Question

Question 3. In the following code, answer these questions:

1. Analyze the code and how it works?
2. How can we know if this code has been overwritten? Justify how?

#include <stdlib.h>

#include <unistd.h>

#include <stdio.h>

int main(int argc, char **argv)

{

int changed = 0;

char buff[8];

while (changed == 0){

gets(buff);

if (changed !=0){

break;}

else{

    printf("Enter again: ");

    continue;

}

}

     printf("the 'changed' variable is modified\n %d", changed);

}

Answer 1

a. Analysis of the code:

• This program is taking parameters as arguments while running the program.
• The while loop will run for the infinite time as the value of "changed" is never changing.
• The final print statement will never execute.
• Please look at the comments of the program for more clarity.

b. We can say the code has been overwritten because of below reasons:

• Since the value of "changed" is always 0, so there is no need to write
  if (changed !=0) { break; }, it is never going to execute. Always, else statement will get executed.
• Also, there is a clear case of an infinite loop in the program, the while loop in the program will never get terminated and because of it, the last print statement will also never get executed.
• So, here we clearly see that it is a case of overwritten code.

Please look at the comments of the program for more clarity.

#include <stdlib.h>
#include <unistd.h>
#include <stdio.h>

int main(int argc, char **argv) // Here basically this program is accepting parameters as arguments
{
    int changed = 0; // Initializing the variable
    char buff[8];  // Initializing a char array of size 8
    while (changed == 0)  // Staring of while loop
    { 
        /*
            This loop will continue running if changed == 0.
            if the value of "changed" is not changing then this while loop will run infinite time.
        */
        gets(buff); // Taking input in buff
        if (changed !=0)  
        { 
            /*
                The program will enter this "if" statement when the value of "changed" is not equal to 0
                This is never going to happen as when "changed" is something other than 0, it will not even enter the while loop.
                If it will not enter the while loop, how can it enter the if statement.
            */
            break;
        }
        else{
            printf("Enter again: "); // It will enter when changed == 0, so it will always execute.
            continue;
            }
    }
    printf("the 'changed' variable is modified\n %d", changed); // If "while loop" ends then the program will come at this point and this print statement will execute.
}

Please let me know in the comments in case of any confusion. Also, please upvote if you like.