Java heap only stores objects, and stack only stores primitive data and object reference.
Consider A.a = B.b , where A.a and B.b are int.
In my understanding, a JVM will first GET the value of A.a from the heap to the stack, and then PUT the value to B.b which is on the heap. It seems that the only way to change data on heap is to PUT the value from the stack.
My question is: is there some way to operate data on Java heap without stack? E.g., copy the value of A.a direct to B.b without stack operation.
If you say "it depends on the implementation of JVM", then my question is about Dalvik.
As far as the abstract machine called JVM (not to be confused with the various pieces of software which go by the same name and implement that abstract machine by mapping it onto real hardware) is concerned, A.a = B.b does indeed load the value of B.b on the stack, then stores it to A.a.1
However, as the name abstract machine may tell you, this is only a way of thinking about semantics. An implementation may do whatever it pleases, as long as it preserves the effect of the program. As you should know, most implementations don't actually interpret JVM instructions most of the time, but instead compile it to machine code for the CPU they run on. If you're concerned about performance or memory traffic, you need to go deeper.
When compiling, the JVM's stack is mostly discarded in favor of the registers most physical CPUs use. If there aren't enough registers available, the hardware stack (which is distinct from the JVM stack!) may also be used. But I digress. On most architectures, there's no instruction for moving from one memory location to the other (see Assembly: MOVing between two memory addresses). However, the hardware stack is also memory, so it's actually impossible to go heap -> stack -> heap. Instead, you'll find that the code loads the value from memory into a register, then stores it to memory from a register.
Finally, if the objects A and B are short-lived and aren't aliased, they may even be elided with their fields ending up on the stack or in registers. Then, this operation becomes even simpler (or may even be removed entirely if it has no effect).
1 These two steps actually take several JVM instructions each, but that's not important here.
When considering JIT, things get complicated. I think my question is actually about Java compiler, not JVM
If you are thinking of the javac you should assume that it does almost no optimisation and gives in byte code almost a literal translation of the code, which is very stack based.
In fact the byte code will be using the stack more than your example suggests. It does operations like
push B
getAndPushField b
push A
popAndSetField a
i.e. instead of one stack operation, there is notionally 3.
The JIT on the other hand might optimise this away so it is not even using a register for the value. Depending on the processor
// R3 contains A, R7 contains B,
// a starts at the 14th byte
// b start at the 16th byte
MOVI [R3+14], [R7+16]
It doesn't depend on the implementation of the JVM. It depends on the Java Virtual Machine Specification, which doesn't provide any other way than via the stack.
Related
Would the data in the following statement be stored as automatic memory allocation, or dynamic memory allocation or both
myFunction(new MyClass());
Thank You!
The terms “automatic memory allocation” and “dynamic memory allocation” make no sense in the context of Java. In Java, all memory is managed by the execution environment.
In other programming languages, the terms “automatic storage” and “dynamic storage” are used to distinguish between storage, which is automatically deallocated when going out of scope, and storage, which requires an explicit deallocation action performed by the application. In Java, there are no explicit deallocations at all. You will find that people and literature continue to distinguish between stack and heap, where only the latter contains objects, whose lifetime may exceed the execution of the method in which they were created. This, however, is only a logical separation, which might not reflect how a particular JVM implementation actually works.
The Java® Language Specification doesn’t mandate much details about the workings of this. There are only two spots at all:
15.12.4.5. Create Frame, Synchronize, Transfer Control
A method m in some class S has been identified as the one to be invoked.
Now a new activation frame is created, containing the target reference (if any) and the argument values (if any), as well as enough space for the local variables and stack for the method to be invoked and any other bookkeeping information that may be required by the implementation (stack pointer, program counter, reference to previous activation frame, and the like). If there is not sufficient memory available to create such an activation frame, a StackOverflowError is thrown.
17.4.1. Shared Variables
Memory that can be shared between threads is called shared memory or heap memory.
All instance fields, static fields, and array elements are stored in heap memory. In this chapter, we use the term variable to refer to both fields and array elements.
Local variables (§14.4), formal method parameters (§8.4.1), and exception handler parameters (§14.20) are never shared between threads and are unaffected by the memory model.
Note that this is the only place where the term “heap” is used as memory type and how it’s actually defined here…
The sections §15.9.4. Run-Time Evaluation of Class Instance Creation Expressions and §15.10.2. Run-Time Evaluation of Array Creation Expressions are much more vague, saying that “space is allocated” and an OutOfMemoryError is thrown if not enough space is available, and nothing more.
So if you go the route of distinguishing between stack and heap, you can say that your code myFunction(new MyClass()); will cause a heap allocation for an instance of MyClass, followed by a stack allocation for the activation frame of the actual implementation of the myFunction method. Not that it matters for any practical purpose.
If you want to dig more into the way, JVMs may implement it, you may refer to the Java® Virtual Machine Specification instead:
2.5.2. Java Virtual Machine Stacks
Each Java Virtual Machine thread has a private Java Virtual Machine stack, created at the same time as the thread. A Java Virtual Machine stack stores frames (§2.6). A Java Virtual Machine stack is analogous to the stack of a conventional language such as C: it holds local variables and partial results, and plays a part in method invocation and return. Because the Java Virtual Machine stack is never manipulated directly except to push and pop frames, frames may be heap allocated. The memory for a Java Virtual Machine stack does not need to be contiguous.
2.5.3. Heap
The Java Virtual Machine has a heap that is shared among all Java Virtual Machine threads. The heap is the run-time data area from which memory for all class instances and arrays is allocated.
The heap is created on virtual machine start-up. Heap storage for objects is reclaimed by an automatic storage management system (known as a garbage collector); objects are never explicitly deallocated. The Java Virtual Machine assumes no particular type of automatic storage management system, and the storage management technique may be chosen according to the implementor's system requirements. The heap may be of a fixed size or may be expanded as required by the computation and may be contracted if a larger heap becomes unnecessary. The memory for the heap does not need to be contiguous
Note, how these definitions only differ in their purpose while not having significant difference in their constraints, i.e. might be fixed-size or resizable and might be contiguous or not, not to speak of the explicit mentioning of the possibility to allocate stack frames on the heap.
There are three groups of memory in Java
Heap - this is where Objects are created and located when you call new MyClass();
Stack - The stack contains stack frames, and stack frames have space allocated for primitives and pointers to Objects in the Heap. Stack frames are allocated on method call and deallocated on method return
String constants: String literals defined at compile time will be added to a String constant pool. At runtime, it is possible to add strings to the pool using String.intern()
That's it
There are two things being allocated in your example;
The expression, new MyClass() allocates an instance of MyClass on the object heap.
The result of the new expression is an object reference. The object reference is saved in the activation record for a call to myFunction(). Activation records are allocated on the calling thread's call stack.
When you create a new object, it often allocated on the heap however with escape analysis, it can be unpacked onto the stack as if it was a local variable.
The only allocation is via new or a capturing lambda (or some native method in rare cases) There isn't any explicit distinction between different ways of allocating an object.
It's well known that GCs will sometimes move objects around in memory. And it's to my understanding that as long as all references are updated when the object is moved (before any user code is called), this should be perfectly safe.
However, I saw someone mention that reference comparison could be unsafe due to the object being moved by the GC in the middle of a reference comparison such that the comparison could fail even when both references should be referring to the same object?
ie, is there any situation under which the following code would not print "true"?
Foo foo = new Foo();
Foo bar = foo;
if(foo == bar) {
System.out.println("true");
}
I tried googling this and the lack of reliable results leads me to believe that the person who stated this was wrong, but I did find an assortment of forum posts (like this one) that seemed to indicate that he was correct. But that thread also has people saying that it shouldn't be the case.
Java Bytecode instructions are always atomic in relation to the GC (i.e. no cycle can happen while a single instruction is being executed).
The only time the GC will run is between two Bytecode instructions.
Looking at the bytecode that javac generates for the if instruction in your code we can simply check to see if a GC would have any effect:
// a GC here wouldn't change anything
ALOAD 1
// a GC cycle here would update all references accordingly, even the one on the stack
ALOAD 2
// same here. A GC cycle will update all references to the object on the stack
IF_ACMPNE L3
// this is the comparison of the two references. no cycle can happen while this comparison
// "is running" so there won't be any problems with this either
Aditionally, even if the GC were able to run during the execution of a bytecode instruction, the references of the object would not change. It's still the same object before and after the cycle.
So, in short the answer to your question is no, it will always output true.
Source:
https://docs.oracle.com/javase/specs/jls/se8/html/jls-15.html#jls-15.21.3
The short answer is, looking at the java 8 specification: No.
The == operator will always perform object equality check (given that neither reference is null). Even if the object is moved, the object is still the same object.
If you see such an effect, you have just found a JVM bug. Go submit it.
It could, of course, be that some obscure implementation of the JVM does not enforce this for whatever strange performance reason. If that is the case, it would be wise to simply move on from that JVM...
TL;DR
You should not think about that kind of stuff what so ever, It's a dark place.
Java has clearly stated out it's specifications and you should not doubt it, ever.
2.7. Representation of Objects
The Java Virtual Machine does not mandate any particular internal structure for objects.
Source: JVMS SE8.
I doubt it! If you may doubt this very basic operator you may find yourself doubt everything else, getting frustrated and paranoid with trust issues is not the place you want to be.
What if it happens to me? Such a bug should not be existed. The Oracle discussion you supplied reporting a bug that happened years ago and somehow discussion OP decided to pop that up for no reason, either without reliable documentation of such bug existed now days. However, if such bug or any others has occurred to you, please submit it here.
To let your worries go away, Java has adjusted the pointer to pointer approach into the JVM pointer table, you can read more about it's efficenty here.
GCs only happen at points in the program where the state is well-defined and the JVM has exact knowledge where everything is in registers/the stack/on the heap so all references can be fixed up when an object gets moved.
I.e. they cannot occur between execution of arbitrary assembly instructions. Conceptually you can think of them occuring between bytecode instructions of the JVM with the GC adjusting all references that have been generated by previous instructions.
You are asking a question with a wrong premise. Since the == operator does not compare memory locations, it isn’t sensible to changes of memory location per se. The == operator, applied to references, compares the identity of the referred objects, regardless of how the JVM implements it.
To name an example that counteracts the usual understanding, a distributed JVM may have objects held in the RAM of different computers, including the possibility of local copies. So simply comparing addresses won’t work. Of course, it’s up to the JVM implementation to ensure that the semantics, as defined in the Java Language Specification, do not change.
If a particular JVM implementation implements a reference comparison by directly comparing memory locations of objects and has a garbage collector that can change memory locations, of course, it’s up to the JVM to ensure that these two features can’t interfere with each other in an incompatible way.
If you are curious on how this can work, e.g. inside optimized, JIT compiled code, the granularity isn’t as fine as you might think. Every sequential code, including forward branches, can be considered to run fast enough to allow to delay garbage collection to its completion. So garbage collection can’t happen at any time inside optimized code, but must be allowed at certain points, e.g.
backward branches (note that due to loop unrolling, not every loop iteration implies a backward branch)
memory allocations
thread synchronization actions
invoking a method that hasn’t been inlined/analyzed
maybe something special, I forgot
So the JVM emits code containing certain “safe points” at which it is known, which references are currently held, how to replace them, if necessary and, of course, changing locations has no impact on the correctness. Between these points, the code can run without having to care about the possibility of changing memory locations whereas the garbage collector will wait for code reaching a safe point when necessary, which is guaranteed to happen in finite, rather short time.
But, as said, these are implementation details. On the formal level, things like changing memory locations do not exist, so there is no need to explicitly specify that they are not allowed to change the semantics of Java code. No implementation detail is allowed to do that.
I understand you are asking this question after someone says it behaves that way, but really asking if it does behave that way isn't the right approach to evaluating what they said.
What you should really be asking (primarily yourself, others only if you can't decide on an answer) is whether it makes sense for the GC to be allowed to cause a comparison to fail that logically should succeed (basically any comparison that doesn't include a weak reference).
The answer to that is obviously "no", as it would break pretty much anything beyond "hello, world" and probably even that.
So, if allowed, it is a bug -- either in the spec or the implementation. Now since both the spec and the implementation were written by humans, it is possible such a bug exists. If so, it will be reported and almost certainly fixed.
No, because that would be flagrantly ridiculous and a patent bug.
The GC takes a great deal of care behind the scenes to avoid catastrophically breaking everything. In particular, it will only move objects when threads are paused at safepoints, which are specific places in the running code generated by the JVM for threads to be paused at. A thread at a safepoint is in a known state, where the positions of all the possible object references in registers and memory are known, so the GC can update them to point to the object's new address. Garbage collection won't break your comparison operations.
Java object hold a reference to the "object" not to the memory space where the object is stored.
Java do this because it allow the JVM to manage memory usage by its own (e.g. Garbage collector) and to improve global usage without impacting the client program directly.
As instance for improvement, the first X int (I don't remember how much) are always allocated in memory to execute for loop fatser (ex: for (int i =0; i<10; i++))
And as example for object reference, just try to create an and try to print it
int[] i = {1,2,3};
System.out.println(i);
You will see that Java returning a something starting with [I#. It is saying that is point on a "array of int at" and then the reference to the object. Not the memory zone!
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I am taking some classes and object lessons in Java having C++ background. I want to know the reason why we cannot choose the objects to be declared on the stack memory? Why must everything go on the heap except for the primitive types?
Here's something to clarify what I was asking.
Essentially, if we have:
Scanner input = new Scanner(System.in);
Then why cannot we have it on stack in the first place?
One of the strongest attractors of the original Java design (in mid-1990s) was simplicity. Supporting heap-based objects is essential, whereas stack-based ones are an optimization. Java is not alone here: many languages take that approach (LISP, Haskell, JavaScript, Ruby, etc.). Stack-based allocation does happen in Java, but only as an internal optimization trick and not something that the user can control.
Especially keep in mind that there is an essential difference in how a pointer to an object passed to a function ("a reference passed to a method" in Java-speak) can be treated by the callee: it is not allowed to retain the pointer if it's stack-based. This alone creates huge complications and bug opportunities.
Finally, stack-based objects bring much less to a garbage-collected language than to manually-managed languages like C and C++.
The data on the stack, say a C struct, disappears after the function call has returned. Hence one would need copying and correction of pointers.
Think of the hidden extra functionality needed here:
struct S* f() {
struct S s = ...;
g(&s);
return &s;
}
Java was meant as simplification, having its own management of memory, and doing things immediately on the heap seemed more direct, less convoluted.
This in view of C++, with its copy constructors, pointers and aliases.
Java does not allow explicit on stack allocation of objects. The language is not competing with low level languages such as C, and the creators of the language made this choice as a simplification.
However times change, and Java has grown since its humble beginnings. As the JVM becomes more sophisticated, automatic allocation of objects to the stack has become possible. The rationale for this is similar to the 'register' keyword in C; let the compiler manage the low level detail. It has become better at doing it than humans. In Java automatic allocation of objects onto the stack has been hampered by two factors, firstly the Sun/Oracle JVM is very old and very complex now. It is difficult to change, and Oracle has been careful about preventing backwards breaks. Secondly, so far their work on stack allocations has not yielded the large benefits that were expected. It did improve some situations, but the JVM has its own trade offs and behaviours. So this comes down to a question of time/pay-off and priorities. I believe that work to improve the benefits of automatic allocation continues behind the scenes; but there are no plans to make it explicit.
To put it simple, the key advantage of objects on stack is that the memory is automatically managed for you. When function puts objects on stack, they are cleaned from memory on function exit.
Since java already has automated garbage collection, this key advantage doesn't bring that much.
Sure there is a speed of access performance price that you might pay by being unable to allocate objects on stack directly, but as Marko mentioned, there are internal optimizations that might do just that.
Why must everything go on the heap except for the primitive types?
This statement is not accurate. Primitive types can go on the heap as well if they are part of a class instance. A local variable is stored on the stack, where as class variables are on the heap.
As for why objects are stored on the heap. It's after all a design decision. One reason is that it is a managed area in the JVM that is subject to garbage collection. As a managed area in the JVM, it may be organized in generations and may grow or shrink in size. See this section from the JVM specification:
The Java Virtual Machine has a heap that is shared among all Java Virtual Machine threads. The heap is the run-time data area from which memory for all class instances and arrays is allocated.
The heap is created on virtual machine start-up. Heap storage for objects is reclaimed by an automatic storage management system (known as a garbage collector); objects are never explicitly deallocated.
I am trying to fully understand how the Java works with its memory arrangement. The discussions on Internet are very confused, and sometimes contradicts each other, so I found no one I can trust. This thing can be very complicate if it mixed with static, static method, local variable, thread, volatile and so on. So I am thinking if there is a way I can study that myself my doing some Java coding experiment. A class MemoryTrack does something like this,
public myMethod(){
int i = 0;
MemoryTrack.show(new myClass()); //print out "Heap at address 111"
MemoryTrack.show(new myClass()); //print out "Heap at address 222"
MemoryTrack.show(i); //print out "Stack at address 333"
MemoryTrack.show("a static method"); //print out "stack at address 444"
}
The use of memory is described in Section 2.5 of the Java Virtual Machine Specification. The stack stores stack frames (containing local variables and partial results). The heap is where all class instances and arrays come from. Stack frames can also be allocated from the heap (and then pushed onto the stack). There are also method areas and runtime constant pool memory. The details are spelled out in the spec.
As far as monitoring memory usage, several profilers have tools for that. For monitoring from within the program, take a look at the MemoryMXBean class (and related classes in the java.lang.management package). It's very easy to use. While it probably doesn't give you everything that it sounds like you want, but it's probably the best thing available.
The rule is pretty simple: heap contains objects, stack contains local variables and method parameters. Object fields are inside the objects, in the heap. Don't know about static fields. Methods and constructors are not stored in the stack, nor in the heap. Threads and volatile don't matter.
Method calls are on the stack. Each has a space reserved to it, the stack frame, that contains the local variables and the parameters.
Two things you need to know
its not as complicated as you think.
it doesn't matter 99% of the time.
Variables are always on the stack.
Objects are always on the heap. (There are exceptions but I wouldn't worry about them)
Methods and classes are always in the perm gen.
In the HotSpot/OpenJDK JVM, static fields are collected in a singleton object for the class. You can see the instance if you do a heap dump. Other JVMs may do this differently.
A class MemoryTrack does something like this,
Such a method wouldn't do anything useful as the argument will always be on the stack and object it refers to will always be on the heap. You can't get the memory location of an Object in a standard way and it unlikely to be useful if it did as it can change at any time.
You may think all variables in Java are on heap. Actually it is implementation dependent.
Why do we use Heap Memory, Can we use Stack memory for the same?
EDITED
One more question came in my mind after reading answers
1) is there any other kind of memory which we can think of alternative to Heap and Stack?
Edited
I came across the string pool, is that memory associated with the heap or Stack?
Well, you have to use the heap if you want to use objects. Objects are inherently pointers on the stack (or inside other objects) to memory chunks in the heap.
Unlike C++ where you can put objects on the stack or heap, Java does things differently.
Even in C++, it's a good idea to use the heap for objects that must outlive the function they were created in. You probably can avoid it but you may find yourself with a performance problem with all the copy constructors.
As to your edit:
Is there any other kind of memory which we can think of alternative to Heap and Stack?
Sure there is: Static data members, the ones where there's only one per class rather than one per instantiated object must go somewhere. These can't be on the stack since they may disappear when you exit from a function. And they can't belong to an particular object.
These (at least in the Sun/Oracle JVM) go into the Method area.
In addition, you should not think of there being one stack. Every thread gets its own stack on creation.
There's also the runtime constant pool and stacks for native (non-Java) calls.
There's lots of information on the JVM internals here regarding that aspect but keep in mind there may be a distinction between the logical machine and the implementation.
Heap memory is central to Java. All objects are allocated on the heap, and local variables hold only references (essentially pointers) to them. C# allows you to have objects (value types) that live on the stack. Java chose not to borrow this from C++, partly in order to simplify the language.
Across languages, heap memory is the way to provide long-lived objects of arbitrary size.
Can we use Stack memory for the same?
Basically no.
In Java, stack memory is used solely for parameters and local variables of methods (and some hidden book-keeping info which is not relevant to this discussion). And for those variables, only primitive types use just stack memory. Any object or array is represented as a reference to something on the heap, and Java provides no way to allocate an object or array on the stack.
[Aside: it is theoretically possible to perform any calculation in Java using just primitive types and recursive static methods. However, it would be horribly messy. Even writing a simple message to the console will entail your application causing objects to be allocated on the heap. So we can dismiss this as being totally impractical.]
In the JVM, instead of a thread local stack for objects it uses a Thread Local Allocation Buffer or (TLAB) This gives much of the performance benifits of a Stack, without the developer needing to worry about wether an object is on the stack or not.
Heap memory is used to store objects in Java. No matter, where object is created in code.
e.g. as member variable, local variable or class variable, they are always created inside heap space in Java.
If there is no memory left in stack for storing function call or local variable, JVM will throw java.lang.StackOverFlowError,
While if there is no more heap space for creating object, JVM will throw
java.lang.OutOfMemoryError Java Heap Space.
Another important factor to consider is scoping. If all objects were to be allocated on stack then they go out of context as stack frame gets rolled out. In layman terms, think of stack as a dump that stores all the variable values and object references that are local to a subroutine/method which is currently in scope. As soon as it finishes executing (or the method goes out of scope), then all in it would be lost.
It makes it easier for compiler to manage large &/or dynamic sized variables too- they still take small constant storage on call stack- that of 4 bytes ( a heap pointer).