Given this MIPS machine language code fragment in memory 0x0

Given this MIPS machine language code fragment in memory: 0x00400024 0x3411000b 0x00400028 0x3cl01001| 0x0040002c 0x2a21000b 0x00400030 0x14200007 0x00400034 0xae080004 0x00400038 0x2231ffff 0x0040003c 0x01129020 0x00400040 0x2210fffc 0x00400044 0x2a210002 0x00400048 0xl020fffb Disassemble and show the MIPS instruction. For constants m I-format instructions, you may leave them m hexadecimal, or translate them into decimal. Don\'t worry about labels. Show clearly target addresses for all branches and jumps.

Solution

Answer:

MIPS instructions after dissembling the given code fragment:

0: 00 04 00 add BYTE PTR [eax+eax*1],al
3: 02 40 34 add al,BYTE PTR [eax+0x34]
6: 11 00 adc DWORD PTR [eax],eax
8: 0b 00 or eax,DWORD PTR [eax]
a: 04 00 add al,0x0
c: 02 80 3c 10 10 01 add al,BYTE PTR [eax+0x110103c]
12: 00 04 00 add BYTE PTR [eax+eax*1],al
15: 02 c0 add al,al
17: 2a 21 sub ah,BYTE PTR [ecx]
19: 00 0b add BYTE PTR [ebx],cl
1b: 00 04 00 add BYTE PTR [eax+eax*1],al
1e: 03 00 add eax,DWORD PTR [eax]
20: 14 20 adc al,0x20
22: 00 07 add BYTE PTR [edi],al
24: 00 04 00 add BYTE PTR [eax+eax*1],al
27: 03 40 ae add eax,DWORD PTR [eax-0x52]
2a: 08 00 or BYTE PTR [eax],al
2c: 04 00 add al,0x0
2e: 04 00 add al,0x0
30: 03 80 22 31 ff ff add eax,DWORD PTR [eax-0xcede]
36: 00 04 00 add BYTE PTR [eax+eax*1],al
39: 03 c0 add eax,eax
3b: 01 12 add DWORD PTR [edx],edx
3d: 90 nop
3e: 20 00 and BYTE PTR [eax],al
40: 04 00 add al,0x0
42: 04 00 add al,0x0
44: 22 10 and dl,BYTE PTR [eax]
46: ff (bad)
47: fc cld
48: 00 04 00 add BYTE PTR [eax+eax*1],al
4b: 04 40 add al,0x40
4d: 2a 21 sub ah,BYTE PTR [ecx]
4f: 00 02 add BYTE PTR [edx],al
51: 00 04 00 add BYTE PTR [eax+eax*1],al
54: 04 80 add al,0x80
56: 10 20 adc BYTE PTR [eax],ah
58: ff (bad)
59: fb sti