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Overflow example and order of memory allocation

This is a discussion on Overflow example and order of memory allocation within the C Programming forums, part of the General Programming Boards category; Small code to show overflow But when I compile and run - buffer_one is not being overwritten when the byte ...

  1. #1
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    Overflow example and order of memory allocation

    Small code to show overflow

    But when I compile and run - buffer_one is not being overwritten when the byte size of buffer_two overflows - can someone please explain why?

    Code:
    #include <stdio.h>#include <string.h>
    
    int main(int argc, char *argv[]) {
      int value = 5;
      char buffer_one[8], buffer_two[8];
    
      strcpy(buffer_two, "two");    //Put "one" into buffer_one
      strcpy(buffer_one, "one");    //Put "two" into buffer_two
    
      printf("[BEFORE] buffer_two is at %p and contains \'%s\'\n", buffer_two, buffer_two);
      printf("[BEFORE] buffer_one is at %p and contains \'%s\'\n", buffer_one, buffer_one);
      printf("[BEFORE] value is at %p and is %d (0x%08x)\n", &value, value, value);
    
      printf("\n[STRCPY] copying %zu bytes into buffer_two\n\n", strlen(argv[1]));
      strcpy(buffer_two, argv[1]);  //Copy first argument into buffer_two.
    
      printf("[AFTER] buffer_two is at %p and contains \'%s\'\n", buffer_two, buffer_two);
      printf("[AFTER] buffer_one is at %p and contains \'%s\'\n", buffer_one, buffer_one);
      printf("[AFTER] value is at %p and is %d (0x%08x)\n", &value, value, value);
    }
    user@ubuntu:~/examples$ ./overflow_example 1234567890
    [BEFORE] buffer_two is at 0x7fff918ff210 and contains 'two'
    [BEFORE] buffer_one is at 0x7fff918ff200 and contains 'one'
    [BEFORE] value is at 0x7fff918ff1fc and is 5 (0x00000005)

    [STRCPY] copying 10 bytes into buffer_two

    [AFTER] buffer_two is at 0x7fff918ff210 and contains '1234567890'
    [AFTER] buffer_one is at 0x7fff918ff200 and contains 'one'
    [AFTER] value is at 0x7fff918ff1fc and is 5 (0x00000005)
    *** stack smashing detected ***: ./overflow_example terminated
    Aborted (core dumped)
    Last edited by sjmp; 07-11-2013 at 03:14 PM.

  2. #2
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    Quote Originally Posted by sjmp View Post
    But when I compile and run - buffer_one is not being overwritten when the byte size of buffer_two overflows - can someone please explain why?
    The standard makes no guarantee that local variables appear on the stack in the order they are declared. Some systems/implementations do this. Many modern systems randomize variable locations specifically to help make buffer overflow attacks difficult. The only thing the standard does guarantee is that writing outside the bounds of an array will result in undefined behavior, which can almost be viewed as a complete lack of guarantee.
    Quote Originally Posted by sjmp View Post
    Also - why is the int variable argc declared in this example. I don't see it ever used.
    You have two choices for main: no arguments, or argc and argv. You may not use argc, but it has to be there if you want to use argv. It gives you the length of argv, to make iterating through all the arguments in argv easier.
    Salem likes this.

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    Thank you - very clear.

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    Note the addresses from your print statements. The order of the variables were reversed on the stack, so buffer_one is "below" buffer_two, and "above" buffer_two is the portion of the stack used for main(), including the return address of whatever code that called main.

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