Answer: Denial of Service
Explanation:
the term is self-explanatory
Answer:
He forgot to give credit to the sources at the end.
If you take something from a different source and you quote it, you must give credit at the end or it can go down as plagiarism.
Answer:
int sumid=0; /* Shared var that contains the sum of the process ids currently accessing the file */
int waiting=0; /* Number of process waiting on the semaphore OkToAccess */
semaphore mutex=1; /* Our good old Semaphore variable ;) */
semaphore OKToAccess=0; /* The synchronization semaphore */
get_access(int id)
{
sem_wait(mutex);
while(sumid+id > n) {
waiting++;
sem_signal(mutex);
sem_wait(OKToAccess);
sem_wait(mutex);
}
sumid += id;
sem_signal(mutex);
}
release_access(int id)
{
int i;
sem_wait(mutex);
sumid -= id;
for (i=0; i < waiting;++i) {
sem_signal(OKToAccess);
}
waiting = 0;
sem_signal(mutex);
}
main()
{
get_access(id);
do_stuff();
release_access(id);
}
Some points to note about the solution:
release_access wakes up all waiting processes. It is NOT ok to wake up the first waiting process in this problem --- that process may have too large a value of id. Also, more than one process may be eligible to go in (if those processes have small ids) when one process releases access.
woken up processes try to see if they can get in. If not, they go back to sleep.
waiting is set to 0 in release_access after the for loop. That avoids unnecessary signals from subsequent release_accesses. Those signals would not make the solution wrong, just less efficient as processes will be woken up unnecessarily.
Import java.util.Scanner;
class hola
{
public static void main(String[]args)
{
Scanner x=new Scanner(System.in);
int a=x.nextInt();
int b;
if(a>20&&a<100)
{
b=a%12;
if(b%2==0){
System.out.print("es par"+b);
}
else{
System.out.print("es impar"+b);
}
}
}
}
Answer:
punctuation_chars = ["'", '"', ",", ".", "!", ":", ";", '#', '@']
def strip_punctuation(strWord):
for charPunct in punctuation_chars:
strWord = strWord.replace(charPunct, "")
return strWord
Explanation:
The function is defined with a single argument.
A for loop is ran to check each character of the the word.
If a punction mark is present as a character in the word, it is removed.
The same word is returned but without the punctuation marks.
