代码如下，想要完成10个学生信息的输入输出，请高手帮忙看看问题出在哪里？
data segment
student_id db 10 dup(5,?,5 dup('$')) ;学号4位
student_name db 10 dup(6,?,6 dup('$')) ; 姓名5位字符
student_sex db 10 dup (?)
student_score db 10 dup (2,?,2 dup('$'))
table dw 4 dup(?)
mess1 db "please input student's ID that you want to delete:",0ah,0dh,'$'
data ends
code segment
assume ds:data,cs:code,es:data
start:
mov ax,data
mov ds,ax
mov es,ax
;-------------------------------------------宏定义
pianyi macro num,ad ;opr1是cx，opr2是数组首地址
push cx
mov ax,10 ;循环3次
sub ax,num ;计算每个数组的偏移地址，存入AX
mul al,10 ;每个数组是10个元素
add ad,ax
pop cx
endm
;---------------------------------------------宏定义换行回车
endl macro
push dx
mov dl,0dh
mov ah,2
int 21h
mov dl,0ah
mov ah,2
int 21h
pop dx
endm
;-------------------------------------------- ;定义地址表
mov table ,offset student_id
mov table+2,offset student_name
mov table+4,offset student_sex
mov table+6,offset student_score
mov bx,offset table ;地址表首地址放入bx
;--------------------------------------------
call input
call delete
call print
mov ax,4c00h
int 21h
input proc near
mov cx,10
input_again:
mov dx,[bx] ;输入学号
pianyi cx,dx
mov ah,10
int 21h
endl
mov dx,[bx+2] ;输入姓名
pianyi cx,dx
mov ah,10
int 21h
endl
mov di,[bx+4] ;准备输入性别
mov ax,10 ;计算student_sex数组的偏移
sub ax,cx
add di,ax
mov ah,1
int 21h
mov byte ptr [di],al
endl
mov dx,[bx+6] ;输入成绩
pianyi cx,dx
mov ah,10
int 21h
endl
loop input_again ;循环10次
ret
input endp
print proc near
mov cx,10
mov dh,0
print_again:
mov dx,[bx]+2 ;输出学号
pianyi cx,dx
mov ah,9
int 21h
endl
mov dx,[bx+2]+2 ;输出姓名
pianyi cx,dx
mov ah,9
int 21h
endl
mov di,[bx+4] ;输出性别
mov ax,10 ;计算student_sex数组的偏移
sub ax,cx
add di,ax