汇编的程序出错了(帮我看看吧)
程序语法没错,但是出来的结果不对,
.model small
.stack 40h
.data
namepar label byte
maxnlen db 21
namelen db ?
namefld db 21 dup(?)
telpar label byte
telmaxlen db 9
tellen db ?
telfld db 9 dup(?)
crlf db 13,10, '$ '
endaddr dw ?
messg1 db 'Input name: ', '$ '
messg2 db 'Input a telephone number: ', '$ '
messg3 db 'Do you want a telephone number?(Y/N) ',13,10, '$ '
messg4 db 'name? ', '$ '
messg5 db 'name telphone ',13,10, '$ '
namectr db 0
tel_tab db 50 dup(28 dup( ' '))
namesav db 20 dup(?),13,10, '$ '
telsav db 8 dup(?),13,10, '$ '
na_telsav db 28 dup(?),13,10, '$ '
swapped db 0
;******************************************************************************
.code
begin proc far
mov ax,@data
mov ds,ax
mov es,ax
cld
lea di,tel_tab
a:
call input_name
call inphone
cmp namelen,0
jz a30
cmp namectr,50
je a30
call stor_name
call stor_tel
jmp a
a30:
cmp namectr,1
jbe a40
call name_sort
call println
call name_search
a40:
mov ax,4c00h
int 21h
;*****************************************************************************
input_name proc near
mov ah,09
lea dx,messg1
int 21h
mov ah,0ah
lea dx,namepar
int 21h
mov ah,09
lea dx,crlf
int 21h
mov bh,0
mov bl,namelen
mov cx,21
sub cx,bx
b: mov namefld[bx],20h
inc bx
loop b
ret
input_name endp
;******************************************************************************
stor_name proc near
inc namectr
cld
lea si,namefld
mov cx,10
rep movsw
ret
stor_name endp
;********************************************************************************
inphone proc near
mov ah,09
lea dx,messg2
int 21h
mov ah,0ah
lea dx,telpar
int 21h
mov ah,09
lea dx,crlf
int 21h
mov bh,0
mov bl,tellen
mov cx,9
sub cx,bx
c: mov telfld[bx],20h
inc bx
loop c
ret
inphone endp
;***************************************************************************************
stor_tel proc near
cld
lea si,telfld
mov cx,4
rep movsw
ret
stor_tel endp
;**************************************************************************************
name_sort proc near
sub di,48
mov endaddr,di
g20:
mov swapped,0
lea si,tel_tab
g30:
mov cx,28
mov di,si
add di,28
mov ax,di
mov bx,si
repe cmpsb
jbe g40
call chg
g40:
mov si,ax
cmp si,endaddr
jbe g30
cmp swapped,0
jnz g20
ret
name_sort endp
;***************************************************************************************
chg proc near
mov cx,14
lea di,na_telsav
mov si,bx
rep movsw
mov cx,14
mov di,bx
rep movsw
mov cx,14
lea si,na_telsav
rep movsw
mov swapped,1
ret
chg endp
;************************************************************************************************
name_search proc near
mov ah,09
lea dx,messg3
int 21h
mov ah,01
int 21h
cmp al, 'N '
jz exit
cmp al, 'n '
jz exit
mov ah,09
lea dx,messg4
int 21h
call input_name
lea di,tel_tab
comp:
lea si,namefld
mov cx,20
repz cmpsb
jz equal
add di,8
jmp comp
equal:
mov cx,8
cld
mov si,di
lea di,telsav
rep movsb
mov ah,09
lea dx,telsav
int 21h
exit:
mov ax,4c00h
int 21h
name_search endp
;***************************************************************************************
println proc near
mov ah,09
lea dx,messg5
int 21h
lea si,tel_tab
k20:
lea di,na_telsav
mov cx,14
rep movsw
mov ah,9
lea dx,na_telsav
int 21h
dec namectr
jnz k20
ret
println endp
;************************************************************************************************
end begin
[解决办法]
这是因为, 搜索 name 时, 虽然 name 总长度为 20, 但 repe cmpsb 不是等所有的 20 个字节都比较完了才终止的, 而是一旦遇到有不同的就终止了. 所以, repe cmpsb 后, di 不一定是指向了 name 的最后, 这样在匹配下一个记录时, 就不能简单的 add di, 8 了. 简单的还是先对 di 进行保存, 然后恢复, 增量以整个记录的长度:
comp:
push di ;*+ 保存 di
lea si,namefld
mov cx,20
repz cmpsb
pop di ;*+ 恢复 di 指向记录的开始
jz equal
add di,28 ;* 指向下一个记录的开始
jmp comp
equal:
mov cx,8
cld
lea si, [di+20] ;* 指向记录的 '号码 ' 域
...
