The Little C Function From Hell

http://blog.regehr.org/archives/482

by gus_massa • 14 years ago

273 113 14 years ago

blog.regehr.org

scott_s4• 14 years ago

If this sort of thing is interesting to you, the author and his colleagues wrote an excellent PLDI 2011 paper about their research which systematically checks various compilers for exactly these kinds of bugs: "Finding and Understanding Bugs in C Compilers": http://www.cs.utah.edu/~regehr/papers/pldi11-preprint.pdf

Perhaps my favorite part of that paper is the fact that there is no ground truth (section 2.6). They discover bugs by testing random programs against multiple compilers. If the result from any of the compilers disagree, then there must be a bug. (They guarantee that the inputs are legal.) In theory, it's possible that all of the compilers could be wrong in the same way, which means they wouldn't discover a bug. In practice, this is extremely unlikely. But you can't know for sure. (In practice, they never saw an instance where there were three different results from three compilers; at least two of the compilers always agreed.)

Deestan• 14 years ago

> They discover bugs by testing random programs against multiple compilers. If the result from any of the compilers disagree, then there must be a bug.

How does this make sense?

If the result differs from the specification, it is a bug.

If the result is unspecified in the specification, the different compilers can differ as much as they want without any of them being considered buggy.

loumf• 14 years ago

I took "They guarantee that the inputs are legal" to mean that they limited it to programs with specified behavior. They don't know what the behavior is -- just that it is specified.

If they can do this, it finds a subset of bugs, with no false positives.

Wilya• 14 years ago

From the linked paper: "Although this compiler-testing approach has been used before [6, 16, 23], Csmith’s test-generation techniques substantially advance the state of the art by generating random programs that are expressive—containing complex code using many C language features—while also ensuring that every generated program has a single interpretation. To have a unique interpretation, a program must not execute any of the 191 kinds of undefined behavior, nor depend on any of the 52 kinds of unspecified behavior, that are described in the C99 standard."

scott_s4• 14 years ago

Both loumf and Wilya are correct. In support of their answers, remember that the specification does not specify the results of interesting programs. It says "if you do this, this must be the result." But if you limit yourself to only testing such simple cases, you're not going to find any interesting bugs - because such simple programs are likely to have already been tested.

MidwestMuster• 14 years ago

Isn't this called "fuzzing"

cube13• 14 years ago

>If the result differs from the specification, it is a bug.

A large part of the C standard is implementation defined(see acqq's post here: http://news.ycombinator.com/item?id=4131828 ), so the result could be different on multiple compilers, not a bug, and STILL completely within spec.

silentbicycle2• 14 years ago

C compilers can be buggy, particularly when you start working with vendor-supplied compilers for embedded platforms. A colleague was furious when he realized that his board's compiler didn't support function pointers.

DanielBMarkham• 14 years ago

I confess to just scanning the code and being cold on C right now, but isn't ++x a post increment? That is, it occurs after the rest of the expression has been evaluated? (as opposed to x++, which is a pre-increment) Just guessing that the perhaps the issue is when the actual overflow is occurring.

dag11• 14 years ago

++x evaluates to x's new (incremented) value.

x++ evaluates to x's old (pre-incremented) value.

vhf2• 14 years ago

It is the opposite, like in many other languages. e.g.

  int i = 0;
  printf("%i %i", i++, ++i); // prints "0 2"

Same goes in C, C++, Java, PHP, ...

[EDIT] Turns out this is a bad example, as "the order in which function arguments are evaluated is undefined" (cf. below) Correct is :

  int i = 0;
  printf("%i", i++); // prints 0
  printf("%i", ++i); // prints 2

simias• 14 years ago

I believe you trigger an undefined behaviour there. You modify i twice in the same statement.

excuse-me• 14 years ago

Is that defined behaviour?

Is there any rule saying that args to a function have to be evaluated in a particular order - ie is ',' a function point?

decklin• 14 years ago

Actually, the order in which function arguments are evaluated is undefined. http://c-faq.com/expr/comma.html

You can use a comma in other expressions to introduce a sequence point: http://c-faq.com/~scs/cgi-bin/faqcat.cgi?sec=expr#seqpoints

jim_kaiser• 14 years ago

The standard says, evaluation order of function arguments is undefined.. but in general, printf evaluates right to left. This was a common puzzle as I remember.

CodeMage• 14 years ago

It's precisely the other way around: ++x is a pre-increment and x++ is a post-increment.

acqq4• 14 years ago

Mnemonic: read as when you would read the text, left to right:

   ++x increment than use the value x. 
   x++ use the value x then increment.

loeg• 14 years ago

Nope, you've got the two backwards.

acqq4• 14 years ago

Before starting reading everything note that author assumed when writing the article that (provided sizeof( int ) > sizeof( char )):

    char c = (char)128;
    printf( "%d\n", c );

should always be -128, whereas his commenter Mans (March 4, 2011 at 4:15 am) points out that this conversion is implementation-deﬁned according to the standard, that is, compiler authors are free to decide what the result of such conversion should be.

dchest2• 14 years ago

Should the implementation-defined behavior be always the same? That is,

    char a = (char)128;
    char b = (char)128;

Will a == b for every implementation?

lmm• 14 years ago

Yes; that's the distinction between "implementation-defined" and "undefined" behaviour.

TazeTSchnitzel• 14 years ago

I'd be surprised if it's non-deterministic, so yes.

dllthomas2• 14 years ago

That depends on what you mean by "should". Is it required by the standard? AFAIK, no. Would it be a braindead decision on the part of the compiler writers to do otherwise? Almost certainly. In practice, the answer to

> Will a == b for every implementation?

is, I suspect, yes.

caf3• 14 years ago

It's true, but it doesn't affect the analysis. When x is CHAR_MAX and char is signed, the result of ++x is an implementation-defined value with type char - but it doesn't matter what this implementation-defined value is, it must always be less than or equal to CHAR_MAX, so ++x > CHAR_MAX is false no matter what the resolution of the implementation-defined behaviour is.

acqq4• 14 years ago

If the ++x is performed using int, there must be a downcast to store it in the char x if you even need to store it (and in the given function you don't), but nowhere I see that the standard insists that that very int can't be used in the comparison, so 128 > 127 is OK. Note that the right side of comparison can be int from the start, even if the right side can fit the char. The only thing wrong was the older clang.

So the whole big article can be shortended just to: "clang in 2.7 had a bug but newer version doesn't, all other compilers are OK."

raverbashing6• 14 years ago

A little C tip I learned from hard experience

NEVER, EVER, NOT IN A MILLION YEARS use a signed int/char etc, unless you are 200% certain you're doing the right thing (that is, it's for something specific you need it)

You WILL have problems, period.

"Oh it's just a matter of knowing the C spec" then please go ahead as I grab the popcorn.

Roboprog• 14 years ago

There's a certain aspect of "functions have domains, not just ranges" at work here as well -- e.g. - restricting the (math) tan() function to the domain of -90 to 90 degrees (exclusive), unless you really get off on watching it cycle over madness. If you are going to be playing around the edges of something, it behooves you to put some kind of pre-condition in with an assert of similar mechanism.

In fairness, I guess a function like this is a good example of why you should put in preconditions, as well as a good demonstration that "not all the world is a VAX" (nor MS C 7, nor GCC version N) :-)

timbre• 14 years ago

I'm rusty on C, and confused. What should I use instead of signed int?

rcfox4• 14 years ago

unsigned int :)

barrkel4• 14 years ago

I've seen more problems from unsigned ints than signed ints (in particular, people doing things like comparing with n-1 in loop boundary conditions). There's a reason Java, C# etc. default to signed integers. Unsigned chars, I have no quibble (and Java, C#, use an unsigned byte here).

rcfox4• 14 years ago

Unsigned integer overflow has defined behaviour in C, while signed overflow doesn't. Is it really better to protect people from a simple logical error by exposing them to possible undefined behaviour?

With signed integers, you'll run into the same problem with comparing to n+1 at INT_MAX or n-1 at INT_MIN.

raverbashing6• 14 years ago

"(in particular, people doing things like comparing with n-1 in loop boundary conditions"

There's your problem: that's like saying cars are more dangerous than motorcycles because your finger can get squeezed by the door.

"There's a reason Java, C# etc. default to signed integer"

Legacy? And in Java/C#, and you usually use ints, not so much chars, shorts, etc and casts are probably more picky

I stand by my point, you should only use signed if you know what you're doing and for a specific use only (like math)

jbooth• 14 years ago

Java actually uses a signed byte type, which was I guess to keep a common theme along with all the other types, but in practice leads to a lot of b & 0xFF, upcasting to int etc when dealing with bytes coming from streams.

ajross• 14 years ago

"For something specific you need it" meaning ... a negative number, like an array or memory address offset? I mean, sure, I agree that you should be doing anything sensitive to 2's complement behavior on unsigned quantities. And if you know the values are positive-definite, unsigned has much cleaner behavior. And I'd even entertain the argument that unsigned should have been the default in the language spec and that signed quantities are the ones that should have specially declared types.

But... your advice as written is just insane. They are real, and required routinely. You can't just apply this as a "for dummies" rule without essentially ruling out half of the code they'll need to write.

raverbashing6• 14 years ago

". a negative number, like an array or memory address offset? "

A negative number yes, but not really a memory offset (you shouldn't mix negative numbers and memory offset, really)

But yeah, if you're doing "math" go for it, but it's on rare occasions where you need negative numbers (subtraction yes)

The most common case I remember may be sound samples, where you have signed chars.

For all other cases you would be using floating point or decimal numbers

tedunangst• 14 years ago

signed usually promotes to unsigned in nice ways, such that if you really want to store -1 in your unsigned, it will just work. I've found using all unsigned types, with the occasional signed value jammed into them, is less error prone than mixing and matching types throughout the program. ymmv, and of course, here be the odd dragon or two.

regularfry2• 14 years ago

It's not just the C spec you've got to watch. I saw a wonderful bug last week where the author hadn't spotted that write(2) returned a ssize_t, not a size_t (or didn't appreciate the difference), so was gleefully checking an unsigned variable for a -1 error result.

nitrogen3• 14 years ago

How did the bug manifest itself? You can store 0xffffffff(ffffffff) in a 32(64)-bit unsigned int, or a 32(64)-bit signed int. In the one case you'll get UINT_MAX, -1 in the other, but they should compare as equal. If you have -Wextra turned on, gcc will give a warning, though.

Here's some sample C code tested on a 32-bit Linux system:

  #include <stdio.h>
  
  int main(int argc, char *argv[])
  {
          unsigned int val1 = 0xffffffff;
  
          printf("val1 == -1: %d\n", val1 == -1);
  
          return 0;
  }

The result:

  val1 == -1: 1

cpeterso• 14 years ago

Google's C++ style guide strongly recommends using signed ints over unsigned ints:

https://google-styleguide.googlecode.com/svn/trunk/cppguide....

duaneb2• 14 years ago

Actually, a colleague recently convinced me to start using signed ints for, e.g., for loops instead of unsigned ints. His reasoning was that if you're overflowing with signed integers, you'll probably overflow with unsigned integers too (we work with very large numbers), but it's easier to notice if you have a negative number rather than silently wrapping around to a still-valid value.

cperciva9• 14 years ago

99% of the time, you should not be using any sort of int for a loop variable. Loop variables are almost always array indexes, and array indexes should be size_t.

dchest2• 14 years ago

Just tried with Plan 9-derived compilers shipped with Go, and there's no bug (provided that my re-implementation of the test case is correct).

Arcticus• 14 years ago

Dan Saks did a great set of presentations called "C and C++ Gotchas" on these types of things at the 2006 Embedded Systems Conference in San Jose.

Sorry couldn't find a link that wasn't behind a paywall but here is one for reference.

http://eetimes.com/electrical-engineers/education-training/t...

sltkr2• 14 years ago

To be absolutely accurate, the program still invokes implementation-defined behaviour. From the C standard on casting integers: “if the new type is signed and the value cannot be represented in it; either the result is implementation-defined or an implementation-defined signal is raised”.

Therefore, the author's conclusion that “the behavior is well-defined and every correct compiler must emit this output” is plain wrong. A correct compiler might emit a signal instead of outputting anything.

(However, printing 1 for the last case is still wrong, because there is no possible way for ++x to yield a value greater than INT_MAX, so this cannot be consistent with any implementation-defined behaviour.)

sltkr2• 14 years ago

^ In the above, I meant CHAR_MAX instead of INT_MAX. Oops!

malkia2• 14 years ago

That's undefined behavior, the compiler can do whatever it pleases to. It can even print 666, and be done.

AceJohnny2• 14 years ago

No it's not. I had the same thought, but the author carefully points out from the standard: "The usual arithmetic conversions ensure that two operands to a “+” operator of type signed char are both promoted to signed int before the addition is performed."

He does assume sizeof(int) > sizeof(char), which is true on all platforms he has tried. It would be undefined on an AVR or other microcontroller where sizeof(int) == sizeof(char) though.

malkia2• 14 years ago

Thanks! Should read the article more carefully.

nova• 14 years ago

undefined on an AVR or other microcontroller where sizeof(int) == sizeof(char) though.

Just noting that ints are 16 bits wide in AVR-GCC, unless you use the -mint8 option which violates C standards.

TwoBit• 14 years ago

Summary: ++x for signed char is supposed to convert x to int before the ++, but some compilers get it wrong.

Dove• 14 years ago

You know, not long ago I took somebody's "how well do you know C?" quizzes, and it was fully of this sort of question -- what C does with overflows and underflows in various circumstances. And I must admit, I felt like I had been asked what happens, exactly and specifically, to a particular piece of memory after dividing by zero. "I don't know, I try to avoid doing that!"

I don't know. I can admire the analysis, but I don't understand the motive. Do people really write code that relies on this sort of behavior? Or is it just trivia for trivia's sake?

egonschiele• 14 years ago

I do similar deep digging for some topics. It may seem useless, and a lot of times it is. But sometimes digging into a problem helps me understand the bigger picture better, and then it seems worthwhile. Lately I've become more picky about what I'll dig into though.

pgeorgi• 14 years ago

It's interesting when trying to do automated code analysis, and C is an important target for that, since it's used everywhere. But computer programs have a hard time to know when type limits are important and when they're not (or in fact not anticipated by the developer) - so they have to cover all cases.

(IMHO) For obscure cases, there's ideally some clearer version of the same behaviour that could be recommended to the developer - either helping them to find a potential error or to have them use a less ambiguous and easier to understand notation.

rcfox4• 14 years ago

If it was a week or two ago, the quiz was probably from the same person as this article.

He's not just doing it for fun; he's a professor at the University of Utah, and he's researching this area, looking for bugs in compilers. In fact, he's developed a tool for this: http://embed.cs.utah.edu/csmith/

These tiny bits of strange code are condensed versions of what you might see in the wild, especially after preprocessing.

Nobody's doing ++x > y, but they do something that looks reasonable like foo(x) > bar(x), where foo() and bar() return chars.

nitrogen3• 14 years ago

Nobody's doing ++x > y, but they do something that looks reasonable like foo(x) > bar(x), where foo() and bar() return chars.

I might write something like "++x > y"; preincrement followed by comparison is a common operation.

simcop2387• 14 years ago

It's most likely this kind of code is encountered on embedded systems where you end up with smaller integers and things. This particular case I don't know if I've ever done it though.

pmjordan2• 14 years ago

C's type system is a minefield. (and by extension, C++'s and Objective-C's) I've been bitten so many times by implicit signed/unsigned conversions and int-promotion that I'm now practically paranoid about the whole thing. Chars are promoted to int practically anywhere you use them for anything other than just copying, so if the compilers produce faulty code in those situations, there's no way you can win.

Oh, and speaking of chars: Objective-C's BOOL is really just a char. Yes, it's signed, and yes it gets int-promoted a lot. I dread to think how many bugs are lurking out there in Objective-C code because of that. I wonder if you could catch some of those by comparing the code generated by compiling with the usual BOOL = char typedef, and the same code but with BOOL typedef'd to _Bool (a real boolean type).

jlarocco• 14 years ago

Nobody purposely writes code that relies on that sort of behavior.

Much more likely is some code relying on it without realizing it, and getting "random" bugs for some input values.

WalterBright• 14 years ago

The Digital Mars C compiler returns the correct answer with or without optimization turned on.

robot• 14 years ago

The user must know about overflows and act accordingly. Compiler behavior may naturally change based on optimisations since it is undefined territory.

krollew4• 14 years ago

"That’s a lot of trouble being caused by a two-line function. C may be a small language, but it’s not a simple one." I disagree. C is very simple. You need to know C is not suppoused to work the same on platform and I guess behaviour you have tested is not defined by standard. I think every good C programmer know when he have to be causious because behaviour may be platform dependent. In matter of fact you did as well. I guess case you've studdied is not common to be used in real code. If so, there is very easy solution:

  #if ((char)CHAR_MAX) + 1 > ((char)CHAR_MAX)
   /* some code here */
  #else
   /* some code here */
  #endif

There is no problem to me.

cperciva9• 14 years ago

Every problem has a solution which is simple, elegant, and doesn't work. This is it.

Section 6.10.1, paragraph 4: "... For the purposes of this token conversion and evaluation, all signed integer types and all unsigned integer types act as if they have the same represenation as, respectively, the types intmax_t and uintmax_t..."

Inside preprocessor directives, your chars aren't chars any more.

krollew4• 14 years ago

So what they are? Why it doesn't work? Your quote don't clarify.

The Little C Function From Hell

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