Java Community | Help. Code. Learn.•7mo ago

Are all Java programs reproducible?

I am learning about deterministic and reproducible builds and come from a more c/c++ background. Assuming JDK and JVM/JRE are the same, does compiling Java bytecode always yield the same results despite building on different machines? Is Java usually more deterministic compared to c/c++ and other compiled languages?

36 Replies

JavaBot•7mo ago

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dan1st•7mo ago

No it isn't always deterministic. If you use multiple threads, it stops being deterministic. If you use Random or similar, it also stops bring deterministic (because that's literally what you want with randomness). and if you rely on things like the system time or maybe some other OS things or native code, it easily becomes nondeterministic as well But other than that, it's pretty deterministic (way more than many other programming languages like C/C++) in terms of the result you get - and this is even the case when using different JVMs as the specs don't leave much room for implementation-defined or even undefined behaviour ofc this doesn't include how long the program takes to execute or how the JIT optimizes it (the actually executed machine code)

Gareth RaderOP•7mo ago

I see, the multithreading stuff makes sense I was just wondering in terms of the executable that’s built from compiling How often do you need to configure the JVM/JRE? Is it a good practice to have explicitly defined configurations? If that makes sense ^

dan1st•7mo ago

In C/C++, you have many sources of undefined or implementation-defined behavior (the latter means there are multiple allowed options) coming from the language. For example, in C/C++, the && operator doesn't define the order in which the arguments are executed. In Java, the program must behave exactly as if these are executed from left to right (except with multithreading).

Gareth RaderOP•7mo ago

Gotcha

dan1st•7mo ago

Do you mean the bytecode or machine-code

Gareth RaderOP•7mo ago

I am going to guess that I probably meant bytecode, but I am also interested in hearing more about the machine-code as well

dan1st•7mo ago

It can be useful especially for optimization and diagnosing issues. But it mostly doesn't have an affect on the determinism we were talking about. But this has some exceptions as well. For example, you can enable/disable cryptographic algorithms (this configuration is something where different JDKs may actually differ) or allowing access to modules/packages that shouldn't be accessible. The bytecode is fairly close to the Java code but the specification doesn't say what exact bytecode must be produced. The specs say what bytecode instructions must do (JVM spec) and they say what the source code/language must do (JLS) but not how the language is transformed to bytecode. But in practice, different compilers result in almost the same bytecode. And these compilers (javac, EJC) are deterministic AFAIK so they produce the same bytecode Now with machine-code, it's a different story. The JIT is responsible for transforming bytecode to machine-code while the application is running and it does so in a way where it thinks that it results in the best performance. This process is nondeterministic but it must adhere to the rules set by the specification. It can produce machine-code in the way it wants to but it must not change the meaning of the program. While the application is running, it collects a lot of profiling data which it uses for better performance. It makes assumptions about the program (e.g. if something never happened the last 10000 times, it will probably not happen) but it makes sure that the program still works correctly if these assumptions are violated. You can see the specifications here: https://docs.oracle.com/en/java/javase/24/docs/specs/index.html The most important (in most cases) are the Java Language specification (JLS) and the Java VM specification (JVM spec) and also the javadocs (Java SE & JDK spec)

daysling•7mo ago

The logical AND operator is evaluated left-to-right in C/C++ as well?

dan1st•7mo ago

not by spec, at least in some C versions AFAIK though maybe it was &

daysling•7mo ago

Probably & C Standard guarantees LTR and Short circuit behaviour

dan1st•7mo ago

note: I meant the order in which operands are evaluated like i++

daysling•7mo ago

True

int i = 0;
result = (++i) & (i * 2);

int i = 0;
result = (++i) & (i * 2);

This is a UB

dan1st•7mo ago

actually I think it might be something like "if both are true, the right side may be evaluated before the left side"

daysling•7mo ago

Afaik almost all operands and some operators don't guarantee LTR or sequence points Most of those stuff undergoes compiler optimisation and is non-deterministic

dan1st•7mo ago

int i=1;
if(i++>0 && (i=5)>0){}

int i=1;
if(i++>0 && (i=5)>0){}

for example but idk whether the order is guaranteed here by the spec yeah something like that

daysling•7mo ago

This is safe

dan1st•7mo ago

sure it's safe - but is it also deterministic? It's implementation defined behaviour* undefined behaviour would be "anything could happen"

daysling•7mo ago

True. As the evaluation of those statements vary from compilers and architectures and I don't think they document these snippets very well

dan1st•7mo ago

I think I called it implementation specific behaviour before - that's my fault

daysling•7mo ago

It should be. && guarantees LTR and so it'll be executed like this. I) I++>0, it evaluates to false when I=0, so it'll just shortcircuit Uhh deterministicity (I cannot spell) is a huge topic and has a lot of debate around stuff Esp when it comes to low level

dan1st•7mo ago

ah my fault, the example is wrong (edited) Is the value of i allowed to have one of multiple values? it would probably always be the same in practice but I am not sure about the spec yes ik but I care about it being deterministic by the spec/whether the spec allows multiple options

daysling•7mo ago

It should also be deterministic, the execution is a bit different. i) i = 1, i > 0 -> true, moves to ROE, post-increment makes i=2. ii) ROE: (i=5)>0, sets i=5, and then it evalutes to true. i is not allowed to have one of multiple possible values at any given step in a non-deterministic way. Uhh, something like this has an unspecified order of evaluation and other side effects..

int val(int& x) {
    x++;
    return x;
}

int main() {
    int a = 0;

    // unknown order of evaluation + side effects
    if (val(a) && val(a) || val(a)) {
        std::cout << "Condition met, a=" << a << std::endl;
    } else {
        std::cout << "Condition not met, a=" << a << std::endl;
    }

    // UB
    if ((a = 5) && a++ && ++a && (a = a++)) {
        std::cout << "UB branch taken, a = " << a << std::endl;
    }

    return 0;
}

int val(int& x) {
    x++;
    return x;
}

int main() {
    int a = 0;

    // unknown order of evaluation + side effects
    if (val(a) && val(a) || val(a)) {
        std::cout << "Condition met, a=" << a << std::endl;
    } else {
        std::cout << "Condition not met, a=" << a << std::endl;
    }

    // UB
    if ((a = 5) && a++ && ++a && (a = a++)) {
        std::cout << "UB branch taken, a = " << a << std::endl;
    }

    return 0;
}

#include <stdio.h>

int f1(int *x) {
    *x = 5;
    return 1; // Returns true-like value
}

int f2(int *x) {
    *x = 10;
    return 0; // Returns false-like value
}

int main() {
    int i = 0;
    if (f1(&i) && f2(&i)) {
        printf("Both true");
    } else {
        printf("At least one false");
    }
    printf("Value of i: %d", i);
    return 0;
}

#include <stdio.h>

int f1(int *x) {
    *x = 5;
    return 1; // Returns true-like value
}

int f2(int *x) {
    *x = 10;
    return 0; // Returns false-like value
}

int main() {
    int i = 0;
    if (f1(&i) && f2(&i)) {
        printf("Both true");
    } else {
        printf("At least one false");
    }
    printf("Value of i: %d", i);
    return 0;
}

Both the examples before were deterministic by spec.

dan1st•7mo ago

Why does your second example have unspecified order? I don't want to think about C++/your first example lol

daysling•7mo ago

It's unspecified by C standard and is open to compiler optimisation

dan1st•7mo ago

and the function calls are necessary for that? But yes that's the thing I meant I think

daysling•7mo ago

Not exactly

int x = 1 // x is modified more than once without a sequence point
if ((x = x++) && (x = ++x)) {
  printf("UB %d", x);
} else {
  printf("No %d", x);
}

int x = 1 // x is modified more than once without a sequence point
if ((x = x++) && (x = ++x)) {
  printf("UB %d", x);
} else {
  printf("No %d", x);
}

This is also a UB

int y = -16;
if (y >> 2) {
    printf("impl defined %d", y >> 2);
} else {
    printf("impl defined shifted to zero/negative");
}

int y = -16;
if (y >> 2) {
    printf("impl defined %d", y >> 2);
} else {
    printf("impl defined shifted to zero/negative");
}

dan1st•7mo ago

Isn't that literally what my example did? Or did the = within the conditional introduce a sequence point? anyway you still mean implementation defined behaviour anyway these things are strictly defined in Java

daysling•7mo ago

Your example only had one assigment in a sequence point.

dan1st•7mo ago

whatever I don't want to deal with shenanigans in low level programming languages anyway

daysling•7mo ago

True lol

dan1st•7mo ago

@DaMango Summary: It's complicated in C/C++

daysling•7mo ago

Like for example:

int i = 1;
if (i++ > 0 && i++ > 0) {
}

int i = 1;
if (i++ > 0 && i++ > 0) {
}

this is a UB (cite: c11 standard), but in C++14 and layer, it's not. that's why you just use rust and enjoy life

dan1st•7mo ago

no, we use Java here

daysling•7mo ago

🤣 Just kidding, I love java, I've been working with it for more than 5 years now.

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Are all Java programs reproducible?

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