5 ms·

I agree with you. However, a quick personal anecdote: I tried to create "assembler", in python, for just one instruction, MOV. And not even 64, or 32-bit, (it

I agree with you. However, a quick personal anecdote: I tried to create "assembler", in python, for just one instruction, MOV. And not even 64, or 32-bit, (it was either 16-bit or 8-bit). After scratching my head with all the corner cases, and handling about 80% of them, my python code looked like a complete mess. There is an article floating around (posted here multiple times) that says MOV is turing complete [0]. [0] https://www.google.com/search?q=MOV+is+turing+complete https://www.google.com/search?q=MOV+is+turing+complete

nikeee 7y ago
One simple way to show that MOV is turing complete is how this project does it: https://github.com/xoreaxeaxeax/movfuscator https://github.com/xoreaxeaxeax/movfuscator This compiler compiles C to MOV instructions only.
blattimwind 7y ago
MOV+ is a mnemonic for at least something like two or three dozen different opcodes though.
vidarh 7y ago
But a compiler does not need to be able to generate every version in a generic way. You only need to generate the variations you're specifically intending to use.
wolfgke 7y ago
> There is an article floating around (posted here multiple times) that says MOV is turing complete [0]. Don't believe this claim - any computer has only access to a finite amount of memory while a Turing machine needs an infinite tape to work. So the proof must be false. In other words: only a mathematically idealized version of MOV might be Turing complete - but this cannot be the MOV that x86 implements.
- chrisseaton 7y ago
  Cone on... anyone reading the claim knows it has a silent ‘(but obviously with a machine-bound tape)’ at the end of it.
  - wolfgke 7y ago
    The class of machines with the ‘(but obviously with a machine-bound tape)’ are even weaker than finite-state machines (because the number of states is bound by a fixed constant instead of "just" assumed to be finite). So if you argue that the silent ‘(but obviously with a machine-bound tape)’ assumption is OK, you argue that it is OK to identify Turing machines with a class of machines that is even weaker than finite-state machines. Any professor for (theoretical) computer science will be horrified by such claims.
    - chrisseaton 7y ago
      If your program finishes without reaching the limit of tape then it doesn’t matter that the tape was limited. It’s not about classes of machines. A machine with a limited tape is a perfect approximation for the proper class of a machine with an unlimited tape if the limit is not reached for all programs you care about. Professors won’t be horrified - they also use the silent parenthetical constraint. Look for example at formal publications like Dolan’s and see how academics talk about it.
- msla 7y ago
  If you want to be pedantic, let's get pedantic. Turing-completeness isn't a property of computational models, it's a property of problems. A problem is Turing-complete when it cannot be solved by any computational model less powerful than a Turing Machine, or other Turing-equivalent computational system, such as the Lambda Calculus. Moreover, a Turing machine does not have an infinite tape. It has an unbounded tape. Its tape is not limitless, it is extensible, such that a computation cannot fail due to running out of tape. Hey, you wanted to get pedantic. Finally, in practical land, the distinction between Turing-equivalent and non-Turing-equivalent is very relevant to real programming languages: Epigram is pointedly not Turing-equivalent, and therefore it's possible to prove programs in that language will halt. Moving away from Turing-equivalence frees you from the tyranny of the Halting Problem. https://scienceblogs.com/goodmath/2007/01/05/turing-equivalent-vs-turing-co https://scienceblogs.com/goodmath/2007/01/05/turing-equivale... https://en.wikipedia.org/wiki/Epigram_(programming_language) https://en.wikipedia.org/wiki/Epigram_(programming_language)
userbinator 7y ago
I'm not sure what you mean by "corner cases" with MOV, unless you mean things like segment registers (which are rarely if ever going to be emitted by a compiler.) There's reg<imm, m/r<imm, and m/r<>reg. That last one makes up the majority of MOV instructions a compiler will emit. Also, the opcodes (and especially the ModRM) make far more sense when written in Octal - see http://www.dabo.de/ccc99/www.camp.ccc.de/radio/help.txt http://www.dabo.de/ccc99/www.camp.ccc.de/radio/help.txt
cpeterso 7y ago
Like the one instruction set computer (OISC) architecture? https://en.wikipedia.org/wiki/One_instruction_set_computer https://en.wikipedia.org/wiki/One_instruction_set_computer

nikeee 7y ago

One simple way to show that MOV is turing complete is how this project does it: https://github.com/xoreaxeaxeax/movfuscator https://github.com/xoreaxeaxeax/movfuscator This compiler compiles C to MOV instructions only.

blattimwind 7y ago

MOV+ is a mnemonic for at least something like two or three dozen different opcodes though.

vidarh 7y ago

But a compiler does not need to be able to generate every version in a generic way. You only need to generate the variations you're specifically intending to use.

wolfgke 7y ago

> There is an article floating around (posted here multiple times) that says MOV is turing complete [0]. Don't believe this claim - any computer has only access to a finite amount of memory while a Turing machine needs an infinite tape to work. So the proof must be false. In other words: only a mathematically idealized version of MOV might be Turing complete - but this cannot be the MOV that x86 implements.

chrisseaton 7y ago
Cone on... anyone reading the claim knows it has a silent ‘(but obviously with a machine-bound tape)’ at the end of it.
- wolfgke 7y ago
  The class of machines with the ‘(but obviously with a machine-bound tape)’ are even weaker than finite-state machines (because the number of states is bound by a fixed constant instead of "just" assumed to be finite). So if you argue that the silent ‘(but obviously with a machine-bound tape)’ assumption is OK, you argue that it is OK to identify Turing machines with a class of machines that is even weaker than finite-state machines. Any professor for (theoretical) computer science will be horrified by such claims.
  - chrisseaton 7y ago
    If your program finishes without reaching the limit of tape then it doesn’t matter that the tape was limited. It’s not about classes of machines. A machine with a limited tape is a perfect approximation for the proper class of a machine with an unlimited tape if the limit is not reached for all programs you care about. Professors won’t be horrified - they also use the silent parenthetical constraint. Look for example at formal publications like Dolan’s and see how academics talk about it.
msla 7y ago
If you want to be pedantic, let's get pedantic. Turing-completeness isn't a property of computational models, it's a property of problems. A problem is Turing-complete when it cannot be solved by any computational model less powerful than a Turing Machine, or other Turing-equivalent computational system, such as the Lambda Calculus. Moreover, a Turing machine does not have an infinite tape. It has an unbounded tape. Its tape is not limitless, it is extensible, such that a computation cannot fail due to running out of tape. Hey, you wanted to get pedantic. Finally, in practical land, the distinction between Turing-equivalent and non-Turing-equivalent is very relevant to real programming languages: Epigram is pointedly not Turing-equivalent, and therefore it's possible to prove programs in that language will halt. Moving away from Turing-equivalence frees you from the tyranny of the Halting Problem. https://scienceblogs.com/goodmath/2007/01/05/turing-equivalent-vs-turing-co https://scienceblogs.com/goodmath/2007/01/05/turing-equivale... https://en.wikipedia.org/wiki/Epigram_(programming_language) https://en.wikipedia.org/wiki/Epigram_(programming_language)

userbinator 7y ago

I'm not sure what you mean by "corner cases" with MOV, unless you mean things like segment registers (which are rarely if ever going to be emitted by a compiler.) There's reg<imm, m/r<imm, and m/r<>reg. That last one makes up the majority of MOV instructions a compiler will emit. Also, the opcodes (and especially the ModRM) make far more sense when written in Octal - see http://www.dabo.de/ccc99/www.camp.ccc.de/radio/help.txt http://www.dabo.de/ccc99/www.camp.ccc.de/radio/help.txt

cpeterso 7y ago

Like the one instruction set computer (OISC) architecture? https://en.wikipedia.org/wiki/One_instruction_set_computer https://en.wikipedia.org/wiki/One_instruction_set_computer