Answer:
- #include <iostream>
- using namespace std;
- class myinteger {
-
- private:
- int value;
-
- public:
- myinteger(int x){
- value = x;
- }
-
- int getValue(){
- return value;
- }
-
- };
- int main()
- {
- myinteger obj(4);
- cout<< obj.getValue();
- return 0;
- }
Explanation:
Firstly, we use class keyword to create a class named myinteger (Line 5). Define a private scope data field named value in integer type (Line 7 - 8).
Next, we proceed to define a constructor (Line 11 - 13) and a getter method for value (Line 15 -17) in public scope. The constructor will accept one input x and set it to data field, x. The getter method will return the data field x whenever it is called.
We test our class by creating an object from the class (Line 23) by passing a number of 4 as argument. And when we use the object to call the getValue method, 4 will be printed as output.
Answer:
a. gpupdate /force
Explanation:
Based on the information provided within the question it can be said that if the administrator does not want to wait she can use the command gpupdate /force. This command allows the individual to update both the local Group Policy settings and Active Directory-based settings. This the force tag makes it so that the policy is immediately update.
Answer:
The program in C is as follows:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main(){
int dice [1000];
int count [6]={0};
srand(time(0));
for (int i = 0; i < 1000; i++) {
dice[i] = (rand() %(6)) + 1;
count[dice[i]-1]++;
}
for (int i = 0; i < 6; i++) {
printf("%d %s %d %s",(i+1)," occurs ",count[i]," times");
printf("\n");
}
return 0;
}
Explanation:
This declares an array that hold each outcome
int dice [1000];
This declares an array that holds the count of each outcome
int count [6]={0};
This lets the program generate different random numbers
srand(time(0));
This loop is repeated 1000 times
for (int i = 0; i < 1000; i++) {
This generates an outcome between 1 and 6 (inclusive)
dice[i] = (rand() %(6)) + 1;
This counts the occurrence of each outcome
count[dice[i]-1]++; }
The following prints the occurrence of each outcome
for (int i = 0; i < 6; i++) {
printf("%d %s %d %s",(i+1)," occurs ",count[i]," times");
printf("\n"); }
