EEPW论坛

哎,这么多高手怎么就不随便指教一下呢，我好伤心哦
费了好大的劲,通过别的路径,终于找到我认为可以是满意解释问题1的答案,现抄录如下:

;when we want to use directing addresing
;we may write down as following
;ld #x,dp
;~~~~~~~~~~~~~
;st a,@x
;then the addres of transformation is the address of x
;that equal as transform a to at x
;That is very important in the directing addresing.
;********************************************************************************************************
;*****that instruction la #x,dp is means that sent the number of page of x in data memory to dp
;*****and the caculation of page number is devided by 128 words from 0x0000 to 0xfff
;***** so every page is 128 words
;***** the page is devided as flowing:
;*****0x0000~0x007f
;*****0x0080~0x00ff
;*****0x0100~0x017f
;*****0x0180~0x01ff
;*****0x0200~0x027f
;*****~~~~~~~~~~~~~~~~~~~~~~~~~
;*****for example:
;*****if x was devote to the addres of 0x0fff,so the x is in NO. 3 page in data memory
;*****and then the instruction la #x,dp is sent 3 to dp,and equal to 0x0180
;*****but notice the next address is derived by the NO.4 page in data memory
;*****@x is the address of x

问题2还请高手不吝赐教
还有你们有没有可以交流的好论坛呀