收获蛮大的 对程序能从更底层去理解了
具体到每个寄存器的操作 程序流程的跳转 变量的定义空间分配and so on
register变量 — 快速 但是不能定义太多或者太大的register
;remark1 的那几句相当于 for (register int i=stdyear;i<year;i++){} 吧
中断 — 程序运行时交还控制给用户的方式之一（唯一吗？） 输入输出时都要利用中断
;remark2 一个典型的getchar()
;remark3 printf("%s",s);?
通过标号控制流程 — 学Pascal C/C++时喜欢用goto的朋友有福气啦~ 你们应该能更快适应和掌握汇编的 循环 分支设计……
真的是无出不在………
简单总结一下（以后补……）
变量生存期 — 遗憾的是这个项目stack用的太少了 不然对这个的理解会更深的
;remark4 唯一用到stack的地方了… 相当于调用函数时把原函数的变量压栈“保护现场”吧？
汇编的函数还得多写点才行 嗯嗯~
附上code.....
格式不齐也太浪费资源 不过还是有好处的
自从nga down掉回档后让我明白 技术帖不要仅保存链接！！
data segment
year dw 0
month dw 0
day dw 0
stdyear equ 1900
status db 0
count dw 0
SY db 0dh,0ah,'Pleas input the year $'
SM db 0dh,0ah,'Month? $'
SD db 0dh,0ah,'And the day $'
SMon db 0dh,0ah,'It is Monday$'
STue db 0dh,0ah,'It is Tuesday$'
SWed db 0dh,0ah,'It is Wednesday$'
SThu db 0dh,0ah,'It is Thursday$'
SFri db 0dh,0ah,'It is Friday$'
SSat db 0dh,0ah,'It is Saturday$'
SSun db 0dh,0ah,'It is Sunday$'
SAsk db 0dh,0ah,'Do you want to inquire one more time? (y/n)$'
SBye db 0dh,0ah,'Thx for using~$'
SHello db 0dh,0ah,'This small program will tell what day is it',0dh,0ah,'Let`s begin$'
data ends
stack segment
dw 10 dup(0)
stack ends
Show macro addr ;remark3 printf("%s",s)
lea dx,addr
mov ah,9h
int 21h
endm
code segment
assume cs:code,ds:data,ss:stack
main proc far
start: push ds
sub ax,ax
push ax
mov ax,data
mov ds,ax
mov ax,stack
mov ss,ax
Show Shello
step_begin:
sub ax,ax
mov year,ax
mov month,ax
mov day,ax
mov count,ax
Show SY
inputY:
mov ax,0 ;remark2 getchar()
mov ah,1 ;remark2
int 21H ;remark2
cmp al,0DH
je IM1
sub al,30h
mov bl,al
mov bh,0
mov ax,10D
mul year
mov year,ax
add year,bx
jmp inputY
IM1: Show SM
inputM: mov ax,0
mov ah,1
int 21H
cmp al,0DH
je ID1
sub al,30h
mov bl,al
mov bh,0
mov ax,10D
mul month
mov month,ax
add month,bx
jmp inputM
ID1: Show SD
inputD: mov ax,0
mov ah,1
int 21H
cmp al,0DH
je step_year
sub al,30h
mov bl,al
mov bh,0
mov ax,10D
mul day
mov day,ax
add day,bx
jmp inputD
step_year:
mov ax,stdyear ;remark1 for ("register int i=stdyear";i<year;i++){}
year_add: ;remark1 for (register int i=stdyear;i<year;i++){""}
cmp ax,year ;remark1 for (register int i=stdyear;"i<year";i++){}
jae step_month
push ax ;remark4
call far ptr leapyear
cmp status,1
je y_leap
add count,365
jmp y_not_leap
y_leap:
add count,366
y_not_leap:
mov ax,count
mov dx,0
mov cx,7
div cx
mov count,dx
pop ax ;remark4
inc ax ;remark1 for (register int i=stdyear;i<year;"i++"){}
jmp year_add ;remark1 for (register int i=stdyear;i<year;i++){""}
step_month:
dec month
cmp month,0
je step_day
cmp month,1
je month1
cmp month,2
je month2
cmp month,3
je month3
cmp month,4
je month4
cmp month,5
je month5
cmp month,6
je month6
cmp month,7
je month7
cmp month,8
je month8
cmp month,9
je month9
cmp month,10
je month10
cmp month,11
je month11
month1: add count,31
jmp step_day
month2: add count,59
jmp month_leap
month3: add count,90
jmp month_leap
month4: add count,120
jmp month_leap
month5: add count,151
jmp month_leap
month6: add count,181
jmp month_leap
month7: add count,212
jmp month_leap
month8: add count,243
jmp month_leap
month9: add count,273
jmp month_leap
month10:
add count,304
jmp month_leap
month11:
add count,334
month_leap:
call far ptr leapyear
cmp status,1
jne step_day
inc count
step_day:
mov cx,day
add count,cx
mov dx,0
mov ax,count
mov cx,7
div cx
mov count,dx
cmp count,0
je w0
cmp count,1
je w1
cmp count,2
je w2
cmp count,3
je w3
cmp count,4
je w4
cmp count,5
je w5
cmp count,6
je w6
w0: Show SSun
jmp ask
w1: Show SMon
jmp ask
w2: Show STue
jmp ask
w3: Show SWed
jmp ask
w4: Show SThu
jmp ask
w5: Show SFri
jmp ask
w6: Show SSat
jmp ask
ask:
Show SAsk
mov ax,0
mov ah,1
int 21H
cmp al,'y'
je step_begin
cmp al,'Y'
je step_begin
cmp al,'n'
je step_end
cmp al,'N'
je step_end
jmp ask
step_end:
Show SBye
mov ax ,4c00h
int 21h
ret
main endp
leapyear proc far
mov status,0
mov bx,ax
mov dx,0
mov cx,400
div cx
cmp dx,0
je leap
mov ax,bx
mov dx,0
mov cx,4
div cx
cmp dx,0
jne exit
mov ax,bx
mov dx,0
mov cx,100
div cx
cmp dx,0
je exit
leap: mov status,1
exit: ret
leapyear endp
code ends
END start
