帮助从 MIPS 指令中查找十六进制地址
您好，我被困在我的作业中，这需要我编写一个ac程序，该程序读取一个输入文件，如下所示：
Registers
r1 = 0c100009
Instructions
0c100009
3c071001
8ce20011
84e33080
80e48000
ace2ffff
a4e39001
a0e48088
03e00008
并查明该指令是保存还是加载，该指令访问了多少字节，以及该指令的地址内存中这些字节的第一个。
我需要帮助了解这些说明的工作原理。我确实知道我必须将指令转换为二进制，例如
8ce20011
转换为
100011 00111 00010 0000000000010001
二进制
lw $t3 17($s7)
，但我不知道如何计算地址和访问的字节数。我应该忽略所有未保存或加载的指令。
另外，符号扩展是什么意思？
感谢您的帮助。
Hi I am stuck on my assignment, which requires me to write a c program which reads in an input file such as this:
Registers
r1 = 0c100009
Instructions
0c100009
3c071001
8ce20011
84e33080
80e48000
ace2ffff
a4e39001
a0e48088
03e00008
and find out whether the instruction is a save or load, how many bytes accessed by the instruction, and the address of the first of these bytes in memory.
I need help understanding how the instructions work. I do know that I have to translate the instruction to binary, for example
8ce20011
to
100011 00111 00010 0000000000010001
which is
lw $t3 17($s7)
but i don't know how to calculate the address and the number of bytes accessed. I am suppose to ignore all the instructions that are not save or load.
And also, what does it mean by sign extend?
Thank you for your help.
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你走在正确的轨道上。
操作码 8ce20011 反汇编为“lw v0, 17(a3)”或等效的“lw $2,17($7)”
在“lw $t3, 17($7)”示例中，您需要知道 $7 的当前值（即$a3) 之前您可以计算它加载的地址。提示：查看前面的说明。 “lw”指令加载一个字，即 4 个字节。
顺便说一句，具有像 17 这样的偏移量且未在字边界上对齐的“lw”是非法的，并且会在 MIPS 中导致陷阱。
签名扩展
You are on the right track.
The opcode 8ce20011 disassembles to "lw v0, 17(a3)" or equivalently "lw $2,17($7)"
In your "lw $t3, 17($7)" example, you need to know the current value of $7 (i.e. $a3) before you can calculate the address that it is loading from. Hint: look at the previous instruction. A "lw" instruction loads a word, or 4 bytes.
Incidentally, "lw" with an offset like 17 which is not aligned on a word boundary is illegal and causes a Trap in MIPS.
Sign extension