We have been provided with the C code for this challenge;
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[]) {
int id = 0;
char name[16] = "";
printf("Input your name: ");
gets(name);
printf("Your name is %s with ID %d.\n", name, id);
if (id == 1179402567) {
printf("%s\n", argv[1]);
}
return 0;
}
As we can see, we are using the vulnerable gets function. We can use gets to overwrite the id variable which is just above the name variable on the stack. To do this, we need to write 16 bytes into name and then our desired value into id – which is 1179402567. This is easy enough to do. The only friction here is writing 1179402567 into id as we must input the ASCII representation for 1179402567. However, by converting 1179402567 into ASCII we can see that 1179402567 is equal to the ASCII string FLAG, which is easy enough to enter by hand. The final potential problem arises from the fact that strings are written to the stack starting at the lowest memory address, and ending at the highest. This is problematic as this binary is little endian, which means that we will have to enter FLAG backwards – again, this is easy to do. This gives our final payload of:
aaaaaaaaaaaaaaaaGALF
Entering this into the running binary gives us the flag quack{gets_is_vulnerable}.