[1]: https://github.com/rofl0r/musl/blob/d05aaedaabd4f5472c233dbb...
Makes me wonder what kind of compiler flags muslibc uses to prevent this.
Edit: Maybe Clang abuses the bitand with an int 7 to remove the code?
“Pointer arithmetic is well defined only on array type and between array[0] and array[array_size+1]”
And it’s even stricter:
- you can create a pointer to array[array_size+1], but dereferencing it invokes undefined behavior.
- changing a pointer that points into an object so that it points into another object invokes undefined behavior. Modern compilers track “pointer provenance” to sometimes speed up code (and occasionally perplex the programmer)
It’s only special knowledge of the behavior of standard malloc that allows the compiler to assume that pointer is to offset 0 of an allocated array.
Yes, but even in that case, it would have to be an array of `size_t` objects for the pointer cast `(size_t *)` in the musl code not to trigger undefined behavior (especially if malloc would get inlined), right?
Which means calloc() would trigger undefined behavior as soon as one of its callers used it for allocating a non-size_t array (especially if calloc() would get inlined with link-time optimization)?
Hmm, I don't think this is true.
Even if you don't assume special semantics for malloc() and friends (by using the `-ffreestanding` and the `-fno-builtin` flags) and even if the compiler knows nothing about the malloc() implementation, I think if you're doing pointer arithmetic (with a pointer type other than `void *` or `[unsigned] char *`) then the compiler can assume that the pointer is pointing to either: 1) an actual array of objects of the underlying type (or a compatible one), or 2) a pointer to a single object, which is treated as an array with one element.
So if you do `((size_t *) p)[-1]` within `calloc()`, then the compiler can assume that `*p`, if used, must either be a `size_t` (or a compatible integer) or `p` must be pointing to one past the end of a single-element array (which, since `*p` is a dereference, would trigger undefined behavior).
Which means that if you call `calloc()` and it returns `p`, then you can only use it to dereference `size_t` objects or compatible ones (otherwise it would trigger undefined behavior). The effects of the undefined behavior would likely be observed if `calloc()` gets inlined into the caller (due to link-time optimization, for instance).
Or am I wrong?You said "there's no way for the compiler to prove that wrong if it doesn't know the internals of the allocator" and I tried to reason how the compiler can prove that `*p` must be a `size_t` or compatible integer, even if it knows nothing about malloc() or its implementation. Which would trigger undefined behavior in the caller of calloc() if it weren't trying to allocate a size_t (or size_t array, or of compatible integers). Although normally this would only produce a visible effect if calloc() gets inlined into the caller.
> Compatible integer types is easy.
What do you mean?
> And can't you also have any pointers you want into a char array as long as they are aligned correctly?
I think so, but I think the problem is that when you do `(size_t *) p` (or pointer arithmetic with that) then the compiler can assume that p points to something that is not just a char array, which allows it to do more optimizations (due to the undefined behavior rules).
> Also it should work to have a struct with a size_t followed by elements of a different type, shouldn't it?
I think in general it wouldn't, because structure members are not guaranteed to be contiguous (i.e. they may have padding between them), so you may be triggering undefined behavior as well. Heck, they aren't even guaranteed to be laid out in memory in the same order as you define them (even though in practice that's what compilers do).
However, if your struct has just a `size_t` field followed by a flexible array member then I think it would be OK!
Edit: on second thought, even with a flexible array member I'm not sure it doesn't trigger undefined behavior because I think the C standard allows there to be padding at the end of such a struct. Which would be OK if the code was operating on a pointer to such a struct, but not to a pointer to size_t. So I think the code triggers undefined behavior even in this case.
For the sake of clarity, I'm going to call `(size_t *) p` "q" for the rest of this post.
Pointers usually have the type of what they point to. But pointers that are one past the end don't follow that rule. q-1 is a size_t, which means q is one past the end of an array containing one size_t. Therefore the existence of the q pointer does not tell the compiler what type actually exists at memory address p/q. Is it invalid to construct q from p? I wouldn't think so as long as they are in the same allocation, but maybe I'm forgetting a rule.
> I think in general it wouldn't, because structure members are not guaranteed to be contiguous
Surely there's a way to allocate two different types up against each other, though? Even if you don't make an actual struct, you can grab some char and say that X bytes are size_t and Y bytes are something else, can't you?
But then the question is, how would you construct an allocation where both q[-1] and *p are guaranteed to be valid objects of their respective types, using standard C? (i.e. without relying on implementation-specific behavior).
> Surely there's a way to allocate two different types up against each other, though?
If by "against each other" you mean without padding between them, then I think the C standard allows that but there's no reliable/portable/guaranteed way to do it (that I know of).
By that I mean that specific implementations can guarantee that, either with struct packing or, in some cases, by carefully aligning them within a struct or a union. But this depends on the implementation/architecture, so it's not strictly portable/standard C, I think.
> Even if you don't make an actual struct, you can grab some char and say that X bytes are size_t and Y bytes are something else, can't you?
I'm not sure that is possible according to the standard, unless they already were objects of those types. See this[0]:
> It is always presumed that a char* may refer to an alias of any object. It is therefore quite safe, if perhaps a bit unoptimal (for architecture with wide loads and stores) to cast any pointer of any type to a char* type.
> The converse is not true. Casting a char* to a pointer of any type other than a char* and dereferencing it is usually in violation of the strict aliasing rule.
> As noted by Pinskla it is not deferencing a char* per se that is specifically recognized as a potential alias of any object, but any address referring to a char object. This includes an array of char objects
That said, I think for converting a bunch of chars to another type it's possible to get around that with memcpy() in some cases, but I'm not sure it would be helpful in this scenario.
[0] https://cellperformance.beyond3d.com/articles/2006/06/unders...
Once we know we have this struct, are there any remaining barriers to the cast and -1 access?
>> Casting a char* to a pointer of any type other than a char* and dereferencing it is usually in violation of the strict aliasing rule.
How does that logic work with malloc, which unlike calloc takes no item size and simply gives you a specific number of bytes?
I think the only problem is that malloc() wouldn't know the type of the non-size_t field, so it couldn't construct such a struct to do a conforming allocation which would allow calloc() to do those operations. But if the compiler doesn't have access to the malloc() implementation (assuming it doesn't know that malloc() is a special function), then maybe that doesn't invoke undefined behavior? I'm not sure, actually!
But I think that this whole saga is actually a consequence of the standard being flawed, in that I think it's not possible to create an implementation of malloc() using strict standard C (but I could be wrong, I'm not exactly an expert on the C standard).
> How does that logic work with malloc, which unlike calloc takes no item size and simply gives you a specific number of bytes?
The return value of malloc() is treated specially by the standard, I believe. Which is why musl has to use the `-ffreestanding` and `-fno-builtin` flags, I think (to avoid optimizations by the compiler due to special knowledge of standard functions), which are non-standard.
Well, musl also uses '-fno-strict-aliasing' (also a non-standard flag) which probably is what actually allows it to implement malloc() without invoking undefined behavior.
I think this is also why GCC provides a "malloc" attribute which you can use in your own functions, so that they are treated specially (like malloc()), but I believe that this is also non-standard.
Let's assume for the sake of simplicity that `sizeof element` is the same as `sizeof size_t` or a multiple of it. We don't need to assume this, it just makes the math nicer for the moment.
I do: char* buffer = malloc((count + 1) * sizeof element);
Or I do: char* buffer = calloc(count + 1, sizeof element);
How is C supposed to know what goes in that buffer?
Maybe it's an array of elements. Maybe it's an array of size_t[sizeof element / sizeof size_t].
Maybe it's an array of union { element, size_t[sizeof element / sizeof size_t]}.
If I want to store an `element` at a location `sizeof element` bytes into the buffer, how can it say no? If I want to store a `size_t` at a location `sizeof element - sizeof size_t` bytes into the buffer, how can it say no? If I want to do both, how can it say no?
It returns a pointer to a chunk of memory of sufficent fore size_t (unsigned, positive whole number of elements) where each of those elements is size_t (unsigned, positive whole number) in size within the limits of your computationally representable chunk of numbers.
No code "knows" squat about the implementation. You can know. You're the programmer, you can read the code. The computer just does. Even if that causes demons to fly out your nose.
Friends don't let friends write code without checking whether it'll cause demons to fly out their nose first. If they ain't your friend though, there's always checking ERRNO.
I think I was operating under more conservative assumptions about what is actually legal to do in C.
I actually had to go and read the standard because the strict aliasing rules can be very tricky and it's not always clear what is legal to do in every situation. Even then, it's hard to understand the wording of the standard and its implications, especially since there are multiple rules coming into play and there's no clear explanation of how they interact with each other.
In this specific case that we're talking about, I think you can indeed allocate a buffer, store the malloc-returned pointer in a `char *` variable, and then store an element and a size_t wherever you want within the buffer (but if you want to store them or access them using a properly-typed pointer rather than memcpy() then the address needs to respect the alignment of the type, I think).
But I think this is only legal because the addresses within a buffer returned by malloc are not considered to have a declared type and that the type of some address within the buffer becomes established whenever you assign or copy some value to it, so it doesn't actually necessarily become established based on the valid conversions between pointer types like I thought it was.
I'm still not exactly 100% certain about whether the code we're talking about is actually undefined behavior or not, because on one hand, the standard clearly says that using the unary `*` operator on a pointer which points one past the end of an array is undefined behavior (and `p` could be considered such a pointer that would be used with `*`), but on the other hand, it's hard to see how this can be the case when `p` can be a perfectly valid pointer to store some other value within the same allocation.
The rule which says that it's undefined behavior to use `*` on a pointer that points one past the end of an array (which can be a single value) is inside the pointer arithmetic section. So my interpretation is that, if there is a valid and properly-typed value stored at the address of a pointer that also points to one past the end of an array, it's only undefined behavior to use `*` on such a pointer if the address of that pointer was obtained through pointer arithmetic based on a pointer that also pointed to the array.
So basically, I think once you do any pointer arithmetic, the resulting pointer cannot ever point outside the array or value, except one past the value/array. However, in this latter case, you're not allowed to use `*` on that address through such a pointer under any circumstance.
I think this gets trickier if the array is a dynamically-allocated array of chars because this array of chars could contain values of other types, so I'm not 100% sure of the implications there.
Anyway, if my interpretation is correct, I think in this case the rules are subtle: it seems that reading from or writing to `((size_t *) p)[-1]` is legal if there already is a `size_t` stored at that address (I think?), otherwise it wouldn't be. But it doesn't seem to be legal to read from or write to `(((size_t *) p)[-1])[1]` (unless `p` points to some non-first element of a size_t array, of course). However, it would be legal to write to `*((size_t *) p)` as long as the allocation is valid, I think. And it would also be legal to read from there if the address represented by `p` already contains a `size_t` there, I think (??).
Note that "C" (or rather, the compiler) can't necessarily know what is the type of some value at some address, especially if the value was (or could have been) assigned or copied inside some other function. So you could try to exploit that lack of knowledge to do something that wouldn't be legal otherwise.
However, I think that is a very flawed approach and that you should never rely on this kind of reasoning, because if the compiler (or linker) chooses to inline the code of that other function or do some other kind of inter-procedural analysis (e.g. to perform optimizations) then, suddenly, it can prove things that it couldn't prove before and your code breaks [0].
And this can happen when a new compiler or linker version gets released, without any warning. And the effects can be quite subtle, silent and dangerous (with extremely important security implications, even -- a similar issue has already happened with the Linux kernel, for instance, although the circumstances were slightly different).
Although, of course, I think this is more likely to happen if you just enable link-time optimization rather than some new compiler version being released.
[0] Of course, I'm assuming that the C standard doesn't necessarily guarantee that function implementations are opaque to each other, even if they are implemented in different modules. But it's possible there is such a rule in the C standard that I'm unaware of -- I have never read the entire standard.
Do you have a source for this? Couldn't find anything on it with a quick web search.
“If the pointer P points to the ith element of an array, then the expressions P + n, n + P, and P - n are pointers of the same type that point to the i+nth, i+nth, and i-nth element of the same array, respectively. The result of pointer addition may also be a one-past-the-end pointer (that is, pointer P such that the expression P - 1 points to the last element of the array). Any other situations (that is, attempts to generate a pointer that isn't pointing at an element of the same array or one past the end) invoke undefined behavior.”
musl probably uses -fno-builtin to disable any special handling of "builtin" library functions like malloc and memcpy?
Yeah, I also noticed that is undefined behavior!
> Makes me wonder what kind of compiler flags muslibc uses to prevent this.
As far as I can see, the only related flags are `-ffreestanding`, `-fno-builtin` and `-fno-strict-aliasing` but I think none of these flags prevent the undefined behavior if 1) either the malloc() implementation gets inlined into calloc(), or 2) calloc() gets inlined into the caller, so that the compiler realizes that `p` is not actually an array of `size_t` or compatible integers.
So perhaps musl also avoids undefined behavior by forcing `malloc()` and `calloc()` to not be inlined into the caller? Although I'm not entirely sure that actually completely prevents undefined behavior, I think it would just avoid it assuming how compilers usually/currently work?
Edit: On further investigation, I think what I said above is wrong. I think `-ffreestanding` and `-fno-builtin` should be enough to prevent undefined behavior. I wasn't considering the case where `((size_t) *p)[-1]` points to a struct like this:
struct alloc {
size_t something;
unsigned char buffer[]; // flexible array member
}
In this case, then I think the code is not triggering undefined behavior when used with those compiler flags (which prevent the compiler from using its knowledge about malloc's special semantics).
Edit 2: On second thought, even then I think it might trigger undefined behavior because the C standard doesn't guarantee that there's no padding at the end of such a struct?The current code doesn't contain that undefined behavior-triggering expression: https://git.musl-libc.org/cgit/musl/tree/src/malloc/calloc.c
It then does the typical unsatisfying clang thing of unrolling and vectorizing a _lot_ with unclear benefits: https://godbolt.org/z/eG33YM7qs
> 7.24.3.2 The calloc function
> […]
> Returns
> The calloc function returns either a pointer to the allocated space or a null pointer if the space cannot be allocated or if the product nmemb * size would wraparound size_t.
Gah! "Wraparound" is not a verb! You'd think someone writing a computer language specification would try to use (human) language correctly.
> the process by which a value is reduced modulo 2ᴺ, where N is the width of the resulting type
Am I the only one that doesn’t think this is “sweet”?
I mean I know what it does and why and how it works, but IMHO a good language should let you express your intention and algorithm in human terms and logic in which there is no -1 value for an unsigned integer type.
With these functions, it could be written as: if (__builtin_mul_overflow(n, m, &n))
It seems the C committee at some point makes things harder than they should. And true, the C committee is not preventing anyone from using something else
(Another reason why we should retire C for most projects)
This seems like entirely the wrong way of looking at it to me. The C committee is not trying to make things harder than they should be, but they’re really just a committee in charge of C. Their goal isn’t to make C into some kind of beautiful safe language, it’s just to provide incremental improvements to C for the benefit of people stuck using it for whatever reason.
Like, the C committee isn’t trying to stop you from retiring C. They’re just trying to help out the people who aren’t retiring C, for whatever reason—which could be toolchain availability (lots of architectures out there, you may only have a C compiler for some), could be so they can work with a legacy codebase, could be that some other tooling or process they have works with C (the various “safe C” / analyzable subsets or formal verification processes).
However, leaving the overflow check/behavior on the multiplication as "implementation defined" seems like a bad call to me
Not everyone cares about the benchmarks, but enough do.
However in 40 years they already made their point not to care beyond make C a portable macro assembler with more UB issues than writing in raw Assembly.
The UB issues are not going away. Some people who use C need to get good performance out of it. If you try fixing the UB issues, you end up having to remove optimization passes from your C compilers. That’s just not viable—hence, UB issues are here to say.
The correct thing for the C committee is to make slow, careful improvements to C for the people who are stuck using it. And the correct choice for most people is to stop using C in favor of safer languages.
If you try to have the committee fix the safety problems with C, you’re going to end up with something which neither fast nor safe. If you are stuck using C and need it to be safe, you can use formal methods. If you don’t have the budget for formal methods, you need safety, and you’re stuck on a system which only has a C toolchain, then the alternative is to use something which compiles to C.
if (n && m > SIZE_MAX / n)I'd rewrite it as `if (n != 0 && m > SIZE_MAX / n)`. Seems legit to me.
Furthermore, optimisation can't bite you if you test the values before you perform the potentially UB operation (and unsigned operations can never overflow anyway, they are always well-defined).
Edit: inverted meaning since check is for error, not for ok case.
p = m n
Then it would be that (by substitution)
m n > SIZE_MAX
m > SIZE_MAX/n
And (size_t)-1 is a silly way to write SIZE_MAX (but it saves you a dependency).
Shouldn't the case of wraparound be caught by testing the resulting size as following?
s = n*m;
if (n && s >= n && s >= m) /* no overflow */ ...