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Core Java - Interview Questions and Answers for 'Tricky questions' - 4 question(s) found - Order By Newest

 Q1. How to find whether a given integer is odd or even without use of modulus operator in java?Core Java
Ans. public static void main(String ar[])
{
int n=5;
if((n/2)*2==n)
{
System.out.println("Even Number ");
}
else
{
System.out.println("Odd Number ");
}
}

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 Q2. What will be the output of following Code ?

class BuggyBread {
   public static void main(String[] args) {
      String s2 = "I am unique!";
      String s5 = "I am unique!";

      System.out.println(s2 == s5);
   }
}
Core Java
Ans. true, due to String Pool, both will point to a same String object.

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 Q3. What will be the output of following code ?
class BuggyBread2 {       
   private static int counter = 0;     
   void BuggyBread2() {        
      counter = 5;    
   }     
   
   BuggyBread2(int x){
      counter = x;    
   }        

   public static void main(String[] args) {        
      BuggyBread2 bg = new BuggyBread2();        
      System.out.println(counter);    
   } 
}
Core Java
Ans.  Compile time error as it won't find the constructor matching BuggyBread2(). 
Compiler won't provide default no argument constructor as programmer has already defined one constructor. 
Compiler will treat user defined BuggyBread2() as a method, as return type ( void ) has been specified for that. 

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Try 3 Question(s) Test


 Q4. If you have access to a function that returns a random integer from one to five, write another function which returns a random integer from one to seven.Core Java
Ans. We can do that by pulling binary representation using 3 bits ( random(2) ).

getRandom7() {
   String binaryStr = String.valuesOf(random(2))+String.valuesOf(random(2))+String.valuesOf(random(2));
   binaryInt = Integer.valueOf(binaryStr);
   int sumValue=0;
   int multiple = 1;
   while(binaryInt > 0){
      binaryDigit = binaryInt%10;
      binaryInt = binaryInt /10;
      sumValue = sumValue + (binaryDigit * multiple);
      multiple = multiple * 2;
   }
}

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