View bytes at specific place in memory in Java

Question

In C I might say

printf("%p",0x11194);
*((int *) 0x11194) = 30240;

(In this example I know that there is an integer at that memory location)

Please tell me, how would I do the same in Java? Btw I may have completely misunderstood how memory works

Learn how to write printf in C: it should be `printf("%p",...` — Grijesh Chauhan, Aug 16 '13 at 15:41
You can't. direct memory access like that does not exist in Java. — Marc B, Aug 16 '13 at 15:41
Thanks for your prompt comments, I shall use C instead @grijesh If I did not need it, I would not have asked — sun, Aug 16 '13 at 15:50
You shouldn't ever need to use it, but the [`sun.misc.Unsafe`](http://www.docjar.com/docs/api/sun/misc/Unsafe.html) class provides direct memory access. — Sam Marsh, Aug 16 '13 at 15:52
You may well want to use Java and sun.misc.Unsafe and/or ByteBuffer for direct memory access. See my answer for more details etc. — Brian Agnew, Aug 16 '13 at 15:57

score 2 · Accepted Answer · edited May 23 '17 at 11:56

2

You can do this in Java using sun.misc.Unsafe and ByteBuffer.

However for normal Java usage you don't do this at all. Java will give you an object reference, which is handle to an object. It doesn't point to one place in memory (the JVM can move objects in memory without you knowing/caring) and you can't perform arithmetic on them (unlike pointers)

edited May 23 '17 at 11:56

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answered Aug 16 '13 at 15:56

Brian Agnew

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score 1 · Answer 2 · answered Aug 16 '13 at 15:44

You have not misunderstood how memory works; you have merely assumed that Java will let you touch it. It won't. The memory and pointers are still there, but must be accessed through proper Object creation and dereferentiation, as opposed to direct pointer memory access.

On the bright side, you don't really have to worry about freeing memory that you no longer need though.

View bytes at specific place in memory in Java

2 Answers2