本文介绍了为什么地址寄存器后递增为（A0）+的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
基本上以68000的汇编语言来后递增地址寄存器，您必须这样做：
Basically in the assembly language of 68000 to postincrement an address register you have to do:
(A0)+
示例
MOVE (A0)+,D0
这将移至 D0 地址 A0 指向的值，也会使 A0 的值增加1。
This will move into D0 the value pointed by address of A0 and also will increment A0 by 1.
考虑（A0）是 A0 所指向的值如果后增量语法为：
Considering that (A0) is the value pointed by A0 wasn't better if the postincrement syntax was:
(A0+)
？还是我缺少什么？
推荐答案
MOVE.L (A1)+,D0 ; increments A1 by 4, because it is long operation
; and 4 is size of long
的大小我认为当前后递增语法指向此功能，而
（A1 +）进一步表明A1总是递增一个。
I think that current postincrement syntax points at this feature, while(A1+) more suggests that A1 incremens by one always.
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