预期的缓冲区溢出并不总是导致程序崩溃

Question

Consider The following minimal C program:

Case Number 1 :

#include <stdio.h>
#include <string.h>

void foo(char* s)
{
    char buffer[10];
    strcpy(buffer,s);
}

int main(void)
{
    foo("01234567890134567");
}

This doesn't cause a crash dump

If add just one character, so the new main is:

Case Number 2 :

void main()
{
    foo("012345678901345678");
                          ^   
}

The program crashes with a Segmentation fault.

Looks like additionally to the 10 characters reserved in the stack there's an additional room for 8 additional characters. Thus the first program doesn't crash. However, if you add one more character you start accessing invalid memory. My questions are:

1. Why we do have these additional 8 characters reserved in the stack?
2. Is this related somehow with the char data type alignment in the memory?

An other doubt I have in this case is how does the OS (Windows in this case) detects the bad memory access? Normally as per the Windows documentation the default stack size is 1MB Stack Size . So I don't see how the OS detects that the address being accessed is outside the process memory specially when the minimum page size is normally 4k. Does the OS use the SP in this case to check the address?

PD: I'm using the following environment for the testing
Cygwin
GCC 4.8.3
Windows 7 OS

EDIT :

This is the generated assembly from http://gcc.godbolt.org/# but using GCC 4.8.2, I can't see the GCC 4.8.3 in the available compilers. But I guess the generated code should be similar. I built the code without any flags. I hope somebody with Assembly expertise could shed some light about what's happening in the foo function and why the extra char causes the seg fault

    foo(char*):
    pushq   %rbp
    movq    %rsp, %rbp
    subq    $48, %rsp
    movq    %rdi, -40(%rbp)
    movq    %fs:40, %rax
    movq    %rax, -8(%rbp)
    xorl    %eax, %eax
    movq    -40(%rbp), %rdx
    leaq    -32(%rbp), %rax
    movq    %rdx, %rsi
    movq    %rax, %rdi
    call    strcpy
    movq    -8(%rbp), %rax
    xorq    %fs:40, %rax
    je  .L2
    call    __stack_chk_fail
.L2:
    leave
    ret
.LC0:
    .string "01234567890134567"
main:
    pushq   %rbp
    movq    %rsp, %rbp
    movl    $.LC0, %edi
    call    foo(char*)
    movl    $0, %eax
    popq    %rbp
    ret

Answer 1

I believe you understand that you have implemented something that leads to Undefined Behavior. So it is hard to answer why it fails with the extra string and doesn't with the original. It is probably related to the internal compiler implementation + affected by the compilation flags (like alignments, optimizations, etc.).

You can try disassembling the binary or creating assembly code and seeing where exactly the buffer is put on the stack. You can do the same with different optimization levels to inspect the changes in the assembly code and the behavior.

how does the OS (Windows in this case) detects the bad memory access? Normally as per the Windows documentation the default stack size is 1MB Stack Size. So I don't see how the OS detects that the address being accessed is outside the process memory specially when the minimum page size is normally 4k. Does the OS use the SP in this case to check the address?

The OS doesn't monitor the code you execute. The HW (CPU) does (since it executes this code). Once your code tries to access an address which was not allocated for your process (was not mapped by the OS for your program) the OS will get an indication since the HW will fire a #PF (page fault) exception. Another case is that you try to access an address which was allocated for you but with improper permissions (for example you try to execute binary data from a DATA page which has no 'execute' permission) or go to the CODE page but with a wrong offset and the instruction that you read doesn't exist or (even worse) it exists and decodes to something you don't expect (did we say Undefined Behavior before?).

In general your code most likely doesn't fail on strcpy (it can if you write enough data to access some forbidden addresses but most likely it is not the case) - it fails when it returns from the foo function. strcpy just overwrote the next instruction pointer which points to the next instruction after the foo function. So the instruction pointer is filled with the data from the "012345678901345678" string and tries to fetch the next instruction from the 'junky' address and fails due to the mentioned above reasons.

This "method"/bug is called a " buffer overflow attack " and widely used among hackers to make your code (and more often OS/BIOS/VMM/SMM code which is executed with higher privileges) execute malicious code provided by the hacker. Just make sure to overwrite the instruction pointer with the address of the code that you prepared in advance.

Answer 2

The official, system agnostic answer is:

Your code writes data beyond the end of the destination array, the behaviour is undefined, anything can happen, including nothing at all or space probe crashed on Mars surface . Your observing no noticeable effect up to 8 bytes beyond the end of the buffer and a crash with a segmentation fault beyond that are possible effects of undefined behaviour, well within the expected outcome.

The extra implementation details you are interested in:

Actual behaviour will depend on many circumstances, for example which compiler you use, which OS and ABI (Application Binary Interface) etc.

Your program is compiled and executed in a 64 bit Windows environment. In this environement, the stack is kept aligned on 64 bit boundaries, or possibly 16 byte boundaries to allow direct loading and storing of the MMX registers from/to stack locations. The array buffer[10] occupies 16 bytes on the stack. Given how the stack is established on this processor, it will be located just below locations used by function foo to store any saved registers and the return address into the caller function main . Whether the extra 6 bytes are before or after the array is a choice for the compiler to make. It could use this space for other local variables or just ignore it.

Writing beyond the end of buffer may be harmless for up to 6 bytes if the padding is after the array, might not have any noticeable effect for another 8 bytes (clobbering the saved rbp register, which is unused in main after the call), but will start having bad side effects beyond that, because you will be overwriting the return address.

When you overwrite the return address, the processor will not return from function foo to the caller main , but to whatever address is stored on the stack and was corrupted by the offending code. If this corrupted address points to executable code, that code will be executed with potential harmful consequences... Hackers do exactly this: they carefuly craft an exploit that manages to store some harmful code at a known location in executable memory and take advantage of the buffer overflow code to store the address of said code in the stack location for the return address.

In your case, the location pointed to by the corrupted return address might not be executable, triggering the segmentation fault you observe.

I suggest your try and compile your code on this site to see the actual assembly code generated under various compiler options: http://gcc.godbolt.org/#

Answer 3

The next entry in stack is function address witch in 64 bit system must be aligned to 8, thus there is enough space for 16 characters.

You can verify this by declaring an int variable after array. Int will be aligned to 4 and there will be less space for characters so program will crash on lower number.