Exploit the vulnerability and execute function$4$ instead of function$3(32-$bit system$).$ It is important to note that function$4$ never gets called by the main function. Your task is to inject a command line argument such that function$4$ gets called, instead of function$3.$ Include screenshots of the stack exploitation demonstration. #include

#include

#include

int function$1($int x$,$ int y$,$ int z$)$

$\{$

int result$\_$func$1$;

result$\_$func$1=$ x $+$ y $+$ z;

return result$\_$func$1$;

$\}$

int function$2($int x$,$ int y$,$ char$*$ input$\_$string$)$

$\{$

int result$\_$func$2$;

char buffer$[20]$;

strcpy$($buffer$,$ input$\_$string$)$;

printf$("$Your input string $\%$s is copied in the buffer

$",$ input$\_$string$)$;

result$\_$func$2=$ x $-$ y;

return result$\_$func$2$;

$\}$

void function$3($int result$1,$ int result$2)$

$\{$

printf$("$The result of function $1$ is $\%$d

$",$ result$1)$;

printf$("$The result of function $2$ is $\%$d

$",$ result$2)$;

$\}$

void function$4($void$)$

$\{$

printf$("$This function never gets called

$")$;

exit$(-1)$;

$\}$

int main$($int argc, char$*$ argv$[])$

$\{$

int result$1$;

int result$2$;

result$1=$ function$1(5,10,15)$;

result$2=$ function$2(20,8,$ argv$[1])$;

function$3($result$1,$ result$2)$;

$\}$