c.asm

这个程序是我们的大作业,里面有语法分析,词法分析,四原式等等,里面有具体内容和事例,看看就能明白

da segment
	data dw 300 dup(?)
	a dw ?
	b dw ?
	i dw ?
da ends
code segment
	 assume cs:code,ds:da
main proc far
start:mov dx,da
	mov ds,dx
j0:
	mov dx,0
	mov i,dx
j1:
	mov dx,0
	mov a,dx
j2:
	mov dx,0
	mov b,dx
j3:
	mov dx,i
	cmp dx,10
	jbe jj31
	mov dx,0
	jmp jj32
jj31:
	mov dx,1
jj32:
	mov bx,3
	mov ax,4
	push(dx)
	mul bx
	pop(dx)
	mov bx,ax
	mov data[bx],dx
j4:
	mov bx,3
	mov ax,4
	push(dx)
	mul bx
	pop(dx)
	mov bx,ax
	cmp data[bx],1
	jz j5
	jmp j15
j5:
	mov dx,i
	cmp dx,5
	jbe jj51
	mov dx,0
	jmp jj52
jj51:
	mov dx,1
jj52:
	mov bx,5
	mov ax,4
	push(dx)
	mul bx
	pop(dx)
	mov bx,ax
	mov data[bx],dx
j6:
	mov bx,5
	mov ax,4
	push(dx)
	mul bx
	pop(dx)
	mov bx,ax
	cmp data[bx],1
	jz j7
	jmp j10
j7:
	mov dx,a
	add dx,i
	mov bx,7
	mov ax,4
	push(dx)
	mul bx
	pop(dx)
	mov bx,ax
	mov data[bx],dx
j8:
	mov bx,7
	mov ax,4
	push(dx)
	mul bx
	pop(dx)
	mov bx,ax
	mov dx,data[bx]
	mov a,dx
j9:
	jmp j12
j10:
	mov dx,b
	add dx,i
	mov bx,10
	mov ax,4
	push(dx)
	mul bx
	pop(dx)
	mov bx,ax
	mov data[bx],dx
j11:
	mov bx,10
	mov ax,4
	push(dx)
	mul bx
	pop(dx)
	mov bx,ax
	mov dx,data[bx]
	mov b,dx
j12:
	mov dx,i
	add dx,1
	mov bx,12
	mov ax,4
	push(dx)
	mul bx
	pop(dx)
	mov bx,ax
	mov data[bx],dx
j13:
	mov bx,12
	mov ax,4
	push(dx)
	mul bx
	pop(dx)
	mov bx,ax
	mov dx,data[bx]
	mov i,dx
j14:
	jmp j3
j15:
	mov dx,a
	add dx,b
	mov bx,15
	mov ax,4
	push(dx)
	mul bx
	pop(dx)
	mov bx,ax
	mov data[bx],dx
j16:
	mov bx,15
	mov ax,4
	push(dx)
	mul bx
	pop(dx)
	mov bx,ax
	mov bx,data[bx]
	mov ah,02h
	mov cx,4
	mov dl,bh
	shr dl,cl
	call putdl
	mov dl,bh
	shl dl,cl
	shr dl,cl
	call putdl
	mov dl,bl
	shr dl,cl
	call putdl
	mov dl,bl
	shl dl,cl
	shr dl,cl
	call putdl
j17:	mov ah,4ch
	int 21h
main endp
putdl proc near
	mov ah,2
	cmp dl,9
	jbe put1
	sub dl,10d
	add dl,'a'
	int 21h
	ret
put1: 
	add dl,'0'
	int 21h
	ret
putdl endp
	code ends
end start