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Mad_guy · Mar 8th, 2005, 11:55 PM

Or not. This is just a question I want to see if anybody can answer.

The following code alters data, I want someone to tell me why it does this, and what the name of this kind of program exploit is called. This program won't die of segmentation fault either, so you can't say "It'll die of segmentation violation."

Code is as follows:

(Toggle Plain Text)

#include <stdio.h>
#include <malloc.h>
int main() {
     char *buf1=(char *)malloc(16),*buf2=(char *)malloc(16);
     memset(buf2,'A',15);
     printf("Before: %s\n",buf2);
     memset(buf1,'B',23);
     printf("After: %s",buf2);
getchar();
return 0;
}

First one to tell me, and get the right answer gets a prize.
Prize: [undecided]

Mar 8th, 2005, 11:55 PM	#1
Mad_guy Hobbyist Programmer Join Date: Oct 2004 Location: Sandstorm, Techno Club Posts: 239 Rep Power: 5	Win money Or not. This is just a question I want to see if anybody can answer. The following code alters data, I want someone to tell me why it does this, and what the name of this kind of program exploit is called. This program won't die of segmentation fault either, so you can't say "It'll die of segmentation violation." Code is as follows: (Toggle Plain Text) #include <stdio.h> #include <malloc.h> int main() { char buf1=(char )malloc(16),buf2=(char )malloc(16); memset(buf2,'A',15); printf("Before: %s\n",buf2); memset(buf1,'B',23); printf("After: %s",buf2); getchar(); return 0; } #include <stdio.h> #include <malloc.h> int main() { char buf1=(char )malloc(16),buf2=(char )malloc(16); memset(buf2,'A',15); printf("Before: %s\n",buf2); memset(buf1,'B',23); printf("After: %s",buf2); getchar(); return 0; } First one to tell me, and get the right answer gets a prize. Prize: [undecided]

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