I. Compile and execute the following programming exercises:

Figure $3\mathrm{.}30$ : Program shows that value modification by child doesn't affect value for parent process. Print the content of the global variable 'value' in child and parent.

Figure $3\mathrm{.}31$ : Program shows that $8$ processes are created. Add a print statement after each fork$().$

Figure $3\mathrm{.}32($Fig $3\mathrm{.}21$ in some books$)$ : Program shows that $16$ total processes are created. Display their 'pid's.

Figure $3\mathrm{.}33($Fig $3\mathrm{.}22$ in some books$)$: Program shows that "LINE J$"$ after call to 'execlp$()\text{'}$ will not be printed if execlp$()$ is successful.

Figure $3\mathrm{.}34($Fig $3\mathrm{.}23$ in some books$)$ : Program shows the outputs at lines A$,$ B$,$ C$,$ and D$.$

Figure $3\mathrm{.}35($Fig $3\mathrm{.}24$ in some books$)$: Programs show the outputs at lines X and Y$.$

$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

Example of adding print statements to display process id in Figure $3\mathrm{.}32$:

#include

#include

#include

int main$()$

$\{$

int i;

for $($i$=0$; i$<3$; i$++)$

$\{$

fork$()$;

if $($i $==0)$ printf$("$Loop $0$: hello from $\%$d

$",$ getpid$())$;

if $($i $==1)$ printf$("\backslash$tLoop $1$ hello from $\%$d

$",$ getpid$())$;

if $($i $==2)$ printf$("\backslash$t$\backslash$tLoop $2$: hello from $\%$d

$",$ getpid$())$;

$\}$

return $0$;

$\}$

$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

II$.$ Zombie & Orphan Processes

Zombie Process

When a child process terminates, an association with its parent survives until the parent in turn either:

calls wait$()$ or

terminates normally

The child process entry in the process table is therefore not freed up immediately. Although no longer active, the child process is still in the system because the exit code needs to be stored in case the parent subsequently calls wait$().$ The child becomes what is known as defunct, or a zombie process.

Orphan Process

If a parent process did not invoke wait $()$ and instead terminated, its child processes remain as orphan processes

Compile and execute fork$1.$c program. The original process $($the parent$)$ finishes before the child has printed all of its messages, so the next shell prompt appears mixed in with the output. Does the child become an orphan? Why or why not?

Convert the fork$1.$c program to a zombie process. $($Hint: change the number of messages. If the child prints fewer messages than the patent, child will finish first and will exist as a zombie until the parent has finished.$)$

$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

III. Submit the screenshots of the compilation, execution, and results.

If you get "warning: implicit declaration of function 'wait' $"$ change #include to #include

Do not submit the source codes. Points will be deducted, if instructions are not followed.

Submit all screenshots on a single file, sequentially numbered.