Is the creation of Java class files deterministic?

Question

When using the same JDK (i.e. the same javac executable), are the generated class files always identical? Can there be a difference depending on the operating system or hardware? Except of the JDK version, could there be any other factors resulting in differences? Are there any compiler options to avoid differences? Is a difference only possibly in theory or does Oracle's javac actually produce different class files for the same input and compiler options?

Update 1 I'm interested in the generation, i.e. compiler output, not whether a class file can be run on various platforms.

Update 2 By 'Same JDK', I also mean the same javac executable.

Update 3 Distinction between theoretical difference and practical difference in Oracle's compilers.

[EDIT, adding paraphrased question]
"What are the circumstances where the same javac executable,when run on a different platform, will produce different bytecode?"

Answer 1

Let's put it this way:

I can easily produce an entirely conforming Java compiler that never produces the same .class file twice, given the same .java file.

I could do this by tweaking all kinds of bytecode construction or by simply adding superfluous attributes to my method (which is allowed).

Given that the specification does not require the compiler to produce byte-for-byte identical class files, I'd avoid depending such a result.

However, the few times that I've checked, compiling the same source file with the same compiler with the same switches (and the same libraries!) did result in the same .class files.

Update: I've recently stumbled over this interesting blog post about the implementation of switch on String in Java 7. In this blog post, there are some relevant parts, that I'll quote here (emphasis mine):

In order to make the compiler's output predictable and repeatable, the maps and sets used in these data structures are LinkedHashMaps and LinkedHashSets rather than just HashMaps and HashSets. In terms of functional correctness of code generated during a given compile, using HashMap and HashSet would be fine; the iteration order does not matter. However, we find it beneficial to have javac's output not vary based on implementation details of system classes .

This pretty clearly illustrates the issue: The compiler is not required to act in a deterministic manner, as long as it matches the spec. The compiler developers, however, realize that it's generally a good idea to try (provided it's not too expensive, probably).

Answer 2

There is no obligation for the compilers to produce the same bytecode on each platform. You should consult the different vendors' javac utility to have a specific answer.

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I will show a practical example for this with file ordering.

Let's say that we have 2 jar files: my1.jar and My2.jar. They're put in the lib directory, side-by-side. The compiler reads them in alphabetical order (since this is lib), but the order is my1.jar, My2.jar when the file system is case insensitive , and My2.jar, my1.jar if it is case sensitive.

The my1.jar has a class A.class with a method

public class A {      public static void a(String s) {} } 

The My2.jar has the same A.class, but with different method signature (accepts Object):

public class A {      public static void a(Object o) {} } 

It is clear that if you have a call

String s = "x";  A.a(s);  

it will compile a method call with different signature in different cases. So, depending on your filesystem case sensitiveness, you will get different class as a result.