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Core java - Interview Questions and Answers for 'Abs' - 169 question(s) found - Order By Newest
Very frequently asked. Among first few questions in almost all interviews. Among Top 5 frequently asked questions. Frequently asked in Indian service companies (HCL,TCS,Infosys,Capgemini etc based on multiple feedback )
Ans. "equals" is the method of object class which is supposed to be overridden to check object equality, whereas "==" operator evaluate to see if the object handlers on the left and right are pointing to the same object in memory.
x.equals(y) means the references x and y are holding objects that are equal. x==y means that the references x and y have same object.
Sample code:
String x = new String("str");
String y = new String("str");
System.out.println(x == y); // prints false
System.out.println(x.equals(y)); // prints true
Ans. 1. String Pool - When a string is created and if it exists in the pool, the reference of the existing string will be returned instead of creating a new object. If string is not immutable, changing the string with one reference will lead to the wrong value for the other references.
Example -
String str1 = "String1";
String str2 = "String1"; // It doesn't create a new String and rather reuses the string literal from pool
// Now both str1 and str2 pointing to same string object in pool, changing str1 will change it for str2 too
2. To Cache its Hashcode - If string is not immutable, One can change its hashcode and hence it's not fit to be cached.
3. Security - String is widely used as parameter for many java classes, e.g. network connection, opening files, etc. Making it mutable might possess threats due to interception by the other code segment.
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Ans. If the Object value will not change, use String Class because a String object is immutable.
If the Object value can change and will only be modified from a single thread, use StringBuilder because StringBuilder is unsynchronized(means faster).
If the Object value may change, and can be modified by multiple threads, use a StringBuffer because StringBuffer is thread safe(synchronized).
Ans. String is immutable in java and stored in String pool. Once it's created it stays in the pool until unless garbage collected, so even though we are done with password it's available in memory for longer duration and there is no way to avoid it. It's a security risk because anyone having access to memory dump can find the password as clear text.
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Ans. OOPs or Object Oriented Programming is a Programming model which is organized around Objects instead of processes. Instead of a process calling series of processes, this model stresses on communication between objects. Objects that all self sustained, provide security by encapsulating it's members and providing abstracted interfaces over the functions it performs. OOP's facilitate the following features
1. Inheritance for Code Reuse
2. Abstraction for modularity, maintenance and agility
3. Encapsulation for security and protection
4. Polymorphism for flexibility and interfacing
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Ans. 1.Abstraction solves the problem at design level while encapsulation solves the problem at implementation level
2.Abstraction is used for hiding the unwanted data and giving relevant data. while Encapsulation means hiding the code and data into a single unit to protect the data from outside world.
3. Abstraction lets you focus on what the object does instead of how it does it while Encapsulation means hiding the internal details or mechanics of how an object does something.
4.For example: Outer Look of a Television, like it has a display screen and channel buttons to change channel it explains Abstraction but Inner Implementation detail of a Television how CRT and Display Screen are connect with each other using different circuits , it explains Encapsulation.
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Ans. final - constant variable, objects cannot be de-referenced, restricting method overriding, restricting class sub classing.
finally - handles exception. The finally block is optional and provides a mechanism to clean up regardless of what happens within the try block. Use the finally block to close files or to release other system resources like database connections, statements etc.
finalize() - method helps in garbage collection. A method that is invoked before an object is discarded by the garbage collector, allowing it to clean up its state.
Ans. It means that the type of variables are checked at compile time in Java.The main advantage here is that all kinds of checking can be done by the compiler and hence will reduce bugs.
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Ans. Underlying data structure for ArrayList is Array whereas LinkedList is the linked list and hence have following differences -
1. ArrayList needs continuous memory locations and hence need to be moved to a bigger space if new elements are to be added to a filled array which is not required for LinkedList.
2. Removal and Insertion at specific place in ArrayList requires moving all elements and hence leads to O(n) insertions and removal whereas its constant O(1) for LinkedList.
3. Random access using index in ArrayList is faster than LinkedList which requires traversing the complete list through references.
4. Though Linear Search takes Similar Time for both, Binary Search using LinkedList requires creating new Model called Binary Search Tree which is slower but offers constant time insertion and deletion.
5. For a set of integers you want to sort using quicksort, it's probably faster to use an array; for a set of large structures you want to sort using selection sort, a linked list will be faster.
Ans. Hashcode is used for bucketing in Hash implementations like HashMap, HashTable, HashSet etc. The value received from hashcode() is used as bucket number for storing elements. This bucket number is the address of the element inside the set/map. when you do contains() then it will take the hashcode of the element, then look for the bucket where hashcode points to and if more than 1 element is found in the same bucket (multiple objects can have the same hashcode) then it uses the equals() method to evaluate if object are equal, and then decide if contain() is true or false, or decide if element could be added in the set or not.
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Ans. Car Engine is an example of encapsulation and abstraction. You ignite the car using an interface called starter and least bothered about how the tire actually moves (This is abstraction). The engine encapsulates the complete process to itself only and doesn't allow you to start the other components like the radiator etc ( this is excapsulation )
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Ans. Abstract classes can have both abstract methods ( method declarations ) as well as concrete methods ( inherited to the derived classes ) whereas Interfaces can only have abstract methods ( method declarations ).
A class can extend single abstract class whereas it can implement multiple interfaces.
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Ans. String is widely used as parameter for many java classes, e.g. network connection, opening files, etc. Making it mutable might possess threats due to interception by the other code segment or hacker over internet.
Once a String constant is created in Java , it stays in string constant pool until garbage collected and hence stays there much longer than what's needed. Any unauthorized access to string Pool pose a threat of exposing these values.
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Ans. Volatile is an instruction that the variables can be accessed by multiple threads and hence shouldn't be cached. As volatile variables are never cached and hence their retrieval cannot be optimized.
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Ans. 1. public is the access modifier that makes the method accessible from anywhere, static is the keyword that makes it accessible even without creating any object, void means it doesn't return anything , String args[] is the array of argument that the method receives.
2. If we use main without the string args , it will compile correctly as Java will treat it as just another method. It wont be the method "main" which Java looks for when it looks to execute the class and hence will throw
Error: Main method not found in class , please define the main method as:
public static void main(String[] args)
3. Main is not a keyword but a special string that Java looks for while initiating the main thread.
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Ans. Because it doesn't make the change in the existing string but would create a new string by concatenating the new string to previous string. So Original string won't get changed but a new string will be created. That is why when we say
str1.concat("Hello");
It means nothing because we haven't specified the reference to the new string and we have no way to access the new concatenated string. Accessing str1 with the above code will still give the original string.
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