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.
spec sheet is a document that summarizes the performance and other technical characteristics of a product, machine or component.
Answer:
Teams are diverse.
Explanation:
Teams have great advantage in problem solving over single person. A single persons's thinking is one dimensional. He sees and analyse things according to his perspective and understanding which limits his ability to solve problems .On the other hand team consist of multiple people with multiple background and perspective. Everyone has its own thinking process and it's own perspective, there fore in teams if one person is missing some perspective someone else might be looking in to that perspective which greatly enhance teams problem solving
Answer:
Explanation:
The system will be deadlock free if the below two conditions holds :
Proof below:
Suppose N = Summation of all Need(i), A = Addition of all Allocation(i), M = Addition of all Max(i). Use contradiction to prove.
Suppose this system isn't deadlock free. If a deadlock state exists, then A = m due to the fact that there's only one kind of resource and resources can be requested and released only one at a time.
Condition B, N + A equals M < m + n. Equals N + m < m + n. And we get N < n. It means that at least one process i that Need(i) = 0.
Condition A, Pi can let out at least 1 resource. So there will be n-1 processes sharing m resources now, Condition a and b still hold. In respect to the argument, No process will wait forever or permanently, so there's no deadlock.
