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Comment by shasta | Hacker News Reader

Decompilation is undecidable?

   "you cannot write an algorithm that would decompile every possible piece of code"

I'm not sure what claim is actually being made here, but decompilation of any particular compiler is trivially decidable: try every input program until you find the one that produces the observed output.

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How do you know that you'll find a program that works? What if the program wasn't generated by the compiler and that compiler can't generate that particular sequence of instructions? At what point do you know, in general, "this instruction sequence isn't derivable from this compiler!"...?

You're right. I initially took the problem to be "given this (specific) compiler output, determine the corresponding compiler input." That seems like a reasonable interpretation of "decompiling". If the problem is instead "determine whether this chunk of machine code could have come from this compiler", then my proposed algorithm obviously doesn't work.

Obviously?

Iterating over all possible input 'programs' (including buggy ones) is easy, even if the language allows for multiple input files of different types.

=> if you can give an upper bound for the length of the source code of a binary of size N, you are all set. I do not see how to proof that such an upper bound exists, but it seems fairly reasonable that one must exist. I cannot really think of any real-world compiler that would need, say, more than a GB of source code for every bit of output (it is easy to design such a language, but it would be quite esoteric)

Well, I think "obviously" was right, since my algorithm would never terminate if no source is ever found that matches the generated binary (there are infinitely many possible programs for most idealized languages).

For your idea of modifying my algorithm to bail after some number of inputs have been checked, clearly there does exist a bound on the source length necessary to generate all possible binaries of size N, since there are only finitely many such binaries (and thus they can be produced by finitely many sources). But to turn this into an algorithm, you'd need a computable bound. For a language like C, I think such a bound probably exists and is a reasonable function of the input size. For a language like Coq, though, I doubt the bound would be computable. Basically, I think that there are programs that work for reasons that are very difficult to prove.

Also, I think we're far enough from practical that appeals to the real world are somewhat silly. Either we're interested in understanding the theoretical situation, in which case lack of real world examples doesn't help, or we're interested in practical decompilation, in which case this whole conversation is stupid.

It depends on what you want from decompilation. Decompilation by its basic nature is just as decideable as compilation is. It's a translation from one language to another in addition to semantics-preserving translations. Now what a lot of people seem to want from a decompiler that makes them intractable is a guarantee that the output source will look just like the source used to compile the source program for the decompiler.

Program1 -> Compiler -> BinProg1 (compilation)

BinProg1 -> Decompiler -> Program2 (decompilation)

Even if the source language of the Compiler and the target language of the decompiler are the same, you're usually still screwed because the compiler is going to manipulate your program to optimize it or fit it onto your target architecture. The result is going to be a program that looks way more general case than your original program.

Is this undecideable? No way.

We typically reserve the term "undecidable" for the case where there is a mathematical function we are interested in, but it isn't computable by any algorithm. Inverting a many-to-one function is just impossible.

That's not what decidable means.

The code could be self-modifying. Figuring out what the code turns into, to generate a meaningful translation of it, turns into the Halting problem pretty quickly. At the limit, the best you can do is produce assembly output, and even that may be problematic if there are ambiguities in the encoding but the program uses code as data or vice versa.

Self-modifying code doesn't change my argument (see my response to kenjackson for my understanding of the problem) so long as it was produced by the compiler.

Well, you have no guarantee that a sane compiler was used, obfuscation and encryption may also play a part. Things worth reverse engineering are often convoluted in such ways.

The set of all possible inputs is infinite: Halting Problem.

The set of all possible inputs for a compiler is infinite, too. Does that mean that compilers are all harangued by the halting problem as well? Nope. Having an infinite number of possible inputs (infinite input set size) does not prevent you from showing that a program can actually halt. I think most compilers are written in a way that can be shown to halt, as well.

The Halting Problem does apply in the general case, but if you carve up your programs and reason about them, you can still show that you can have a halter. The Halting Problem just states that there does not exist a method that will take an arbitrary program and show it to be a halter.

"The set of all possible inputs for a compiler is infinite, too. Does that mean that compilers are all harangued by the halting problem as well? "

No, because unlike this "try all possible inputs" plan for decompilers, compilers only operate on any particular single input.

"I think most compilers are written in a way that can be shown to halt, as well."

Not C++ compilers (Turing Complete macros) or Lisp dialects (same issue, but even moreso)

The point about the macros being Turing Complete is trivial in this instance because if you wrote a macro that never terminated, you'd never compile anything do decompile ;-).

No, you are incorrect about the point being trivial. First off, the macros used in the original program will of course terminate, but if you simply "try every input to the compiler" like is being suggested, you will be attempting to compile any number of non-terminating macros (and you will of course not know which terminate and which do not (Halting Problem)).

The point is that "compilers terminate for every input" is trivially false.

Non-termination of some compiler inputs is a red herring. The only modification required to the original algorithm is that the search proceed in parallel (such that every input eventually makes arbitrary progress).

I agree with your point in a sister thread that decidable doesn't imply practical, but the claim of the original article was undecidable, which has a technical meaning.

I would qualify that statement better by saying "compilers with Turing Complete macros or type systems terminate for any input is trivially false". The point may not be trivial (which I'm not ready to concede), but it still doesn't matter. The author's means implicitly relies on an evasion where programs aren't written in this (rather strange) way. It's a safe assumption to say that you can eliminate programs written in such a way that will not compile, then the Halting Problem as you've stated it doesn't apply.

I think you may be right about the point your trying to make, but the point I'm making is that your point covers a negligible section of programs that the author would hope to analyze, so it doesn't matter.

My point is not about checking if the given program will halt or not.

The parent comment's suggestion was: (1) for each possible input program, (1.a) compile it, and (1.b) check if the result equals the given compiled code.

Agreed, steps (1.a) and (1.b) terminate deterministically (for a given compiler).

However, the search space for this search procedure is infinite.

Similarly, it would be impossible, in general, to exhaustively test every possible input of a compiler.

And yet, if the output you are decompiling is known to come from a particular compiler, the search will terminate.

If you know the compiler (and every other tools involved in the process), then yes, there exist at least one solution (the original input). The search will eventually find it.

Assuming all possible inputs to the compiler cause it to halt. Not true for many high-profile languages.

And while not strictly related to "is it decidable", in the real world we have to keep in mind the complexity of the solution. The presence of the naive solution for a particular language doesn't say much about how well we can* actually do it. Is the problem actually decidable within the confines of the physical observable universe for real world inputs?

"Assuming all possible inputs to the compiler cause it to halt."

If the compiler did not halt, then we don't have anything to disassemble in the first place.

Formally, it is obvious that a diagonalization-based search will eventually find a correct input, regardless of the halting status of any given input. In practice, none of this matters very much.

As I explained in my other comment, the compiler certainly does halt for the correct input, but that does not mean it will halt for all inputs.

You are however correct that this can be circumvented with diagonalization.

The compiler isn't part of the input. You could get code that was compiled by a compiler not available to you.

When the compiler is part of the input, then yes, the problem is decidable.

I'm still doubtful. I think you're trying to invoke Rice's theorem here, but I don't think it really applies. Compilers don't apply arbitrary transformations that preserve behavior - they only employ transformations that are understood to preserve behavior.

I don't think it is decidable even when you have the compiler available to you - what happens if I decide to insert a bunch of assembly code in my gcc-compiled program using the gcc embedded assembly extensions? How do I decide that a particular sequence of generated op-codes is actually impossible from pure C code, and hence must have been assembly in the original program?

In that case, the GCC extension is part of the input language (in a broader sense). Still, your example clearly shows the difficulties involved in practice.

The problem only makes sense if the compiler is fixed. Otherwise, it's trivial: for any input x I can say the source code is x. It's source code written in a programming language called MachineLanguage. The compiler is called cat.

I think he had in mind that the language is fixed, but not the compiler.

In this context, "programming language" and compiler are really the same thing: a mapping from source code to object code.

In any case, the problem with your original argument is not the choice of compiler, it's that your algorithm migh not halt, as explained by kenjackson.

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