Answer:
This is the complete correct program:
#include <stdio.h>
#include<sys/types.h>
#include<unistd.h>
int value = 128;
int main()
{
pid_t pid;
pid=fork();
if (pid==0) /* child process */
{
value +=8;
return 0; }
else if (pid > 0) {/* parent process */
wait (NULL);
printf ("PARENT: value =%d\n" ,value); /* LINEA */
return 0;
}
}
The output of the LINE A is:
PARENT: value = 128
Explanation:
The fork() function used in the program creates a new process and this process is the child process. The child process is same as the original process having its own address space or memory.
In the child process the value of pid is 0. So the if condition checks if pid==0. Then the child process adds 8 to the value of its variable according to the following statement
value +=8;
Now the original process has value = 128. In else if part the parents process has the value of pid greater than zero and this portion of the program is of the parent process :
else if (pid > 0)
{ wait (NULL);
printf ("PARENT: value =%d\n" ,value);
return 0; }
So the value 128 is printed at the end in the output.
wait(NULL) is used to wait for the child process to terminate so the parent process waits untill child process completes.
So the conclusion is that even if the value of the variable pid is changed in the child process but it will not affect the value in the variable of the parent process.
Yeh it’s becoz they work with hardware that’s why
In general, synchronous communication means you have to wait for the answer all the time. The programming logic is simpler, but the cost that you spend a lot of time waiting.
If the options are:
<span>a. The people communicating don't need to be online at the same time.
b. There is lag time in the communication.
c. The communication occurs in real time.
a is false, you do need to be online to receive the message
b is true, typically you continue only after an acknowledgement
c is true, you wait for acknowledgement that occurs in real time (not necessarily fast though)</span>
