what is stack smashing (C)?

Question

Code:

int str_join(char *a,  const char *b) {
   int sz =0; 
   while(*a++) sz++;  
   char *st = a -1, c;  
   *st = (char) 32;
   while((c = *b++)) *++st = c;  
   *++st = 0;
   return sz;
}

....

char a[] = "StringA"; 
printf("string-1 length = %d, String a = %s\n", str_join(&a[0],"StringB"), a);

Output:

string-1 length = 7, char *a = StringA StringB

*** stack smashing detected **** : /T02 terminated

Aborted (core dumped)

I don't understand why it's showing stack smashing? and what is *stack smashing? Or is it my compiler's error?

Tearoom

Possible duplicate of Stack smashing detected – Stansberry

Answer 1

Well, stack smashing or stack buffer overflow is a rather detailed topic to be discussed here, you can refer to this wiki article for more info.

Coming to the code shown here, the problem is, your array a is not large enough to hold the final concatenated result.

Thereby, by saying

 while((c = *b++)) *++st = c;

you're essentially accessing out of bound memory which invokes undefined behavior. This is the reason you're getting the "stack smashing" issue because you're trying to access memory which does not belong to your process.

To solve this, you need to make sure that array a contains enough space to hold both the first and second string concatenated together. You have to provide a larger destination array, in short.

Peyton

Answer 2

Stack smashing means you've written outside of ("smashed" past/through) the function's storage space for local variables (this area is called the "stack", in most systems and programming languages). You may also find this type of error called "stack overflow" and/or "stack underflow".

In your code, C is probably putting the string pointed to by a on the stack. In your case, the place that causes the stack "smash" is when you increment st beyond the original a pointer and write to where it points, you're writing outside the area the C compiler guarantees to have reserved for the original string assigned into a.

Whenever you write outside an area of memory that is already properly "reserved" in C, that's "undefined behavior" (which just means that the C language/standard doesn't say what happens): usually, you end up overwriting something else in your program's memory (programs typically put other information right next to your variables on the stack, like return addresses and other internal details), or your program tries writing outside of the memory the operating system has "allowed" it to use. Either way, the program typically breaks, sometimes immediately and obviously (for example, with a "segmentation fault" error), sometimes in very hidden ways that don't become obvious until way later.

In this case, your compiler is building your program with special protections to detect this problem and so your programs exits with an error message. If the compiler didn't do that, your program would try to continue to run, except it might end up doing the wrong thing and/or crashing.

The solution comes down to needing to explicitly tell your code to have enough memory for your combined string. You can either do this by explicitly specifying the length of the "a" array to be long enough for both strings, but that's usually only sufficient for simple uses where you know in advance how much space you need. For a general-purpose solution, you'd use a function like malloc to get a pointer to a new chunk of memory from the operating system that has the size you need/want once you've calculated what the full size is going to be (just remember to call free on pointers that you get from malloc and similar functions once you're done with them).

Galiot

Answer 3

Minimal reproduction example with disassembly analysis

main.c

void myfunc(char *const src, int len) {
    int i;
    for (i = 0; i < len; ++i) {
        src[i] = 42;
    }
}

int main(void) {
    char arr[] = {'a', 'b', 'c', 'd'};
    int len = sizeof(arr);
    myfunc(arr, len + 1);
    return 0;
}

GitHub upstream.

Compile and run:

gcc -fstack-protector-all -g -O0 -std=c99 main.c
ulimit -c unlimited && rm -f core
./a.out

fails as desired:

*** stack smashing detected ***: terminated
Aborted (core dumped)

Tested on Ubuntu 20.04, GCC 10.2.0.

On Ubuntu 16.04, GCC 6.4.0, I could reproduce with -fstack-protector instead of -fstack-protector-all, but it stopped blowing up when I tested on GCC 10.2.0 as per Geng Jiawen's comment. man gcc clarifies that as suggested by the option name, the -all version adds checks more aggressively, and therefore presumably incurs a larger performance loss:

-fstack-protector

Emit extra code to check for buffer overflows, such as stack smashing attacks. This is done by adding a guard variable to functions with vulnerable objects. This includes functions that call "alloca", and functions with buffers larger than or equal to 8 bytes. The guards are initialized when a function is entered and then checked when the function exits. If a guard check fails, an error message is printed and the program exits. Only variables that are actually allocated on the stack are considered, optimized away variables or variables allocated in registers don't count.

-fstack-protector-all

Like -fstack-protector except that all functions are protected.

Disassembly

Now we look at the disassembly:

objdump -D a.out

which contains:

int main (void){
  400579:       55                      push   %rbp
  40057a:       48 89 e5                mov    %rsp,%rbp

  # Allocate 0x10 of stack space.
  40057d:       48 83 ec 10             sub    $0x10,%rsp

  # Put the 8 byte canary from %fs:0x28 to -0x8(%rbp),
  # which is right at the bottom of the stack.
  400581:       64 48 8b 04 25 28 00    mov    %fs:0x28,%rax
  400588:       00 00 
  40058a:       48 89 45 f8             mov    %rax,-0x8(%rbp)

  40058e:       31 c0                   xor    %eax,%eax
    char arr[] = {'a', 'b', 'c', 'd'};
  400590:       c6 45 f4 61             movb   $0x61,-0xc(%rbp)
  400594:       c6 45 f5 62             movb   $0x62,-0xb(%rbp)
  400598:       c6 45 f6 63             movb   $0x63,-0xa(%rbp)
  40059c:       c6 45 f7 64             movb   $0x64,-0x9(%rbp)
    int len = sizeof(arr);
  4005a0:       c7 45 f0 04 00 00 00    movl   $0x4,-0x10(%rbp)
    myfunc(arr, len + 1);
  4005a7:       8b 45 f0                mov    -0x10(%rbp),%eax
  4005aa:       8d 50 01                lea    0x1(%rax),%edx
  4005ad:       48 8d 45 f4             lea    -0xc(%rbp),%rax
  4005b1:       89 d6                   mov    %edx,%esi
  4005b3:       48 89 c7                mov    %rax,%rdi
  4005b6:       e8 8b ff ff ff          callq  400546 <myfunc>
    return 0;
  4005bb:       b8 00 00 00 00          mov    $0x0,%eax
}

  # Check that the canary at -0x8(%rbp) hasn't changed after calling myfunc.
  # If it has, jump to the failure point __stack_chk_fail.
  4005c0:       48 8b 4d f8             mov    -0x8(%rbp),%rcx
  4005c4:       64 48 33 0c 25 28 00    xor    %fs:0x28,%rcx
  4005cb:       00 00 
  4005cd:       74 05                   je     4005d4 <main+0x5b>
  4005cf:       e8 4c fe ff ff          callq  400420 <__stack_chk_fail@plt>

  # Otherwise, exit normally.
  4005d4:       c9                      leaveq 
  4005d5:       c3                      retq   
  4005d6:       66 2e 0f 1f 84 00 00    nopw   %cs:0x0(%rax,%rax,1)
  4005dd:       00 00 00 

Notice the handy comments automatically added by objdump's artificial intelligence module.

If you run this program multiple times through GDB, you will see that:

• the canary gets a different random value every time
• the last loop of myfunc is exactly what modifies the address of the canary

The canary randomized by setting it with %fs:0x28, which contains a random value as explained at:

• https://unix.stackexchange.com/questions/453749/what-sets-fs0x28-stack-canary
• Why does this memory address %fs:0x28 ( fs[0x28] ) have a random value?

How to debug it?

See: Stack smashing detected

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