Answer:
If you want to learn coding than get a game and look at the code inside.
Explanation:
I sugest terraria
Answer:
The output is "A"
Explanation:
public class Solution {
public static void main(String args[]) {
mystery(7);
}
public static void mystery(int a) { System.out.println("A"); }
public static void mystery(double a) { System.out.println("B"); }
public static void mystery(int a, double b) { System.out.println("C"); }
public static void mystery(double a, int b) { System.out.println("D"); }
}
In the code above; mystery is defined in four different ways called method overloading. Method overloading is when same method is defined with different parameters.
In the first case; mystery will be called if the argument is int.
In the second case; mystery will be called if the argument is double.
In the third case; mystery will be called if the arguments are int and double.
In the fourth case; mystery will be called if the arguments are double and int.
When mystery(7) is called; the mystery method requiring only int will be called and the output is "A".
Answer:
The answer to this question is given below in the explanation section.
Explanation:
This program is written in C++.
#include <iostream>
using namespace std;
int main()
{
string word;// variable for taking user input
int cond;// to set condition true if user preses the stop and exit from loop
cout<<"Enter word: \n";
cin>>word;
do// start while loop
{
if(word=="stop" || word =="STOP" || word == "Stop")// if user enter word stop, then program exit
{
cond=0;
}
else//otherwise loop continue
{
cout<<" You Entered "+ word +"\n";
cout<<"Enter word: \n";
cin>>word;
cond=1;
}
}
while(cond == 1);// if user don't enter word "stop" loop run continuesly.
cout<<"Program exit";
return 0;
}
Answer:
I have no clue. I wish I could help though.
Explanation:
Sorry
