10 ms·

I don't think you know what it means when someone asks whether a loop will terminate. It's an analysis of the logic of the loop to determine if it will exit.

by zedshaw 12y ago

I don't think you know what it means when someone asks whether a loop will terminate. It's an analysis of the logic of the loop to determine if it will exit. This kind of analysis has been the actual foundation of computer science since there was computer science. I believe this guy Turing was doing some stuff with it. However, you have a logical flaw in your statement. You say the function will not terminate if it is passed an invalid pointer, but processes terminate when they access invalid pointers. If that's true, then it will terminate. However, if processes do not terminate when passed invalid pointers, then this function will still terminate because it will hit the end of its length variable and terminate. So you're wrong on many counts.

mikeash 12y ago
I take it that you're completely unfamiliar with the concept of "undefined behavior" in the context of C. You're not qualified to discuss the language at this level until you understand that. It's possible to invoke undefined behavior in your function. Undefined behavior can mean something never terminates. Thus, it's possible for your function to never terminate. If you don't believe me, I pose the following challenge: give me an implementation of this function. I will then provide a chunk of code that calls it and causes it to enter an infinite loop.
- zedshaw 12y ago
  Ahhh the "undefined behavior" trope, whereby a C "expert" who's memorized a standard trots out the abstract machine to justify his point. An abstract machine that doesn't actually exist and that no computer actually functions as. I'll take you up on that challenge. I'm curious to see how you'd write a program to make a for-loop never exit. Give me a few minutes...
- zedshaw 12y ago
  Alright my friend, here's the gist: https://gist.github.com/zedshaw/c20a69f17578909523c4 https://gist.github.com/zedshaw/c20a69f17578909523c4 The rules: 1. You said you can make that for-loop run forever that can call it and it'll enter an infinite loop. 2. To prove that, you can only alter the main function, then hand me back the code and I'll compile it and run it on my machines. 3. It has to run without stopping for 24 hours. If you can do that then I'll consider that an "infinite loop". 4. You can't call any more functions than what's in there already. So no fancy hacks to keep the OS from allowing segfaults by putting in signal handlers, linking against other libraries, or anything. Very curious how you do this. This is fun! Edit: And, I may not be checking comments so email me to gloat if you figure it out. help@learncodethehardway.org any time. You can also post it here. Just link me the reply so I can go look.
  - mikeash 12y ago
    Here you go: int main(int argc, char *argv[]) { int offset = -63; char input[] = { 1, 1, 1 }; char *output = input + offset; safercopy(3, output, 3, input); return 0; } This is running on a Mac with 10.10.1 and Xcode 6.1.1, compiled without optimizations. The offset value may need to be different on other architectures. With optimizations on, the approach may need to change. Don't give me any guff about the conditions needed, since that's the whole point of undefined behavior: it depends on context that should be irrelevant. There's no need to run it for 24 hours. Just run it, then pause in the debugger and step through a few loops. It'll be evident that nothing changes. If you need help getting it to work properly on your own setup, let me know.
    - zedshaw 12y ago
      Fail. I've ran it repeatedly over and over and it doesn't run forever. It segfaults or exits. But....you did cover a corner case I hadn't considered. Thanks! https://gist.github.com/zedshaw/81edf35857e137ccd7d3 https://gist.github.com/zedshaw/81edf35857e137ccd7d3 is the results.
      - mikeash 12y ago
        Did you adjust the offset like I said you would probably need to?
        mfukar 12y ago
        I can't see any changes in `safercopy` or in the calling code you provided - only the invocations from the shell showing the prog terminating.
        mikeash 12y ago
        My point is that the offset value needed to produce the described behavior depends on various implementation-specific things, so that constant may need to be altered when trying the code on other compilers, OSes, or CPU architectures.

by zedshaw 12y ago

mikeash 12y ago

I take it that you're completely unfamiliar with the concept of "undefined behavior" in the context of C. You're not qualified to discuss the language at this level until you understand that. It's possible to invoke undefined behavior in your function. Undefined behavior can mean something never terminates. Thus, it's possible for your function to never terminate. If you don't believe me, I pose the following challenge: give me an implementation of this function. I will then provide a chunk of code that calls it and causes it to enter an infinite loop.

zedshaw 12y ago
Ahhh the "undefined behavior" trope, whereby a C "expert" who's memorized a standard trots out the abstract machine to justify his point. An abstract machine that doesn't actually exist and that no computer actually functions as. I'll take you up on that challenge. I'm curious to see how you'd write a program to make a for-loop never exit. Give me a few minutes...
zedshaw 12y ago
Alright my friend, here's the gist: https://gist.github.com/zedshaw/c20a69f17578909523c4 https://gist.github.com/zedshaw/c20a69f17578909523c4 The rules: 1. You said you can make that for-loop run forever that can call it and it'll enter an infinite loop. 2. To prove that, you can only alter the main function, then hand me back the code and I'll compile it and run it on my machines. 3. It has to run without stopping for 24 hours. If you can do that then I'll consider that an "infinite loop". 4. You can't call any more functions than what's in there already. So no fancy hacks to keep the OS from allowing segfaults by putting in signal handlers, linking against other libraries, or anything. Very curious how you do this. This is fun! Edit: And, I may not be checking comments so email me to gloat if you figure it out. help@learncodethehardway.org any time. You can also post it here. Just link me the reply so I can go look.
- mikeash 12y ago
  Here you go: int main(int argc, char *argv[]) { int offset = -63; char input[] = { 1, 1, 1 }; char *output = input + offset; safercopy(3, output, 3, input); return 0; } This is running on a Mac with 10.10.1 and Xcode 6.1.1, compiled without optimizations. The offset value may need to be different on other architectures. With optimizations on, the approach may need to change. Don't give me any guff about the conditions needed, since that's the whole point of undefined behavior: it depends on context that should be irrelevant. There's no need to run it for 24 hours. Just run it, then pause in the debugger and step through a few loops. It'll be evident that nothing changes. If you need help getting it to work properly on your own setup, let me know.
  - zedshaw 12y ago
    Fail. I've ran it repeatedly over and over and it doesn't run forever. It segfaults or exits. But....you did cover a corner case I hadn't considered. Thanks! https://gist.github.com/zedshaw/81edf35857e137ccd7d3 https://gist.github.com/zedshaw/81edf35857e137ccd7d3 is the results.
    - mikeash 12y ago
      Did you adjust the offset like I said you would probably need to?
      - mfukar 12y ago
        I can't see any changes in `safercopy` or in the calling code you provided - only the invocations from the shell showing the prog terminating.
        mikeash 12y ago
        My point is that the offset value needed to produce the described behavior depends on various implementation-specific things, so that constant may need to be altered when trying the code on other compilers, OSes, or CPU architectures.