Java equivalents of malloc(), new, free() and delete (ctd)

The C malloc() function is used to allocate an area of memory from a heap of memory available to the program. Does Java have or need an equivalent of malloc and free for allocating and releasing arbitrary blocks of memory?

The Java equivalent of malloc()?

Java memory management works differently to a language such as C. In C, the programmer has much more responsibility for allocating and de-allocating memory when required. In Java, we allocate objects using new and have them automatically released by the garbage collector when they are no longer referenced. So does that mean that we need a Java equivalent to malloc at all? In C/C++, malloc() gives us a pointer to an arbitrary block of memory: whether that memory is then used to store a struct, buffer the contents of a file etc is up to the programmer. So in a sense, asking about the Java equivalent is a bit like asking what is the Java equivalent of pointers: there's strictly no such thing, but there are functional equivalents when we need to achieve the same purpose such as buffering the contents of a file.

Java ewill never give us access to "raw" memory in the sense of a memory address or pointer. But Java does provide the following rough equivalents to an area of memory allocated via malloc():

So, for example, the equivalent of the following C code:

unsigned char *memarea = (char *) malloc(200);
*(memarea + 10) = 200;
*((int *) (memarea + 16) = 123456;
ch = memarea[4];

would be something along the following lines in Java, using a ByteBuffer object:

ByteBuffer bb = ByteBuffer.allocate(200);
bb.put(0, (byte) 200);
bb.putInt(16, 123456);
int ch = bb.get(4) & 0xff;

Notice a subtlety of buffer access in Java is that we must deal with sign conversions ourselves. To read an unsigned byte from the ByteBuffer, we must store it in an int (or some primitive big enough to handle values between 128 and 255— basically, anything but a byte!), and must explicitly "AND off" the bits holding the sign.


Note that accesses to ByteBuffers usually compile directly to MOV instructions at the machine code level. In other words, accessing a ByteBuffer really is usually as efficient as accessing a malloced area of memory from C/C++1.

Direct byte buffers

In the above example, the byte buffer would actually be "backed" by an ordinary Java byte array. It is even possible to call bb.array() to retrieve the backing array. An alternative that is perhaps closer to malloc() is to create what is called a direct ByteBuffer in Java: a ByteBuffer that is backed by a "pure block of memory" rather than a Java byte array2. To do so requires us simply to change the initial call:

ByteBuffer bb = ByteBuffer.allocateDirect(200);

Note that from Java, we still don't see or have control over the actual memory address of the buffer. Direct byte buffers have the advantage of accessibility via native code: although the address is of the buffer isn't available or controllable in Java, it is possible for your Java program to interface with native code (via the Java Native Interface). From the native side, you can:

This very last point is quite significant, as it effectively allows things like device drivers to be written in Java (OK, in "Java plus a couple of lines of C" at any rate...).

Note, however, that you generally shouldn't use direct ByteBuffers "by default":

What to read next

You may be interested in the following topics relating to memory usage in Java:

1. The same is often true of accessing object fields: the JIT compiler can compile accesses to object fields into MOV instructions that "know" the offset of the field in question.
2. You can actually create buffers backed by other types of primitive array, such as an IntBuffer backed by an int array.

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Editorial page content written by Neil Coffey. Copyright © Javamex UK 2021. All rights reserved.