Pointers Are More Abstract Than You Might Expect in C

c

language-lawyer

pointer

A pointer references a location in memory and dereferencing a pointer refers to the lookup of the value of the memory location the pointer references. The value of a pointer is a memory address. The C standard does not define the representation of a memory address. This is crucial since not every architecture makes use of the same memory addressing paradigm. Most modern architectures make use of a linear address space or something similar. Still, even this is not precise enough since you might want to talk about physical or virtual addresses. Some architectures make even use of non-numeric addresses. For example, the Symbolics Lisp Machine 3600 makes use of tuples of the form <array-object, index> as addresses.

The representation of a pointer is not defined by the C standard. However, operations involving pointers are defined—at least more or less. In the following we will have a look at these operations and how they are defined. Lets start with an introductory example:

#include <stdio.h> int main(void) { int a, b; int *p = &a; int *q = &b + 1; printf("%p %p %d

", (void *)p, (void *)q, p == q); return 0; }

If compiled with GCC 8.1.1 and optimization level 1, then a run of the program on a x86-64 Linux system prints:

0x7fff4a35b19c 0x7fff4a35b19c 0

Note that the pointers p and q point to the same memory address. Still the expression p == q evaluates to false which is very surprising at first. Wouldn’t one expect that if two pointers point to the same memory address, then they should compare equal?

The C standard defines the behavior for comparing two pointers for equality as follows:

C11 § 6.5.9 Equality operators ¶ 6 Two pointers compare equal if and only if both are null pointers, both are pointers to the same object (including a pointer to an object and a subobject at its beginning) or function, both are pointers to one past the last element of the same array object, or one is a pointer to one past the end of one array object and the other is a pointer to the start of a different array object that happens to immediately follow the first array object in the address space.

The first question which probably comes up is: What is an object ? Since we consider the language C it has certainly nothing to do with objects as known from object oriented programming languages like C++. The C standard defines an object rather informally as:

C11 § 3.15 object

region of data storage in the execution environment, the contents of which can represent values NOTE When referenced, an object may be interpreted as having a particular type; see 6.3.2.1.

Lets be nit-picky. A 16 bit integer variable in memory is a data storage and can represent 16 bit integer values. Therefore it is an object. Should two pointers compare equal if the first pointer points to the first byte of the integer and the second pointer to the second byte of the integer? Of course this is not what the language committee intended. But at that point we should note that the language is not formally defined and we have to start guessing what the intention of the language committee was.

When the Compiler Gets Into Your Way

Lets get back to our introductory example. Pointer p is derived from object a and pointer q is derived from object b . The latter involves pointer arithmetics and this is defined for the operators plus and minus as follows:

C11 § 6.5.6 Additive operators ¶ 7 For the purposes of these operators, a pointer to an object that is not an element of an array behaves the same as a pointer to the first element of an array of length one with the type of the object as its element type.

Since every pointer which points to a non-array object is virtually lifted to a pointer of type array of length one, the C standard only defines pointer arithmetics for pointers of array types which is finally given in paragraph 8. The interesting part for our case is:

C11 § 6.5.6 Additive operators ¶ 8 […] if the expression P points to the last element of an array object, the expression (P)+1 points one past the last element of the array object […] the evaluation shall not produce an overflow […] When an expression that has integer type is added to or subtracted from a pointer, the result has the type of the pointer operand. If the pointer operand points to an element of an array object, and the array is large enough, the result points to an element offset from the original element such that the difference of the subscripts of the resulting and original array elements equals the integer expression. In other words, if the expression P points to the i-th element of an array object, the expressions (P)+N (equivalently, N+(P) ) and (P)-N (where N has the value n) point to, respectively, the i+n-th and i−n-th elements of the array object, provided they exist. Moreover, if the expression P points to the last element of an array object, the expression (P)+1 points one past the last element of the array object, and if the expression Q points one past the last element of an array object, the expression (Q)-1 points to the last element of the array object. If both the pointer operand and the result point to elements of the same array object, or one past the last element of the array object, the evaluation shall not produce an overflow; otherwise, the behavior is undefined. If the result points one past the last element of the array object, it shall not be used as the operand of a unary * operator that is evaluated.

That means, the expression &b + 1 should evaluate to an address without any problem. Hence p and q should be valid pointers. Recap what the C standard defines for comparing two pointers: Two pointers compare equal if and only if […] one is a pointer to one past the end of one array object and the other is a pointer to the start of a different array object that happens to immediately follow the first array object in the address space (C11 § 6.5.9 ¶ 6). This is exactly the case in our example. Pointer q points one past the end of object b which is immediately followed by object a to which p points. Is this a bug in GCC? The finding has been reported in 2014 as bug #61502 and so far the GCC people argue that this is not a bug and therefore won’t fix it.

The Linux people ran into a similar problem in 2016. Consider the following code:

extern int _start[]; extern int _end[]; void foo(void) { for (int *i = _start; i != _end; ++i) { /* ... */ } }

The symbols _start and _end are used to span a memory region. Since the symbols are externalized, the compiler does not know where the arrays are actually allocated in memory. Therefore, the compiler must be conservative at this point and assume that they may be allocated next to each other in the address space. Unfortunately GCC compiled the loop condition into the constant true rendering the loop into an endless loop as described in this LKML post where they make use of a similar code snippet. It looks like that GCC changed its behavior according to this problem. At least I couldn’t reconstruct the behavior with GCC version 7.3.1 on x86_64 Linux.

Defect Report #260 to the Rescue?

Defect report 260 may apply in our case. The topic of the report is more about indeterminate values, however, there is one interesting response from the committee:

Implementations […] may also treat pointers based on different origins as distinct even though they are bitwise identical.

If we take this literally, then it is sound that p == q evaluates to false, since p and q are derived from distinct objects that are in no relation to each other. It looks like we are getting closer and closer to the truth, or do we? So far we only considered operators for equality but what about relational operators?

Relational Operators to the Final Rescue?

An interesting point is made while defining the semantics of the relational operators < , <= , > , and >= , in order to compare pointers:

C11 § 6.5.8 Relational operators ¶ 5 When two pointers are compared, the result depends on the relative locations in the address space of the objects pointed to. If two pointers to object types both point to the same object, or both point one past the last element of the same array object, they compare equal. If the objects pointed to are members of the same aggregate object, pointers to structure members declared later compare greater than pointers to members declared earlier in the structure, and pointers to array elements with larger subscript values compare greater than pointers to elements of the same array with lower subscript values. All pointers to members of the same union object compare equal. If the expression P points to an element of an array object and the expression Q points to the last element of the same array object, the pointer expression Q+1 compares greater than P . In all other cases, the behavior is undefined.

According to this definition comparing pointers is only defined behavior if the pointers are derived from the same object. Lets demonstrate the idea of this by two examples.

int *p = malloc(64 * sizeof(int)); int *q = malloc(64 * sizeof(int)); if (p < q) // undefined behavior foo();

In this example the pointers p and q point into two different objects which are not related to each other. Hence comparing them is undefined behavior. Whereas in the following example

int *p = malloc(64 * sizeof(int)); int *q = p + 42; if (p < q) foo();

the pointer p and q point into the same object and are therefore related. Hence it is sound to compare them—assuming that malloc does not return the null pointer.

Storage Layout

So far we didn’t examine the standard w. r. t. the storage layout of objects. Lets consider objects of aggregate types first. An aggregate type is either a structure or an array type. The former is a sequentially allocated nonempty set of member objects. The only guarantee we get for members of a structure is that they are sequentially allocated in the given order. Thus, a compiler is not allowed to reorder members. However, nothing is said about the space between adjacent members. There we have that arbitrarily many padding bits may be added. For example, consider the following structure: struct { char a; int b; } x; . On most modern architectures between members a and b several padding bits are introduced—leaving it open how many these are since this depends on the alignment requirements of the type int . Therefore, deriving pointers from x.a and x.b and comparing them for equality results in undefined behavior whereas comparing them for relation as e.g. &x.a < &x.b results in defined behavior.

For array types we have that these describe a contiguously allocated nonempty set of objects. The crucial point is that in contrast to structure members array members are contiguously allocated. Thus not only the ordering of array members is defined but also that adjacent members are allocated without space in-between. This enables us to finally perform pointer arithmetics in a well defined manner on array members.

For all other types, i.e., non-aggregate types, we have that the standard does not define the corresponding storage layout. Hence for our introductory example the storage layout of the variables a and b is not defined. Therefore, deriving pointers from the variables and comparing them results in undefined behavior. GCC exploits this fact and evaluates the expression p == q statically to false. The assembler output for the introductory example is given as follows if compiled with optimization level 1:

.LC0: .string "%p %p %d

" main: sub rsp, 24 mov ecx, 0 lea rdx, [rsp+12] mov rsi, rdx mov edi, OFFSET FLAT:.LC0 mov eax, 0 call printf mov eax, 0 add rsp, 24 ret

The expression p == q is compiled into the assembler instruction mov ecx, 0 .

Different Array Objects

It looks like we are getting closer and closer to the truth ;-) The most problematic part we stumbled across so far was in section 6.5.9 paragraph 6 where it is explicitly allowed to compare two pointers from two different array objects. Lets be philosophical. What are different array objects? According to the wording used in the standard each dimension of a multidimensional array is an array on itself. A modified version of our introductory example containing a multidimensional array is given as follows:

#include <stdio.h> int main(void) { int x[2][1]; int *p = &x[0][1]; int *q = &x[1][0]; printf("%p %p %d

", (void *)p, (void *)q, p == q); return 0; }

Pointer p points one past the last element of an array object which is part of a multidimensional array object. Pointer q points to the first element of an array object which is adjacent to the array object from which p is derived from. Since both arrays are part of a multidimensional array it is defined behavior to compare p and q for equality. Thus p == q always evaluates to true. GCC and Clang evaluate the expression at compile time to true, i.e., emit the assembler instruction mov ecx, 1 for all optimization levels but 0.

The important part in the example is that &x[0] points to a different array object than &x[1] . However, this is not explicitly stated in the C11 standard but is written between the lines.

Wrap-up

We started with an innocent looking example and stumbled across several pitfalls which led to undefined behavior. Our introductory example has the same problem as the example from the Linux people: Comparing two pointers which are derived from to completely unrelated objects invokes undefined behavior. It does not matter if the objects have external or internal linkage, or if they have automatic storage duration or not.

The most problematic part was in section 6.5.9 paragraph 6 where it is explicitly allowed to compare two pointers from two different array objects. At this point in time I would have expected at least a single sentence stating that both pointers must be derived from two arrays which are subaggregates of the same multidimensional array. The wording became even more confusing in section 6.5.8 paragraph 5 where the relational operators are defined. There the standard only speaks of pointers to the same array object.

In my humble opinion speaking of different arrays for each dimension of a multidimensional array is misleading. Philosophically speaking isn’t an element of an array object which is a subaggregate of a multidimensional array object also an element of the multidimensional array object? If so, then two elements e1,e2 of two different array objects a1,a2 which are subaggregates of the same multidimensional array object x are also two elements of the same multidimensional array object x . Then two pointers p1,p2 pointing to the elements e1,e2 also point to different array objects a1,a2 and simultaneously to the same array object x . Thus, same and different become superfluous and confuse more than help.

The overall feeling regarding the wording of the C11 standard is unsatisfactory w. r. t. the presented problem. Since several people already stumbled across this, the question which is left is: Why not make the wording more precise?

The takeaway message is that pointer arithmetic is only defined for pointers pointing into array objects or one past the last element. Comparing pointers for equality is defined if both pointers are derived from the same (multidimensional) array object. Thus, if two pointers point to different array objects, then these array objects must be subaggregates of the same multidimensional array object in order to compare them. Otherwise this leads to undefined behavior.

If you are interested in related work I can recommend this one: Clarifying the C memory object model (n2012)

All experiments were done with GCC 8.1.1 and Clang 6.0.0 using options -std=c11 -O1 on an x86_64 Linux distribution.

Addendum

Pointers One Past the Last Element of an Array

If we lookup the C11 standard and read about pointer arithmetics and comparison we find exceptions for pointers which point one past the last element of an array all over the place. Assume it wouldn’t be allowed to compare two pointers derived from the same array object where at least one pointer points one element past of the array, then code like this

const int num = 64; int x[num]; for (int *i = x; i < &x[num]; ++i) { /* ... */ }

would not work. Via the loop we iterate over the array x consisting of 64 elements, i.e., the loop body should be evaluated exactly 64 times. However, the loop condition gets evaluated 65 times—once more than we have array elements. In the first 64 evaluations, the pointer i always points into the array x whereas the expression &x[num] always points one element past the array. In the 65th iteration the pointer i also points one element past the array x rendering the condition of the loop false. This is a convenient way to iterate over an array which makes the exception for arrays feasible. Note, the standard only defines the behavior of comparing such pointer—dereferencing pointer is another topic.

Can we change the example such that no pointer points one past the last element of array x ? Well, the solution to that is not straight forward. We have to change the loop condition and also make sure that at the end of the loop we do not increment i anymore.

const int num = 64; int x[num]; for (int *i = x; i <= &x[num-1]; ++i) { /* ... */ if (i == &x[num-1]) break; }

This code is rather cluttered with technical details which we do not want to deal with and which distracts us from the actual job we want to accomplish. Despite that it also contains one additional branch inside the loop body. Hence, I think it is reasonable to have exceptions for pointers one past the last element of an array.