Answer:
Explanation:
a. In this scenario, the best solution would have an Object of Traditional Books, CD, Music, Bookstore and Customer.
b. All five objects would be able to be called by the main program loop and the Customer Object would call upon and use either the Books or CD object, While the Bookstore object would call upon all of the other objects.
c. Both the Bookstore object and Customer object will "have" other objects as the Bookstore needs to hold information on every Book or CD in the Inventory. While the Customer object would call upon the Book and CD object that they are purchasing.
d. The Music Object will extend the CD object and use information on the CD object as its parent class.
e. Since the Music Object extends the CD object it is also considered a CD since it is in CD format like the Books on CD and therefore is both objects.
When a user is created in Linux using the useradd command, the <em>immediate state</em> of the created account is locked.
- The useradd command can be invoked in other to create a new user account using the following command
sudo useradd fast_user
<em>#fast_user</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>name</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>account</em><em> </em><em>created</em><em> </em>
- At this point, the account has been created but it will be locked until the required login details are set. This can be achieved using the line ;
sudo passwd fast_user
Therefore, an account created using the useradd command is in a locked state.
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Answer:
Following are the program in the C++ Programming Language.
//set header file
#include <iostream>
//set namespace
using namespace std;
//define class
class format
{
//set access modifier
public:
//set string type variable
string res;
//define function
void names(string first_name, string last_name)
{
//set if-else if condition to check following conditions
if(first_name.length()>0 && last_name.length()>0)
{
res="Name: "+last_name+", "+first_name;
}
else if(first_name.length()>0 and last_name.length()==0)
{
res="Name: "+first_name;
}
else if(first_name.length()==0 and last_name.length()==0)
{
res="";
}
}
//define function to print result
void out(){
cout<<res<<endl;
}
};
//define main method
int main() {
//set objects of the class
format ob,ob1,ob2;
//call functions through 1st object
ob.names("John","Morris");
ob.out();
//call functions through 2nd object
ob1.names("Jhon","");
ob1.out();
//call functions through 3rd object
ob2.names("", "");
ob2.out();
}
<u>Output</u>:
Name: Morris, John
Name: Jhon
Explanation:
<u>Following are the description of the program</u>:
- Define class "format" and inside the class we define two void data type function.
- Define void data type function "names()" and pass two string data type arguments in its parameter "first_name" and "last_name" then, set the if-else conditional statement to check that if the variable 'first_name' is greater than 0 and 'last_name' is also greater than 0 then, the string "Name" and the following variables added to the variable "res". Then, set else if to check that if the variable 'first_name' is greater than 0 and 'last_name' is equal to 0 then, the string "Name" and the following variable "first_name" added to the variable "res".
- Define void data type function "out()" to print the results of the variable "res".
- Finally, we define main method to pass values and call that functions.
Answer: i’m thinking it’s reviewer
Explanation:
