![]() The compiler can decide the order it wants things stored in and may change that order from program to program for non-obvious reasons.Įssentially, the exact location and ordering of local variables on the stack is undefined, and so undefined behavior is why the buffer overflow works in one case but not the other. ![]() Now the main question, Why did this work in the first case but not the second?įor some reason the compiler decided to place x after str in the first example and x before str in the second example.Īs was pointed out in the comments, the exact location of local variables on the stack is not defined by C. Now we can easily see that str comes before x and it is clear to see why writing past the end of str would cause x to be overwritten. (1) Your stack typically grows backwards, i.e. The stack should look something like: esp +25 +31 comes from uncontrolled sources, you are very close to generating a buffer overflow vulnerability. So we know str is at 25(%esp) and x is at 31(%esp). It is not too big of a leap to assume that the 5 character string "Testt" is being stored at 25(%esp) as the distance between 25(%esp) and 31(%esp) is just enough to store 5 characters and a null terminator. The x variable is definitely at 31(%esp) as we see the decimal ASCII value for 'X' being placed there. We can't see the addresses of each variable directly in your output however we can see their locations on the stack in the assembly. Next lets look at why a buffer overflow did happen in the first example. In this case it doesn't matter how far past the end of str we write because x comes before str on the stack. The stack looks something like 0061FF29 0061FF2A 0061FF36 The string "Hello world" is taking up memory addresses 0061FF2A through 0061FF36 We can see that str is above x on the stack. file "hello.c"įirst lets look at why a buffer overflow did not happen in the second example. The generated assembly for the first case (no printf). Why does simply printing the memory address of the x variable have this affect on the buffer overflow situation?Įdit: added in assembly for the two situations So in this situation, the buffer overflow did not overwrite the x variable. This causes the output to be: 0061FF2A: Testt However, if I remove the commented // printf("%p: ", &x) on the third to last line, the buffer overflow does not cause the x variable to be overwritten.įor clarity here is that code (notice the change on the third to last line) char str = "Testt" This situation shows that the buffer overflow did occur, and it caused the value of the x variable to change from 'X' to 'w'. When run, this code produces output that looks like 0061FF29: Testt I then use the function strcpy to cause a buffer overflow situation which would seemingly overwrite the memory content of the character variable x (assuming it is stored in adjacent memory). It makes the code less readable since the typedef is buried in the blob of code and there's potential for conflict with other libraries. Please limit maximum input characters in snprintf().ĭefined buff's maximum size is equal to the sum of the variables's size concatenated with the snprintf().I have a simple program that initializes a c style string and then initializes a character. 1 New contributor 2 typedef uint8t BTYE Take a careful look at that spelling versus BYTE buffer BLOCkSIZE Personally I would use uint8t everywhere and avoid the typedef. In case tmp have maximum size, input read into buf += 1024 + "\n\r" + leng(dt->name). But with data already written in from previous code, we can only write 1024 - off bytes. Sprintf() on line 473 read tmp and dt->name data in format defined before. Off += sprintf(buf + off, "%s\r\n", dt->d_name) Off += sprintf(buf + off, "%s -> %s\r\n", dt->d_name, tmp) // Read tmp to buf with max size is 1024, and buf max size is also 1024 so trigger bufferoverflow ![]() Readlink(dt->d_name, tmp, sizeof(tmp)) //Get tmp with sizeof(tmp), max is 1024 from path in dt->d_name If (lstat(dt->d_name, &sbuf) d_name = '.')Ĭhar buf = // Defined tmp's size is 1024 ![]()
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