Does Java have Pointers?

Question

If Java does not have pointers then what does the the new keyword do in Java?

I am confused, please explain.

Answer 1

Java has pointers in the sense of variables that store references to data in memory. All variables of Object types in Java are pointers in this sense.

However, the Java language does not allow arithmetic operations on the values of pointers, like you'd be able to do in a language like C.

Answer 2

new in Java returns a reference to the newly created object.

Answer 3

new returns reference. it has some similarities with pointers (if you pass to function, reference is passed, same as with pointer), but there is no pointer arithmetics.

Answer 4

Java has references. All objects are accessed through having references to their instances. You create a new instance using new, which returns a reference to the object.

Java references are not like pointers in C, you cannot "look under the hood" at the raw memory that makes up the object.

Answer 5

As pointed out, Java has references. How are these different ?

1. you can't perform arithmetic or other such operations on these
2. they do not point to the memory containing the object (i.e. they are not pointers by another name). The JVM is at liberty to move objects around within the VM memory, and most likely will do during garbage collection. The references however still point to that object, despite its movement within memory.

So they're not like C++ references (pointing directly to an object). Perhaps a better name would be handle.

Answer 6

Java Does not have Pointers. The operator "new" is used to the reference variable in java.

Answer 7

In java we come across only THIS pointer, it is used to refer the variables of same class. The operator new is used as reference to an object.

Answer 8

new does (roughly) the following:

1. Find a contiguous free block of heap memory equal to the instance size of the class you're creating, plus some space for bookkeeping
2. Zero said space & remove it from the free list
3. Run the constructor
4. Return a reference (NOT a pointer, as other posts have explained) to the created instance.

Answer 9

Java doesn't have pointers; Java has references.

It's a fine point, but a pointer has extra operations that you may (or may not) typically use; a reference lacks these operations because the operations may be unsafe.

For example, if you use a pointer to index the first element of an array like so:

int primes[] = {2, 3, 5, 7, 11, 13, 17, 19};
int* intPointer = primes;

you may dereference the pointer and get the value "2", but you may also:

intPointer++

and after you do that, when you dereference the pointer you will get the value "3". This is because the ++ operation moves the pointer one "unit" ahead in memory.

The issue comes from the weaknesses in the C / C++ typechecking system (C++ must maintain compatibilty with C, so it allows the same issues). The pointer stores an address in memory and the ++ operation adds the appropriate number of bytes to the address. On many systems ++ing an int adds four bytes, but if the pointer was a char pointer ++ing it should only add one byte. Note that since the underlying data type of a pointer is an address in memory, the following is legal (but not recommended):

char* charPointer = primes;
charPointer++;

void* voidPointer = primes;
voidPointer++;

Since pointers are addresses in memory, they might represent (correctly) any bit of memory in the computer, but they are only properly dereferenced when the underlying data maches the type and alignment of the pointer. For pointers that aren't managed by lots of code to make them safe, this means you might stray off the data type (or alignment) of the desired information and a dereference might end in disaster. Attempting to fix this issue with custom code tends to slow down one pointers badly enough that you notice performance issues, and it opens the doors for adding errors in the custom "pointer management" code.

Java side steps all of these issues by returning a reference. A reference does not refer to any location in memory; Java maintains an internal "reference to pointer" table. This table takes the reference and returns the data associated with it, wherever that data may reside in memory. This slows down code execution, because two lookups are done for each "dereferencing", one lookup in the reference table, one in the machine's memory.

A big advantage of Java using references is that the memory can be moved around without breaking the would-be pointer addresses. In a C program, if you move data into a new memory location, it is very difficult to know whether some other part of the program has a pointer to the data. Should a stale pointer be dereferenced after the memory is moved, the program will be accessing corrupt data, and typically a crash will be shortcoming.

Ability to move the memory around in a running program allows programs to easily recycle memory. Any program which doesn't need chunks of memory can release the unused memory, but this creates memory holes of unused memory in between chunks of used memory. Internally computers use pages of memory, which are quite large. If a sparsely used page of memory could have the few used bits moved into another page, then a page of memory can be freed. This increases the density of data to memory, improving cache performance. Sometimes this translates into performance improvements that can be quite dramatic.

Java's Garbage Collector takes advantage of the use of references by temporarily blocking access to the data for a set of references. During that blockage of access, it moves the data around (to compact it). After the blockage, the reference to address table has the new memory addresses. Since the "functional" layer of the code never knew the addresses in the first place, this operation will not break a running Java program.