This alone is not a backdoor but it might lead to one.

This assumes a scenario where an attacker is in full control of "num".

For example: A lot of protocols have a structure like "<SIZE><PAYLOAD>" where the sending side (the attacker) specifies SIZE and the receiving side allocates a buffer of that size).

The code calling that function assumes it will never create a buffer that is larger then 256 bytes, so perhaps it is doing something like this:

    char buf[256];
    strcpy (buf, allocatebufs (ATTACKER_SIZE))

If the limit-check fails and the supplied size is in a valid range for malloc, it would be a typical buffer overflow.

Agreed; the explanation does not clarify why the first snippet is considered "totally safe" and the second one is not. How does the second snippet introduce a new security issue? A negative value for num passed into the function will always be forwarded to malloc (after going through the size_t conversion), no?

Without the cast, a negative value for num wouldn't have made it past the limit constraint as the operand `num` would've been implicitly promoted to size_t (an unsigned type) in the comparison. That promotion would've reliably caused the negative value to be outside the constraint range in this case (because C has well-defined rules for signed->unsigned conversions), or in any event (if limit was >SSIZE_MAX) preserving the intended semantics viz-a-viz malloc.

With the cast of limit to int, num is never promoted. A negative value can then reach the malloc expression, where it will be implicitly promoted to unsigned anyhow--because malloc takes a size_t argument. Worse, the constraint check had the side-effect of preventing multiplicative overflow in the malloc expression, which means if an attacker can control the value of num, they can cause malloc to return a block of memory smaller than what the caller expected (though this is contingent on arithmetic bugs in caller code elsewhere in the hypothetical program).

The real bug here is that allocatebufs takes an int argument rather than a size_t argument. And the real issue with C's arithmetic typing isn't promotions in normal expressions (promotion to unsigned almost always preserves the intended semantics of size constraints) so much as the ability to pass arguments of the wrong signedness to functions, which has the side-effect of hiding the fact that caller and callee aren't on the same page wrt value semantics and arithmetic limits.

I'm on the fence regarding C's behavior here because almost everybody in the comments instinctively were okay with casting, just that they would've casted some other expression. But the real problem here was the type of the argument to allocatebufs. It should've never been signed because negative values for object sizes are non-sensical. But because everybody's instinct is to cast, what good does stricter arithmetic type checking do for callers? Especially when simply leaving C's existing promotion rules in place, particularly implicit promotion to unsigned, would've resulted in the correct behavior. This eagerness to simply cast arithmetic values to quiet compiler diagnostics is also a problem in other languages with formally stricter type checking.

> Though this is contingent on arithmetic bugs in caller code elsewhere in the hypothetical program

That was my point. The article claims to show how easy it is to introduce such a bug in the second snippet, but that isn't true. You need to introduce more bugs to get a security vulnerability.

The negative number is multiplied by 64, so it can then underflow and become positive again.

Ah interesting... but then what could happen? Is the risk that the program could try allocating too much memory and crash? Calling this a "backdoor" seems excessive, no? -- at worst it's just a denial of service.

It's hypothetical, but other code may expect a bigger buffer than was actually allocated.

Suppose we're on a 32 bit platform and num is 0xF0000001. When multiplied by 64 (0x40), we'll end up allocating a 64 (0x40) byte buffer. But other code converting num to unsigned may be expecting a buffer large enough for 0xF0000001 64 byte records. After all, that was probably the reason for multiplying by 64.

You would be right if this were an honest mistake, but a malicious actor can take advantage of this for all kinds of heap-based attacks. https://en.wikipedia.org/wiki/Heap_overflow

If you take (1 << 63) + 1 as size (which is a large negative int64), multiply that by 64 and you get 64. So it will call malloc(64) which will succeed. If other parts do the correct unsigned comparison then you have a heap overflow.

This makes sense, but it also implies that there is no actual backdoor in TFA. The code as shown in the article is not exploitable without assuming more (actually exploitable) bugs somwhere else in the callsite (which the article doesn't mention). Or maybe we haven't figured out the actual vulnerability yet...

Well, the rest of the code is also correct on its own though. If I tell the function to allocate X bytes, it should either fail or return a buffer that has space for X bytes. If the function returns less than X bytes if X happens to be a large unsigned int (that is a negative value if interpreted signed) then that is a bug and is exploitable.

But that is not what happens in the code shown in TFA.

Passing a large value into the method shown in the article will do nothing nefarious (assuming sizeof(size_t) >= sizeof(int)). It will either return a large allocation, or, more likely, fail because the amount of requested memory is too large.

If you have a narrowing/casting bug somewhere else in your program, which BTW would produce an obvious warning, that would of course cause trouble as you have described.

And while mixing signed and unsigned arithmetic for buffer sizes is, of course, a recipe for desaster, I think it's incorrect to claim that the allocatebufs method shown in TFA has a "backdoor" because of this. I feel that is a bit like saying memcpy has a backdoor because you might get your pointer arithmetic wrong when calling it.

On Linux malloc doesn't crash if you give it a too large size. The cash only happens later once you're accessing a memory page that the kernel cannot allocate.

While malloc() won't "crash", it can and will, in some circumstances, return NULL. E.g., the following:

  printf("%p\n", malloc(0x7fffffffffffffff));

prints

  (nil)

on my amd64 machine, as there is no way for the kernel to allocate that much virtual address space. (Even if it wouldn't back it w/ physical RAM.)

The scenario being discussed in this subthread — having a negative number get inadvertently converted into a `size_t` and passed to `malloc()` is exactly the sort of way you end up getting a NULL back. But this is also not guaranteed.

It will become positive anyhow, since malloc takes a size_t as argument. My question still stands - how is the ability to coerce a program to malloc an arbitrarily large block a backdoor?

The caller of allocatebufs will probably expect that allocatebufs returns num * 64 bytes, but due to overflow it could return fewer. Thus, it could end up reading or writing out of bounds.

But how does that result in a vulnerability? At best you can call the function with garbage input (negative number) and still receive a valid buffer from malloc, no?

TFA calls this a "backdoor"; so how do you actually "get in" after you managed to get the backdoor through code review and deployed into production?

This depends, obviously, on the code calling allocatebufs. The implication is that the calling code will most likely assume that a buffer of the right size (i.e. "num" elements) has been allocated, while the real underlying buffer can be of any size depending on the low bits of "num".

For example, suppose the code was parsing user input as follows (a fairly common pattern):

    unsigned int count = read_int();
    struct buf *bufs = allocatebufs(count);
    if(!bufs) goto fail;
    for(unsigned int i=0; i<count; i++) {
        bufs[i] = read_buf();
        if(!bufs[i]) break;
    }

This code isn't really safe because it passes an unsigned int to allocatebufs, but by default you won't see a warning for this. In the previous version of the code it would work fine - reject anything above 256. In the new "fixed" code, if count = 0x20000001 (for example) this will allocate 64 bytes and proceed to read up to 34 GB of data into the buffer. (A clever attacker can probably cause read_buf to fail early to avoid running off the end of the heap).

Even if read_buf doesn't fail, an even more clever attacker may be able to still exploit a bug like this in a multithreaded environment: https://googleprojectzero.blogspot.com/2015/03/taming-wild-c...

> This depends, obviously, on the code calling allocatebufs

That was my point. The article claims to introduce a backdoor with the tiny change in the second example, i.e. "commit the change from the second example an you're in". But that just isn't true without assuming some other vulnerable code at the callsite.

And arguably, the assumed bug is a bug of the assumed callsite and not a bug of the allocatebufs method!

"See how easy it is to introduce a backdoor into C code which just one small change that looks completely harmless" might generally be true (debatable), but claiming that the change shown in the article (on it's own) is an example of this is incorrect and looks a bit like fearmongering.

You get a buffer that's too small for whatever data gets written there. That data (presumably controllable by the attacker) overwrites whatever other data is right after the allocated (too small) buffer; in many cases this can be leveraged to gain control of the program in ways that will depend the rest of the code around that backdoor.

That code doesn't need to be buggy - simple, correct code that traverses e.g. a linked list can be abused to gain arbitrary memory writes and reads if you can overwrite the pointers in that linked list with values under your control; and these arbitrary memory writes can be abused to gain arbitrary code execution.

Exploiting a heap overflow is not as straightforward as a stack overflow, but certainly possible, there are many real world code execution vulnerabilities that arise from a buffer overflow in heap.

I don't think that what you are saying is correct.

If you ask the method to allocate a negative number of bytes and the method returns a buffer which is greater than zero, that doesn't seem like a backdoor in the allocation method!

Saying you can get a return buffer that is smaller than whatever amount of bytes you requested is wrong I think. Can you give an input to the second method that will result in a buffer smaller than the input value?

Signed overflow is undefined behavior in C.

It is, but in this case the compiler will assume it never occurs and generate assembly that will function as the parent describes. Which, from a security perspective, is what matters in this case.

Just a small nit: underflow is something else; this is still overflow (or wrap-around).

My favorite example to trot out of how C behavior can lead to security problems is a Cap'n'Proto problem a couple years back, reported here.[1] While not entirely the same, there are some some similarities. The article submitted in that discussion does a good job explaining the problem, and the discussion covers even more possible issues. It pays to be very careful how these checks are done in C.

1: https://news.ycombinator.com/item?id=14163111