MIPS-32链接器如何转换LW和SW地址? [英] How does a MIPS-32 linker convert lw and sw addresses?

问题描述

我正在阅读

解决方案

在指令字中找到的16位偏移在被添加到基址之前，先符号扩展到32位./p>

因此，如果在指令字中找到的偏移量为0x8000，则将得到0x10008000 + 0xFFFF8000 == 0x10000000(当结果被截断为32位时).同样，对于第二个示例.

I am reading Chapter 2.12 of Computer Organization and Design, trying to understand the logic of a MIPS-32 linker. I understand the concept of linking two object files and absolute reference linking. What I don't understand is the quoted paragraph below explaining the logic for calculating absolute addresses.

The load and store addresses are harder because they are relative to a base register. This example uses the global pointer as the base register. Figure 2.13 shows that $gp is initialized to 1000 8000_hex. To get the address 1000 0000_hex (the address of word X), we place 8000_hex in the address field of lw at address 40 0000_hex. Similarly, we place 8020_hex in the address field of sw at address 40 0100_hex to get the address 1000 0020_hex (the address of word Y).
Page 128

According to the MIPS-32 instruction set for instructions lw and sw, the immediate offset operand is added to the register operand.

But
1000 8000_hex + 8000_hex = 1001 0000_hex != 1000 0000_hex
1000 8000_hex + 8020_hex = 1001 0020_hex != 1000 0020_hex

So maybe I am misunderstanding something and the offsets are being subtracted? Even if that was the case:

1000 8000_hex - 8000_hex = 1000 0000_hex == 1000 0000_hex
1000 8000_hex - 8020_hex = FFFF FE0_hex != 1000 0020_hex

What is happening here?

Figure 2.13:

解决方案

The 16-bit offsets found in the instruction words are sign-extended to 32 bits before being added to the base address.

So if the offset found in the instruction word is 0x8000, you get 0x10008000 + 0xFFFF8000 == 0x10000000 (when the result is truncated to 32 bits). And similarly for the second example.

这篇关于MIPS-32链接器如何转换LW和SW地址?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！