asm - 关于MIPS中%hi()与%lo()的一点疑问

问题描述

问 题

在看《See MIPS Run（2nd Edition）》的Chapter 9 Reading MIPS Assembly Language中的9.4 Addressing Modes中，碰见这样一段话：

The constructs %hi() and %lo() represent the high and low 16 bits of the address. This is not quite the straightforward division into low and high halfwords that it looks, because the 16-bit offset field of an lw is interpreted as signed. So if the addr value is such that bit 15 is a 1, then the %lo(addr) value will act as negative, and we need to increment %hi(addr) to compensate:

图片版的原文在此：

文中说：

because the 16-bit offset field of an lw is interpreted as signed. So if the addr value is such that bit 15 is a 1, then the %lo(addr) valuwill act as negative, and we need to increment %hi(addr) to compensate：

不是很懂为什么need to increment %hi(addr) to compensate。。。

解决方案

因为 %lo(addr) 是一个有符号16位整数，也就是说bit#15是符号位，因此 0x8000 是一个负数。
在做相加计算时，其符号位扩展到高16bit，也就变成 0xFFFF8000, 要想得到原来的地址 0x10008000，
%hi(addr) 必须加1， 0x1001.0000 + 0xFFFF.8000 = 0x1000.8000 .
把 lw、sw 指令的 offset 字段设为有符号整数的好处是可以寻址寄存器前后两个方向的地址，如果设计成无符号整数就只能寻址大于等于寄存器值的地址范围了。

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