The hashing function can take any number of key-value pairs and there is no specific limit to it.
<h3>What is hashing?</h3>
Hashing is a file-based algorithm for producing a fixed-length bit string value. A file is essentially a collection of data blocks. The length of the data is reduced by hashing to a fixed number or key that represents the original string.
When hashing is employed, the hash function may plot all of the keys and values to what the real size of the table is, demonstrating that the hashing function can take any number of key-value pairs with no restriction.
However, if the passwords are hashed in encryption, recovering the passwords is extremely difficult.
Thus, the hashing function can take any number of key-value pairs and there is no specific limit to it.
Learn more about the hashing here:
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It’s either b or c, but my final answer would be C
Answer:
Following are the code to the given question:
#include <iostream>//header file
using namespace std;
class Window //defining a class Window
{
private:
int width, height;//defining integer variable
public:
friend ostream& operator << (ostream& stm, Window& width)//defining a friend function that takes two parameters
{
return stm<<"a ("<<width.width<<" x "<<width.height<<") window"; //use return keyword that return its values
}
Window(int width, int height): width(width), height(height)//defining parameterized constructor that inherit width and height in its parameters
{}
};
int main() //Main method
{
Window w(80,90);//calling class constructor
cout<<w;//print object value
return 0;
}
Output:
a (80 x 90) window
Explanation:
In the above code, a class "Window" is defined that uses a friend function "ostream& operator" is declared that uses the "ostrea&" as a data type to hold two-variable "stm and w" in its parameter, and declared the parameterized constructor to hold value by inheriting width and height in its parameters.
Inside the main method, a class object is created that calls the constructor and uses the print method to print object value.
Answer:
A
Explanation:
The value will be the original stored in the int variables. This is because the method swap(int xp, int yp) accepts xp and yp parameters by value (They are passed by value and not by reference). Consider the implementation below:
public class Agbas {
public static void main(String[] args) {
int xp = 2;
int yp = xp;
swap(xp,yp); //will swap and print the same values
System.out.println(xp+", "+yp); // prints the original in values 2,2
int xp = 2;
int yp = 5;
swap(xp,yp); // will swap and print 5,2
System.out.println(xp+", "+yp); // prints the original in values 2,5
}
public static void swap(int xp, int yp){
int temp = xp;
xp = yp;
yp = temp;
System.out.println(xp+", "+yp);
}
}
