Answer:
The solution is as follows.
class LFilters implements Lock {
int[] lvl;
int[] vic;
public LFilters(int n, int l) {
lvl = new int[max(n-l+1,0)];
vic = new int[max(n-l+1,0)];
for (int i = 0; i < n-l+1; i++) {
lvl[i] = 0;
}
}
public void lock() {
int me = ThreadID.get();
for (int i = 1; i < n-l+1; i++) { // attempt level i
lvl[me] = i;
vic[i] = me;
// rotate while conflicts exist
int above = l+1;
while (above > l && vic[i] == me) {
above = 0;
for (int k = 0; k < n; k++) {
if (lvl[k] >= i) above++;
}
}
}
}
public void unlock() {
int me = ThreadID.get();
lvl[me] = 0;
}
}
Explanation:
The code is presented above in which the a class is formed which has two variables, lvl and vic. It performs the operation of lock as indicated above.
Answer:
A.)
arr[0] = 10;
arr[1] = 10;
Explanation:
Given the array:
arr = {1,2,3,4,5}
To set the first two elements of array arr to 10.
Kindly note that ; index numbering if array elements starts from 0
First element of the array has an index of 0
2nd element of the array has an index of 1 and so on.
Array elements can be called one at a time using the array name followed by the index number of the array enclosed in square brackets.
arr[0] = 10 (this assigns a value of 10 to the index value, which replace 1
arr[1] = 10 (assigns a value of 10 to the 2nd value in arr, which replaces 2
Answer:
4. Standards are what guarantee that the different pieces of network are configured to communicate with one another
Explanation:
Networking standards ensure the interoperability of networking technologies by defining the rules of communication among networked devices. Networking standards exist to help ensure products of different vendors are able to work together in a network without risk of incompatibility
Hope it will help you...
Answer:
Hey mate.....
Explanation:
This is ur answer.....
<em>To remove the last n elements from an array, use arr. splice(-n) (note the "p" in "splice"). The return value will be a new array containing the removed elements.</em>
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Hope it helps!
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