER.
call display_interval
答案 1 :(得分：0)
这是伪代码（由你来翻译成汇编）：
count = firstNumber
endCount = secondNumber + 1
total = 0
while count != endCount
total = total + count
count = count + 1
答案 2 :(得分：0)
您可以使用条件跳转来执行此操作。
下面是一个简单的代码段，其中ebx包含要添加的当前数字，ecx包含循环运行的次数（即second_number - first_number +1）。 sum将存储在eax中。
mov eax,0 ; initialise sum with 0
mov ebx,dword[first_number] ;initialise ebx with first_number
mov ecx,dword[second_number]
sub ecx,dword[first_number]
inc ecx ; ecx content will be end_number - start_number + 1, now
calculation: ; after calculation sum can be accessed from eax
add eax,ebx ; sum = sum + ebx content
inc ebx ; ebx = ebx + 1
dec ecx ; ecx = ecx - 1
cmp ecx,0h ; ecx == 0 ?
jne calculation ; if not then once again go through calculation.
答案 3 :(得分：0)
proc display_interval
push BP
mov BP, SP
mov AX, [BP+4]
mov SP, BP
pop BP
print msg4
interval:
mov ax, num1 ;AX = CURRENT NUMBER.
add suma, ax
;CONVERT CURRENT NUMBER TO STRING.
call dollars ;FILL "STR" WITH '$'.
call number2string ;PARAMETER = AX.
mov ah, 9
mov dx, offset str
int 21h
mov ah, 9
mov dx, offset plus
int 21h
inc num1
mov ax, num1
cmp ax, num2
jbe interval ;IF AX <= NUM2
;BACKSPACE (DELETE LAST +).
print back
print equal
call dollars ;FILL "STR" WITH '$'.
mov ax, suma
call number2string ;PARAMETER = AX.
mov ah, 9
mov dx, offset str
int 21h
;-------------------------------------------
;IF SUMA IS EVEN RET WILL JUMP FEW ROWS BELOW(d2)
mov dx,0 ;dividing
mov bl,suma
mov ax,bx
mov cl,2
div cl ;the remainder from division stores in AH
mov dl,ah
add dl,48 ;48 ASCII is 0, for compare
cmp dl,48 ;0?
je d2
;dividing end
pop ax
mov sp,bp
pop bp
ret ;normal RET
d2: pop ax
mov dx,583 ;number whitch it jumps from RET
mov [bp+2],dx
mov sp,bp
pop bp
ret ; jumping RET, if suma is even
endp