Why Java doesnot have Pointer ?
Asked by Satyajit Sethy Last Modified
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The reasons are Pointers are fundamentally unsafe. Java has a robust security model and disallows pointer arithmetic.Java ensures pointer access via indexed array. A big advantage of Java's indexed array access is that it detects and disallows out of bounds array access, which can be a major source...
read more The reasons are Pointers are fundamentally unsafe. Java has a robust security model and disallows pointer arithmetic.Java ensures pointer access via indexed array. A big advantage of Java's indexed array access is that it detects and disallows out of bounds array access, which can be a major source of bugs. Java has the concept of References to objects,but it just doesn't call them pointers. Any normal object reference works as one of these. When you do String s="Hello"; you get what is effectively a pointer to a string object. read less
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Jay Shah
Industry expert and professional lecturer/trainer
To avoid memory hacks and leaks it is an intentional decision taken by JAva Pioneers
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Because Java was built with safety in mind and pointers are a point of complexity(and thus failure) in a lot of programs if the developer is not careful. Java abstracts a number of things away from the developer to make things easier and safer for them, ie, garbage collection.
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Vamshi Krishna
Expert Tutor
Java does not have pointers. This was an intentional decision by the creators of Java, because most people would agree that having pointers creates a lot of potential for bugs in the code -- pointers can be quite confusing, especially to new programmers. Because arrays and strings are provided as class...
read more Java does not have pointers. This was an intentional decision by the creators of Java, because most people would agree that having pointers creates a lot of potential for bugs in the code – pointers can be quite confusing, especially to new programmers. Because arrays and strings are provided as class types in Java, there is no need for pointers to those constructs. By not allowing pointers, Java provides effectively provides another level of abstraction to the programmer. read less
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Career Group Institute Chd.
Do remove the complexity in programming
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1. References store an address. That address is the address in memory of the object. So, when a class is declared like so: "PersonClass y = new PersonClass();", the "y" variable actually stores an address in memory. If you were to look at that address in memory you would see the details of...
read more 1. References store an address. That address is the address in memory of the object. So, when a class is declared like so: "PersonClass y = new PersonClass();", the "y" variable actually stores an address in memory. If you were to look at that address in memory you would see the details of the PersonClass object. Pointers in C++, however, point directly to the object. 2. You can not perform arithmetic operations on references. So, adding 1 to a pointer is not possible, but is possible in C++. read less
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Ganesh Br
Professional Java J2EE Training with certification focus or Maths/English Tuition
Java manages memory in a different way compared to C++. It has automatic memory management and garbage collector. So it does not have a pointer again to confuse programmer and rather ease programming.
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Having pointers creates a lot of changes for bugs in the code -- pointers can be quite confusing, especially to new programmers. By not allowing pointers, Java provides abstraction to the programmer. and also pointers in java are acheived thru reference variables in java whose References store an address...
read more Having pointers creates a lot of changes for bugs in the code – pointers can be quite confusing, especially to new programmers. By not allowing pointers, Java provides abstraction to the programmer. and also pointers in java are acheived thru reference variables in java whose References store an address in memory where we can not perform any arithmetic operations on this reference variable(address) Animal a=new Animal(). a->Reference variable read less
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Because Java does not provide direct memory manipulation rights to it's users. With help of pointers you are allowed to access memory directly which is not allowed in Java. That's why Java does not have any pointers. Java provides special classes to interact with system hardware.
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To reference objects in java you use references. In other languages like C/C++ you use pointers. Pointers and references are used to point to objects in a program. In C/C++ you declare a pointer to point to a object or data type. Once you are done and no longer require the datatype or object, you...
read more To reference objects in java you use references. In other languages like C/C++ you use pointers. Pointers and references are used to point to objects in a program. In C/C++ you declare a pointer to point to a object or data type. Once you are done and no longer require the datatype or object, you must release the pointer, otherwise it is pointing to something that no longer exists. This can occur during pointer arithmetic, or just forgetting to deallocate the pointer to null. This causes problems for memory management, and is more dangerous during run-time which can cause the program to crash; sometimes this can go undetected at compile time. Sun microsystems created and enforced references to make "pointing" safer to use in collaboration with the garbage collector memory manager that runs in the Java Virtual Machine. You do not have to explicitly declare or release objects, even if you made the error of forgetting to in your code, it is handled by the garbage collector. A reference can never point to nothing, it is either null or points to an object. The garbage collector, manages the allocation and release of memory and makes sure there is no referencless objects. This makes it safer. read less
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Related Questions
public class X { public static void main(String [] args) { X x = new X(); X x2 = m1(x); /* Line 6 */ X x4 = new X(); x2 = x4; /* Line 8 */ doComplexStuff(); } static X m1(X mx) { mx = new X(); return mx; } } After line 8 runs. how many objects are eligible for garbage collection? 
Only one. After line 8 the original object referred to by x2 is dereferenced and ready for garbage collection. In m1(), mx ignores the object passed as argument and refers to a new object which is returned.
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