The user inputs any number between 1 and 100, and the decimal result is output
The general idea is to read user input, process it into decimal numbers, and output it to the screen. First, the user’s input needs to be read. Initialize the memory and registers, clear the data in them, and read the first character
init: ;Initialization
xor ax, ax ;clear
xor bx, bx ;clear
xor cx, cx ;clear
xor dx, dx ;clear
mov ah, 1 ; read a character mode
int 21h ;interrupt
No illegal characters were read, reading in a loop
input: ;Start loop input
cmp al, 30h; judge '0'
jb check; smaller than 0, jump
cmp al, 39h; judge '9'
ja check; greater than 9, jump
sub al, 30h;ASCII processing
shl bx, 1 ;bx shift left
mov cx, bx ;assignment
shl bx, 1 ;bx shift left
shl bx, 1 ;bx shift left
add bx, cx ;Add
add bl, al ;add
mov ah, 1 ; read a character mode
int 21h ;interrupt
jmp input ;End of loop input
When you read an illegal character, check whether it is a carriage return. If yes, jump to Save Results section
check: ;Check input
cmp al, 0dh ; Enter and press Enter
je save; jump to save the result
jmp init ; Input error, jump to initialization
save: ;Save the result
mov ax, bx ;ax is the result
pop dx; pop off the stack
pop cx ; pop off the stack
pop bx ; pop off the stack
pop bp ;pop
Put the value in the register into the memory data segment
mov num, ax; store in data segment memory
push num ;push
Next is the output part. The hexadecimal number is still converted and decomposed, stored in the stack, and popped out of the stack in order to obtain the input number. Hexadecimal conversion part
divide: ;base conversion
xor dx, dx ;clear
div bx; divide by 10 to get the remainder, dx:ax / bx = ax...dx
add dl, 30h ;ASCII conversion
push dx; Business is pushed into the stack
inc cx ;Number of digits + 1
cmp ax, 0 ; quotient is 0, end
jne divide ;quotient is not 0, continue dividing
Loop output
output: ;Start loop output
pop dx; pop up the results one by one
mov ah, 2 ;output mode
int 21h ;interrupt
loop output ;end of loop output
Complete code
stk segment
stk ends
data segment
numdw?
data ends
code segment
assume cs:code, ss:stk, ds:data
start:
mov ax, data ;Data segment base address
mov ds, ax ; assign value to base address register
;------------------------------------------------- ---------------
; Input part
push bp; push onto stack
mov bp, sp ; stack top pointer
push bx ;bx is pushed onto the stack
push cx ;cx pushed onto stack
push dx ;dx push onto the stack
init: ;Initialization
xor ax, ax ;clear
xor bx, bx ;clear
xor cx, cx ;clear
xor dx, dx ;clear
mov ah, 1 ; read a character mode
int 21h ;interrupt
input: ;Start looping input
cmp al, 30h; judge '0'
jb check; smaller than 0, jump
cmp al, 39h; judge '9'
ja check; greater than 9, jump
sub al, 30h;ASCII processing
shl bx, 1 ;bx shift left
mov cx, bx ;assignment
shl bx, 1 ;bx shift left
shl bx, 1 ;bx shift left
add bx, cx ;Add
add bl, al ;add
mov ah, 1 ; read a character mode
int 21h ;interrupt
jmp input ;end of loop input
check: ;Check input
cmp al, 0dh ; Enter and press Enter
je save; jump to save the result
jmp init ; Input error, jump to initialization
save: ;Save the result
mov ax, bx ;ax is the result
pop dx; pop off the stack
pop cx ; pop off the stack
pop bx ; pop off the stack
pop bp; pop off the stack
; End of input section
;------------------------------------------------- ---------------
mov num, ax; store in data segment memory
push num; push onto stack
;------------------------------------------------- ---------------
; Output section
push bp; push onto stack
mov bp, sp ; stack top pointer
push ax ;ax pushed onto the stack
push bx ;bx is pushed onto the stack
push cx ;cx pushed onto stack
push dx ;dx push onto the stack
xor cx, cx ;cx clear
mov bx, [bp + 4] ; put num into bx
mov ax, bx ;assignment
mov bx, 10 ;decimal
divide: ;base conversion
xor dx, dx ;clear
div bx; divide by 10 to get the remainder, dx:ax / bx = ax...dx
add dl, 30h ;ASCII conversion
push dx; Business is pushed into the stack
inc cx ;Number of digits + 1
cmp ax, 0 ; quotient is 0, end
jne divide ;The quotient is not 0, continue to divide
output: ;start loop output
pop dx; pop up the results one by one
mov ah, 2 ;output mode
int 21h ;interrupt
loop output; loop output ends
pop dx; dx pops off the stack
pop cx ;cx pops off the stack
pop bx ;bx pops off the stack
pop ax ;ax pops off the stack
pop bp; bp pops off the stack
; End of output section
;------------------------------------------------- ---------------
mov ah, 4ch
int 21h
code ends
end start
Q & amp;A
How to print the carriage return?
0DH is a line feed, 0AH is a carriage return, put this hexadecimal number into the ax register and execute an interrupt to input characters to the screen , you can complete the line break
How to distinguish between reading and writing?
01H is for writing, 02H is for reading. Put this hexadecimal number into the ax register and execute an interrupt. You can input characters to the computer. or output characters from computer
Experience
Through this homework exercise, I have deepened my understanding of loops and jumps, and become more familiar with the underlying registers and memory calls of the computer. I also reviewed the knowledge of base conversion.
Reference materials
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Use of gcc compiler
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data segment
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Assembly: 8086 memory management method and data addressing method
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Intel8086 processor-segment register ES/DS/CS/SS and addressing
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8086 Assembly_Common Instructions
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8086 assembly, decimal conversion to hexadecimal
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8086 series (20): Hexadecimal to decimal conversion program
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8086 assembly: input and output numbers, characters, and string functions
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8086 assembly learning code segment, data segment, stack segment and segment address register
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Detailed explanation of the 14 registers in 8086CPU
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Assembly language (10) – conditional judgment instructions
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Assembly language–cmp instruction
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Assembly language CMP instructions
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Display 26 English letters”>Assembly->Display 26 English letters
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Assembly Language (3rd Edition)