实验4 汇编应用编程和c语言程序反汇编分析

assume cs:code, ds:data
data segment
str db 'welcome to masm!'
data ends
code segment
start:
mov ax,0b800h；目标段地址
mov ds,ax
mov bx,data；源段地址
mov es,bx
mov cx,16
mov di,0
mov si,0720h；屏幕中间的位置的偏移地址
s:jcxz sss；如果cx为零跳转下一个循环
mov ax,es:[di]
movds:[si],ax；将当前的字符复制给目标位置
mov byte ptr ds:[si+1],02h；给当前字符添加颜色属性
add si,2
inc di
loop s
sss:mov cx,16
mov di,0
mov si,07c0h
s1:jcxz ss1
mov ax,es:[di]
movds:[si],ax
mov byte ptr ds:[si+1],24h
add si,2
inc di
loop s1
ss1:mov cx,16
mov di,0
mov si,0860h
s2:jcxz ss2
mov ax,es:[di]
movds:[si],ax
mov byte ptr ds:[si+1],71h
add si,2
inc di
loop s2
ss2:mov ah,4ch
int 21h
code ends
end start

assume cs:code, ds:data
data segment
str db 'another try', 0
data ends
code segment
start:
mov ax, data
mov ds, ax
mov si, offset str
mov al, 4
call printStr
mov ah, 4ch
int 21h
printStr:
push bx
push cx
push si
push di
mov bx, 0b800H
mov es, bx
mov di, 0
s:mov cl, [si]
mov ch, 0
jcxz over
mov ch, al
moves:[di], cx
inc si
add di, 2
jmp s
over:pop di
pop si
pop cx
pop bx
ret
code ends
end start

assume cs:code, ds:data
data segment
x dw 1234
str db 16 dup(0)
data ends
code segment
start:
mov ax, data
mov ds, ax
mov ax, x
mov di, offset str
call num2str
mov si,offset str
mov al,4
call printStr
mov ah, 4ch
int 21h
num2str:
push ax
push bx
push cx
push dx
mov cx, 0
mov bl, 10
s1:
div bl
inc cx
mov dl, ah
push dx
mov ah, 0
cmp al, 0
jne s1
s2:
pop dx
or dl, 30h
mov [di], dl
inc di
loop s2
pop dx
pop cx
pop bx
pop ax
ret
printStr:push bx
push cx
push si
push di
mov bx, 0b800H
mov es, bx
mov di, 0
s:mov cl, [si]
mov ch, 0
jcxz over
mov ch, al
moves:[di], cx
inc si
add di, 2
jmp s
over:pop di
pop si
pop cx
pop bx
ret
code ends
end start

assume cs:code, ds:data
data segment
str db 80 dup(?)
data ends
code segment
start:
mov ax, data
mov ds, ax
mov si, 0
s1:
mov ah, 1
int 21h
mov [si], al
cmp al, '#'
je next
inc si
jmp s1
next:
mov cx, si
mov si, 0
s2:mov ah, 2
mov dl, [si]
int 21h
inc si
loop s2
mov ah, 4ch
int 21h
code ends
end start

#include <stdio.h>
int sum(int, int);
int main() {
int a = 2, b = 7, c;
c = sum(a, b);
return0;
}
int sum(int x, int y) {
return (x + y);
}