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本帖最后由 Windy5657 于 2012-07-25 17:04 编辑
- c03: 57 push %edi
- > c04: 56 push %esi
- > c05: 53 push %ebx
- > c06: 81 ec 1c 24 00 00 sub $0x241c,%esp
- > c0c: 8b 55 14 mov 0x14(%ebp),%edx
- > c0f: 8b 45 08 mov 0x8(%ebp),%eax
- > c12: 8d b5 1d dc ff ff lea 0xffffdc1d(%ebp),%esi
- > c18: e8 fc ff ff ff call c19 <validLicenseFound+0x19>
- > c1d: 81 c3 02 00 00 00 add $0x2,%ebx
- > c23: 89 f7 mov %esi,%edi
- > c25: 89 95 14 dc ff ff mov %edx,0xffffdc14(%ebp)
- > c2b: 65 8b 15 14 00 00 00 mov %gs:0x14,%edx
- > c32: 89 55 e4 mov %edx,0xffffffe4(%ebp)
- > c35: 31 d2 xor %edx,%edx
- > c37: f7 c6 01 00 00 00 test $0x1,%esi
- > c3d: 89 85 10 dc ff ff mov %eax,0xffffdc10(%ebp)
- > c43: 0f b6 45 0c movzbl 0xc(%ebp),%eax
- > c47: 88 85 0f dc ff ff mov %al,0xffffdc0f(%ebp)
- > c4d: 66 ba 00 20 mov $0x2000,%dx
- > c51: 0f 85 e9 02 00 00 jne f40 <validLicenseFound+0x340>
- > c57: f7 c7 02 00 00 00 test $0x2,%edi
- > c5d: 0f 85 fd 02 00 00 jne f60 <validLicenseFound+0x360>
- > c63: 89 d1 mov %edx,%ecx
- > c65: 31 c0 xor %eax,%eax
- > c67: c1 e9 02 shr $0x2,%ecx
- > c6a: f6 c2 02 test $0x2,%dl
复制代码 想问下这段汇编的具体意思是什么?Linux平台下的。objdump 反汇编出来的。对汇编一窍不通,特来请教!!另外在想问一下:这个是反汇编的.a静态库文件,我修改其汇编代码后早在让他重新生成.a文件,我应该怎么做呢?? |
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