Attempt to execute non-executable address ...

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最新推荐文章于 2023-06-14 18:47:35 发布
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本文通过一起Windows XP启动时出现错误0xC000021A的实际案例，详细解析了数据执行保护（DEP）如何防止非执行内存区域被恶意利用。分析了winlogon进程崩溃的原因，发现第三方DLL修改了kernel32.dll中的CreateProcessW函数，导致尝试执行非执行内存区域的代码。

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出问题的一种可能。


转载一发，以免原文没了。


原文链接：https://blogs.msdn.microsoft.com/ntdebugging/2008/07/23/data-execution-protection-in-action/


Hello, my name is Graham, and I’m an escalation engineer on the Platforms Global Escalation Team.  I recently worked a case where a group of Windows XP machines were hitting a bugcheck on boot, error 0xC000021A.   This error occurs when a critical usermode process such as winlogon or csrss crashes.  I had access to a failing machine, so I attached the kernel debugger to find out why winlogon was crashing.  I found the cause, and a little bit more about Data Execution Prevention (DEP) in the process.


 


The initial debugger spew gave me this information:


 


*** An Access Violation occurred in winlogon.exe:


 


The instruction at 10030F90 tried to write to an invalid address, 10030F90


 


 *** enter .exr 0006F4AC for the exception record


 *** enter .cxr 0006F4C8 for the context


 *** then kb to get the faulting stack


 


 


So I followed its cue and got the exception record and context record:


 


1: kd> .exr 0006F4AC


ExceptionAddress: 10030f90


   ExceptionCode: c0000005 (Access violation)


  ExceptionFlags: 00000000


NumberParameters: 2


   Parameter[0]: 00000008


   Parameter[1]: 10030f90


Attempt to execute non-executable address 10030f90


 


Ahh, OK, so we know this is a DEP crash now.


 


1: kd> .cxr 0006F4C8


eax=00000400 ebx=00000000 ecx=00000000 edx=00010000 esi=00000000 edi=00084370


eip=10030f90 esp=0006f794 ebp=0006f81c iopl=0         nv up ei pl nz na pe nc


cs=001b  ss=0023  ds=0023  es=0023  fs=003b  gs=0000             efl=00010206


001b:10030f90 33c0            xor     eax,eax


 


 


Let’s check out the crashing stack to see what’s going on:


 


1: kd> kb


  *** Stack trace for last set context – .thread/.cxr resets it


ChildEBP RetAddr  Args to Child             


0006f81c 010297c1 00084370 01010ab4 00000000 3rdparty!nosymbols


0006fcfc 010312a6 00072364 7c80b6a1 00000000 winlogon!ExecSystemProcesses+0x14d


0006ff50 0103d4d0 01000000 00000000 00072364 winlogon!WinMain+0x2b6


0006fff4 00000000 7ffd7000 000000c8 000001ec winlogon!WinMainCRTStartup+0x174


 


 


The first thing I decided to look for was how we got to this address.  To begin, I unassembled the code right before the return address to winlogon!ExecSystemProcesses.


 


kd> ub 010297c1


winlogon!ExecSystemProcesses+0x12e


010297a2 6a02            push    2


010297a4 ffb594fbffff    push    dword ptr [ebp-46Ch]


010297aa 6880000000      push    80h


010297af 56              push    esi


010297b0 56              push    esi


010297b1 68b40a0101      push    offset winlogon!`string’ (01010ab4)


010297b6 ffb5a0fbffff    push    dword ptr [ebp-460h]


010297bc e891fcffff      call    winlogon!StartSystemProcess (01029452)


 


 


According to the stack, winlogon!ExecSystemProcesses didn’t call the function currently running.  So, I suspected some hooking was going on.  Using !chkimg, I verified this was the case.  Note that chkimg requires a valid copy of the binary in the symbol path.


 


1: kd> !chkimg -db kernel32


10 errors : kernel32 (7c802332-7c80236b)


7c802330  90  90 *e9 *59 *ec *82 *93  6a  00  ff  75  2c  ff  75  28  ff …Y…j..u,.u(.


…


7c802360  28  00  90  90  90  90  90 *e9 *d4 *eb *82 *93  6a  00  ff  75 (………..j..u


1: kd> u 7c802330 


kernel32!WriteProcessMemory+0x10d:


7c802330 90              nop


7c802331 90              nop


kernel32!CreateProcessW


7c802332 e959ec8293      jmp     3rdparty!nosymbols (10030f90)


 


 


Aha! Something has hooked CreateProcessW to jump to our current instruction.  Now that we know how we got there, let’s understand why we crashed.  Since DEP fired, that means this address is non-executable.  I verified this by dumping out the PTE for the address.


 


1: kd> !pte 10030F90


               VA 10030f90


PDE at 00000000C0600400    PTE at 00000000C0080180


contains 000000004E102867  contains 800000004E021867


pfn 4e102 —DA–UWEV    pfn 4e021 —DA–UW-V


 


Notice that in the protection flags for the PTE, the ‘E’ bit isn’t set, saying this page isn’t executable.   So, where is this address we were trying to execute?  Many times with DEP crashes this will be in stack or heap memory.  But not this time.  In this case, the address is actually in a module’s memory mapped address space, as shown by the ‘lm’ command


 


1: kd> lm m 3rdparty


10000000 1003c000   3rdparty C (export symbols)       3rdparty.dll


 


Hmm…  So the address falls in this module. Why isn’t it executable?   Usually when I think of image files, I think of running code.  But, remembering back to how the PE images are laid out, a module is broken into subsections, with different types of data in each one, and different protection levels.  There’s a place in the image for code, and for data, such as global variables and static data.  So, let’s dump the image header and find which section offset 0x30F90 is in.


 


1: kd>!dh 3rdparty


 


<snip>


SECTION HEADER #3


   .data name


   1EE3C virtual size


   1A000 virtual address   //  (1A000+1EE3C=0x38e3c so mem range for section is 1A000 to 0x38e3c)


    3000 size of raw data


   1A000 file pointer to raw data


       0 file pointer to relocation table


       0 file pointer to line numbers


       0 number of relocations


       0 number of line numbers


C0000040 flags


         Initialized Data


         (no align specified)


         Read Write  // no Execute !


 


 


This is our section, since the virtual address starts at 0x1A000 and is 0x1EE3C in size, putting the end of the section at 0x38e3c.  Our address of 0x30F90 falls between them.


Sure enough, this section is labeled as “Initialized Data”, and the protection flags show Read and Write, but no Execute!  So, this address is not in a code section of the module, and DEP will not allow it to run. 


 


Knowing this, I was able to find an update on the 3rd party manufacturer’s site that modified their DLL to prevent this from occurring.  Mystery solved!

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Edit the name/email address string in the main() function to your student number, student email and student name. The format of the student ID line is CSV (Comma Separated Variables) with NO SPACES- student_id,student_email,student_name When the program is run without any operands i.e. simply the name of the executable such as: lab1_1234567.exe the program MUST print student ID string in Comma Separated Values (CSV) format with no spaces. For example the following text should be outputted to the console updated with your student details: "1234567,s1234567@student.rmit.edu.au,FirstName_LastName" G3. All outputs are a single error character or a numerical number, as specified by the FUNCTIONAL REQURMENTS, followed by a linefeed ( endl or \n). G4. DO NOT add more than what is specified to the expected console output. Do NOT add additional information, text or comments to the output console that are not defined within the SPECIFICATIONS/FUNCTIONAL REQURMENTS. G5. 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For situation where TWO command line arguments are passed to your program: M2.1 Your program must perform an addition operation, taking the first two arguments as the operands and display only the result to the console with a new line character. Example1: lab1_1234567.exe 10 2 which should calculate 10 + 2 = 12, i.e. the last (and only) line on the console will be: 12 M3. For situations where THREE command line arguments are passed to your program: M3.1 If the third argument is &#39;a&#39;, your program must perform an addition operation, taking the first two arguments as the operands and display only the result to the console with a new line character. M3.2 If the third argument is &#39;s&#39;, your program must perform a subtraction operation, taking the first two arguments as the operands and display only the result to the console with a new line character. The second input argument should be subtracted from the first input argument. M4. 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Example2: lab1_1234567.exe 10 2 s which should calculate 10 - 2 = 8, i.e. the last (and only) line on the console will be: 8 Example3: lab1_1234567.exe 10 2 which should calculate 10 + 2 = 12, i.e. the last (and only) line on the console will be: 12 Example4: lab1_1234567.exe 10 4 a which should calculate 10 + 4 = 14, i.e. the last (and only) line on the console will be: 14 ___________________________________________________________________________________________ ___ FUNCTIONAL REQUIREMENTS (to get over approximately 50%): E1. For situations where THREE command line arguments (other than &#39;a&#39; or &#39;s&#39;) are passed to your program: E1.1 If the third argument is &#39;m&#39;, your program must perform a multiplication operation, taking the first two arguments as the operands and display only the result to the console with a new line character. E1.2 If the third argument is &#39;d&#39;, your program must perform a division operation, taking the first two arguments as the operands and display only the result to the console with a new line character. E1.3 If the third argument is &#39;p&#39;, your program must perform an exponential operation, taking the first argument as the base operand and the second as the exponent operand. The result must be display to the console with a new line character. Hint: Consider using the pow() function, which has the definition: double pow(double base, double exponent); Example5: lab1_1234567.exe 10 2 d which should calculate 10 / 2 = 5, i.e. the last (and only) line on the console will be: 5 Example6: lab1_1234567.exe 10 2 p which should calculate 10 to power of 2 = 100, i.e. the last (and only) line on the console will be: 100 NOTE1: DO NOT use the character ^ in your command line arguments as your user input. Question: Why don&#39;t we use characters such as + - * / ^ ? to determine the operation? Answer: Arguments passed via the command line are processed by the operating system before being passed to your program. During this process, special characters such as + - * / ^ ? are stripped from the input argument stream. Therefore, the input characters: + - * / ^ ? will not be tested for by the autotester. See sections G6 and E7. NOTE2: the pow() and powl() function/s only work correctly for given arguments. Hence, your code should output and error if there is a domain error or undefined subset of values. For example, if the result does not produce a real number you code should handle this as an error. This means that if the base is negative you can&#39;t accept and exponent between (but not including) -1 and 1. If you get this then, output a MURPHY&#39;s LAW error: "Y" and return 0; NOTE3: zero to the power of zero is also undefined, and should also be treated MURPHY&#39;s LAW error. So return "Y" and return 0; In Visual Studio, the 0 to the power of 0 will return 1, so you will need to catch this situation manually, else your code will likely calculate the value as 1. ___ REQUIRED ERROR HANDLING (to get over approximately 70%): The following text lists errors you must detect and a priority of testing. NB: order of testing is important as each test is slight more difficult than the previous test. All outputs should either be numerical or upper-case single characters (followed by a new line). Note that case is important: In C, &#39;V&#39; is not the same as &#39;v&#39;. (No quotes are required on the output). E2. Valid operator input: If the third input argument is not a valid operation selection, the output shall be &#39;V&#39;. Valid operators are ONLY (case sensitive): a addition s subtraction m multiplication d division p exponentiation i.e. to the power of: 2 to the power of 3 = 8 (base exponent p) E3. Basic invalid number detection (Required): Valid numbers are all numbers that the "average Engineering graduate" in Australia would consider valid. Therefore if the first two arguments are not valid decimal numbers, the output shall be &#39;X&#39;. For example: -130 is valid +100 is valid 1.3 is valid 3 is valid 0.3 is valid .3 is valid ABC123 is not valid 1.3.4 is not valid 123abc is not valid ___ ERROR HANDLING (not marked by the autotester): E4. Intermediate invalid number detection (NOT TESTED BY AUTOTESTER - for your consideration only): If the first two arguments are not valid decimal numbers, the output shall be &#39;X&#39;. Using comma punctuated numbers and scientific formatted numbers are considered valid. For example: 0000.111 is valid 3,000 is valid - NB: atof() will read this as &#39;3&#39; not as 3000 1,000.9 is valid - NB: atof() will read this as &#39;1&#39; not as 1000.9 1.23e2 is valid 2E2 is valid -3e-0.5 is not valid (an integer must follow after the e or E for floating point number to be valid) 2E2.1 is not valid e-1 is not valid .e3 is not valid E5. Advanced invalid number detection (NOT TESTED BY AUTOTESTER - for your consideration only): If the first two arguments are not valid decimal numbers, the output shall be &#39;X&#39;. 1.3e-1 is valid 1,00.0 is valid - NB: if the comma is not removed atof() will read this as &#39;1&#39; not as 100 +212+21-2 is not valid - NB: mathematical operation on a number of numbers, not ONE number 5/2 is not valid - NB: mathematical operation on a number of numbers, not ONE number HINT: consider the function atof(), which has the definition: double atof (const char* str); Checking the user input for multiple operators (i.e. + or -) is quite a difficult task. One method may involve writing a &#39;for&#39; loop which steps through the input argv[] counting the number of operators. This process could also be used to count for decimal points and the like. The multiple operator check should be considered an advanced task and developed once the rest of the code is operational. E6. Input number range checking: All input numbers must be between (and including) +2^16 (65536) or -2^16 (-65536). If the operand is out of range i.e. too small or too big, the output shall be &#39;R&#39;. LARGE NUMBERS: is 1.2e+999 acceptable input ? what happens if you enter such a number ? try and see. Hint: #INF error - where and when does it come up ? SMALL NUMBERS: is 1.2e-999 acceptable input ? what happens if you enter such a number ? try and see. Test it by writing your own test program. E7. ERROR checks which will NOT be performed are: E7.1 Input characters such as: *.* or / or \ or : or any of these characters: * / ^ ? will not be tested for. E7.2 Range check: some computer systems accept numbers of size 9999e999999 while others flag and infinity error. An infinity error becomes an invalid input Therefore: input for valid numbers will only be tested to the maximum 9.9e99 (Note: 9.9e99 is out of range and your program should output &#39;R&#39;) E8. Division by zero should produce output &#39;M&#39; E9. Error precedence: If multiple errors occur during a program execution event, your program should only display one error code followed by a newline character and then exit (using a return 0; statement). In general, the precedence of the error reported to the console should be displayed in the order that they appear within this proforma. However to clarify the exact order or precedence for the error characters, the precedence of the displayed error code should occur in this order: &#39;P&#39; - Incorrect number of input command line arguments (see M4) &#39;X&#39; - Invalid numerical command line argument &#39;V&#39; - Invalid third input argument &#39;R&#39; - operand (command line argument) value out of range &#39;M&#39; - Division by zero &#39;Y&#39; - MURPHY&#39;S LAW (undefined error) Therefore if an invalid numerical command line argument and an invalid operation argument are passed to the program, the first error code should be displayed to the console, which in this case would be &#39;X&#39;. Displaying &#39;V&#39; or &#39;Y&#39; would be result in a loss of marks. E10. ANYTHING ELSE THAT CAN GO WRONG (MURPHY&#39;S LAW TEST): If there are any other kinds of errors not covered here, the output shall be &#39;Y&#39;. Rhetorical question: What for example are the error codes that the Power function returns ? If this happens then the output shall be &#39;Y&#39;. See section E1.3, NOTE2. ___________________________________________________________________________________________ ___ HINTS: - Use debug mode and a breakpoint at the return statement prior to program finish in main. - What string conversion routines, do you know how to convert strings to number? Look carefully as they will be needed to convert a command line parameter to a number and also check for errors. - ERROR CHECKING: The basic programming rules are simple (as covered in lectures): 1) check that the input is valid. 2) check that the output is valid. 3) if any library function returns an error code USE IT !!! CHECK FOR IT !!! - Most conversion routines do have inbuilt error checking - USE IT !!! That means: test for the error condition and take some action if the error is true. If that means more than 50% of your code is error checking, then that&#39;s the way it has to be. ____________________________________________________________________________________________ */ // These are the libraries you are allowed to use to write your solution. Do not add any // additional libraries as the auto-tester will be locked down to the following: #include <iostream> #include <cstdlib> #include <time.h> #include <math.h> #include <errno.h> // leave this one in please, it is required by the Autotester! // Do NOT Add or remove any #include statements to this project!! // All library functions required should be covered by the above // include list. Do not add a *.h file for this project as all your // code should be included in this file. using namespace std; const double MAXRANGE = pow(2.0, 16.0); // 65536 const double MINRANGE = -pow(2.0, 16.0); // All functions to be defined below and above main() - NO exceptions !!! Do NOT // define function below main() as your code will fail to compile in the auto-tester. // WRITE ANY USER DEFINED FUNCTIONS HERE (optional) // all function definitions and prototypes to be defined above this line - NO exceptions !!! int main(int argc, char *argv[]) { // ALL CODE (excluding variable declarations) MUST come after the following &#39;if&#39; statement if (argc == 1) { // When run with just the program name (no parameters) your code MUST print // student ID string in CSV format. i.e. // "studentNumber,student_email,student_name" // eg: "3939723,s3939723@student.rmit.edu.au,Yang_Yang" // No parameters on command line just the program name // Edit string below: eg: "studentNumber,student_email,student_name" cout << "3939723,s3939723@student.rmit.edu.au,Yang_Yang" << endl; // Failure of your program to do this cout statement correctly will result in a // flat 10% marks penalty! 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