The C++ Programming Language 3rd edition by Stroustrup says that,
Subtraction of pointers is defined only when both pointers point to
elements of the same array (although the language has no fast way of
ensuring that is the case). When subtracting one pointer from another,
the result is the number of array elements between the two pointers
(an integer). One can add an integer to a pointer or subtract an
integer from a pointer; in both cases, the result is a pointer value.
If that value does not point to an element of the same array as the
original pointer or one beyond, the result of using that value is
undefined.
For example:
void f ()
{
int v1 [10];
int v2 [10];
int i1 = &v1[5] - &v1[3]; // i1 = 2
int i2 = &v1[5] - &v2[3]; // result undefined
}
I was reading about unspecified behavior on Wikipedia. It says that
In C and C++, the comparison of pointers to objects is only strictly defined if the pointers point to members of the same object, or elements of the same array.
Example:
int main(void)
{
int a = 0;
int b = 0;
return &a < &b; /* unspecified behavior in C++, undefined in C */
}
So, I am confused. Which one is correct? Wikipedia or Stroustrup's book? What C++ standard says about this?
Correct me If I am misunderstanding something.
Note that pointer subtraction and pointer comparison are different operations with different rules.
C++14 5.6/6, on subtracting pointers:
Unless both pointers point to elements of the same array object or one past the last element of the array object, the behavior is undefined.
C++14 5.9/3-4:
Comparing pointers to objects is defined as follows:
If two pointers point to different elements of the same array, or to subobjects thereof, the pointer to the element with the higher subscript compares greater.
If one pointer points to an element of an array, or to a subobject thereof, and another pointer points one past the last element of the array, the latter pointer compares greater.
If two pointers point to different non-static data members of the same object, or to subobjects of such members, recursively, the pointer to the later declared member compares greater provided the two members have the same access control and provided their class is not a union.
If two operands p and q compare equal (5.10), p<=q and p>=q both yield true and p<q and p>q both yield false. Otherwise, if a pointer p compares greater than a pointer q, p>=q, p>q, q<=p, and q<p all yield true, and p<=q, p<q, q>=p, and q>p all yield false. Otherwise, the result of each of the operators is unspecified.
I ran into a new warning message after a compiler upgrade.
warning: ordered comparison of pointer with integer zero [-Wextra]
if (inx > 0)
As it turns out inx is a pointer. Normally I would expect to see this old code compared against 0, or NULL. This got me to thinking about signed and unsigned values, and possible risk.
A bit of research suggests:
A pointer greater than zero in c++, what does mean?
Can a pointer (address) ever be negative?
memory address positive or negative value in c?
malloc returns negative value
These seem to suggest that an address (returned by malloc) can never be zero
Which took me to my old copy of the standard.
4.10 Pointer conversions
1 A null pointer constant is an integral constant expression (5.19) prvalue of integer type that evaluates to zero
or a prvalue of type std::nullptr_t. A null pointer constant can be converted to a pointer type; the result
is the null pointer value of that type and is distinguishable from every other value of pointer to object or
pointer to function type. Such a conversion is called a null pointer conversion. Two null pointer values of the
same type shall compare equal. The conversion of a null pointer constant to a pointer to cv-qualified type is
a single conversion, and not the sequence of a pointer conversion followed by a qualification conversion (4.4).
A null pointer constant of integral type can be converted to a prvalue of type std::nullptr_t.
It specifically states that two null pointers compare equal.
With that in mind, is that little piece of code undefined behavior? or is there another piece to the puzzle I am missing?
It's not undefined behaviour, but the result is unspecified if inx is not null.
C++11 5.9/2: If two pointers p and q of the same type point to different objects that are not members of the same object or elements of the same array or to different functions, or if only one of them is null, the results of p<q, p>q, p<=q, and p>=q are unspecified.
So you can be sure that the the conditional code won't execute if inx is null - but not that it will if it's not null. The comparison should probably be inx != 0, which is well defined to be true if and only if inx is non-null.
You're looking at pointer conversions, but you should be looking at pointer comparisons.
Specifically, comparisons between pointers that don't refer to (subobjects of) the same array or object.
Section 5.9, paragraphs 3 and 4, this wording is found in C++14 drafts.
Comparing pointers to objects is defined as follows:
If two pointers point to different elements of the same array, or to subobjects thereof, the pointer to the element with the higher subscript compares greater.
If one pointer points to an element of an array, or to a subobject thereof, and another pointer points one past the last element of the array, the latter pointer compares greater.
If two pointers point to different non-static data members of the same object, or to subobjects of such members, recursively, the pointer to the later declared member compares greater provided the two members have the same access control (Clause 11) and provided their class is not a union.
If two operands p and q compare equal (5.10), p<=q and p>=q both yield true and p<q and p>q both yield
false. Otherwise, if a pointer p compares greater than a pointer q, p>=q, p>q, q<=p, and q<p all yield true
and p<=q, p<q, q>=p, and q>p all yield false. Otherwise, the result of each of the operators is unspecified.
In your case, no "pointer compares greater than" relationship is defined, and therefore the operators act according to their "otherwise" behavior, giving unspecified results. This comparison won't directly crash the program, but it could take either branch through the if, assuming that inx is non-null.
I know the pointer comparisons with < is allowed in C standard only when the pointers point at the same memory space (like array).
if we take an array:
int array[10];
int *ptr = &array[0];
is comparing ptr to array+10 allowed? Is the array+10 pointer considered outside the array memory and so the comparison is not allowed?
example
for(ptr=&array[0]; ptr<(array+10); ptr++) {...}
Yes, a pointer is permitted to point to the location just past the end of the array. However you aren't permitted to deference such a pointer.
C99 6.5.6/8 Additive operators (emphasis added)
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.
And, specifically for comparision operations on pointers:
C99 6.5.8/5 Relational operators
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.
Yes, that is allowed, and C++ relies heavily on it (C doesn't use it quite as much, but in C++, a very common way to denote ranges is by have a pointer (or more generally, an iterator) pointing to the first element, and another pointing one past the end of the range.
It is legal for such a pointer to exist, and to compare it against the rest of the array.
But it is not legal to ever dereference the pointer.
Some C or C++ programmers are surprised to find out that even storing an invalid pointer is undefined behavior. However, for heap or stack arrays, it's okay to store the address of one past the end of the array, which allows you to store "end" positions for use in loops.
But is it undefined behavior to form a pointer range from a single stack variable, like:
char c = 'X';
char* begin = &c;
char* end = begin + 1;
for (; begin != end; ++begin) { /* do something */ }
Although the above example is pretty useless, this might be useful in the event that some function expects a pointer range, and you have a case where you simply have a single value to pass it.
Is this undefined behavior?
This is allowed, the behavior is defined and both begin and end are safely-derived pointer values.
In the C++ standard section 5.7 ([expr.add]) paragraph 4:
For the purposes of these operators, a pointer to a nonarray object 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.
When using C a similar clause can be found in the the C99/N1256 standard section 6.5.6 paragraph 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.
As an aside, in section 3.7.4.3 ([basic.stc.dynamic.safety]) "Safely-derived pointers" there is a footnote:
This section does not impose restrictions on dereferencing pointers to memory not allocated by ::operator new. This maintains the ability of many C++ implementations to use binary libraries and components written in other languages. In particular, this applies to C binaries, because dereferencing pointers to memory allocated by malloc is not restricted.
This suggests that pointer arithmetic throughout the stack is implementation-defined behavior, not undefined behavior.
I believe that legally, you may treat a single object as an array of size one. In addition, it is most definitely legal to take a pointer one past the end of any array as long as it's not de-referenced. So I believe that it is not UB.
It is not Undefined Behavior as long as you don't dereference the invalid iterator.
You are allowed to hold a pointer to memory beyond your allocation but not allowed to dereference it.
5.7-5 of ISO14882:2011(e) states:
When an expression that has integral 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 integral 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.
Unless I overlooked something there, the addition only applies to pointers pointing to the same array. For everything else, the last sentence applies: "otherwise, the behaviour is undefined"
edit:
Indeed, when you add 5.7-4 it turns out that the operation you do is (virtually) on an array, thus the sentence does not apply:
For the purposes of these operators, a pointer to a nonarray object
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.
In general it would be undefined behaviour to point beyond the memory space, however there is an exception for "one past the end", which is valid according to the standard.
Therefore in the particular example, &c+1 is a valid pointer but cannot be safely dereferenced.
You could define c as an array of size 1:
char c[1] = { 'X' };
Then the undefined behavior would become defined behavior.
Resulting code should be identical.
I have seen it asserted several times now that the following code is not allowed by the C++ Standard:
int array[5];
int *array_begin = &array[0];
int *array_end = &array[5];
Is &array[5] legal C++ code in this context?
I would like an answer with a reference to the Standard if possible.
It would also be interesting to know if it meets the C standard. And if it isn't standard C++, why was the decision made to treat it differently from array + 5 or &array[4] + 1?
Yes, it's legal. From the C99 draft standard:
§6.5.2.1, paragraph 2:
A postfix expression followed by an expression in square brackets [] is a subscripted
designation of an element of an array object. The definition of the subscript operator []
is that E1[E2] is identical to (*((E1)+(E2))). Because of the conversion rules that
apply to the binary + operator, if E1 is an array object (equivalently, a pointer to the
initial element of an array object) and E2 is an integer, E1[E2] designates the E2-th
element of E1 (counting from zero).
§6.5.3.2, paragraph 3 (emphasis mine):
The unary & operator yields the address of its operand. If the operand has type ‘‘type’’,
the result has type ‘‘pointer to type’’. If the operand is the result of a unary * operator,
neither that operator nor the & operator is evaluated and the result is as if both were
omitted, except that the constraints on the operators still apply and the result is not an
lvalue. Similarly, if the operand is the result of a [] operator, neither the & operator nor the unary * that is implied by the [] is evaluated and the result is as if the & operator
were removed and the [] operator were changed to a + operator. Otherwise, the result is
a pointer to the object or function designated by its operand.
§6.5.6, paragraph 8:
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.
Note that the standard explicitly allows pointers to point one element past the end of the array, provided that they are not dereferenced. By 6.5.2.1 and 6.5.3.2, the expression &array[5] is equivalent to &*(array + 5), which is equivalent to (array+5), which points one past the end of the array. This does not result in a dereference (by 6.5.3.2), so it is legal.
Your example is legal, but only because you're not actually using an out of bounds pointer.
Let's deal with out of bounds pointers first (because that's how I originally interpreted your question, before I noticed that the example uses a one-past-the-end pointer instead):
In general, you're not even allowed to create an out-of-bounds pointer. A pointer must point to an element within the array, or one past the end. Nowhere else.
The pointer is not even allowed to exist, which means you're obviously not allowed to dereference it either.
Here's what the standard has to say on the subject:
5.7:5:
When an expression that has integral
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
integral 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.
(emphasis mine)
Of course, this is for operator+. So just to be sure, here's what the standard says about array subscripting:
5.2.1:1:
The expression E1[E2] is identical (by definition) to *((E1)+(E2))
Of course, there's an obvious caveat: Your example doesn't actually show an out-of-bounds pointer. it uses a "one past the end" pointer, which is different. The pointer is allowed to exist (as the above says), but the standard, as far as I can see, says nothing about dereferencing it. The closest I can find is 3.9.2:3:
[Note: for instance, the address one past the end of an array (5.7) would be considered to
point to an unrelated object of the array’s element type that might be located at that address. —end note ]
Which seems to me to imply that yes, you can legally dereference it, but the result of reading or writing to the location is unspecified.
Thanks to ilproxyil for correcting the last bit here, answering the last part of your question:
array + 5 doesn't actually
dereference anything, it simply
creates a pointer to one past the end
of array.
&array[4] + 1 dereferences
array+4 (which is perfectly safe),
takes the address of that lvalue, and
adds one to that address, which
results in a one-past-the-end pointer
(but that pointer never gets
dereferenced.
&array[5] dereferences array+5
(which as far as I can see is legal,
and results in "an unrelated object
of the array’s element type", as the
above said), and then takes the
address of that element, which also
seems legal enough.
So they don't do quite the same thing, although in this case, the end result is the same.
It is legal.
According to the gcc documentation for C++, &array[5] is legal. In both C++ and in C you may safely address the element one past the end of an array - you will get a valid pointer. So &array[5] as an expression is legal.
However, it is still undefined behavior to attempt to dereference pointers to unallocated memory, even if the pointer points to a valid address. So attempting to dereference the pointer generated by that expression is still undefined behavior (i.e. illegal) even though the pointer itself is valid.
In practice, I imagine it would usually not cause a crash, though.
Edit: By the way, this is generally how the end() iterator for STL containers is implemented (as a pointer to one-past-the-end), so that's a pretty good testament to the practice being legal.
Edit: Oh, now I see you're not really asking if holding a pointer to that address is legal, but if that exact way of obtaining the pointer is legal. I'll defer to the other answerers on that.
I believe that this is legal, and it depends on the 'lvalue to rvalue' conversion taking place. The last line Core issue 232 has the following:
We agreed that the approach in the standard seems okay: p = 0; *p; is not inherently an error. An lvalue-to-rvalue conversion would give it undefined behavior
Although this is slightly different example, what it does show is that the '*' does not result in lvalue to rvalue conversion and so, given that the expression is the immediate operand of '&' which expects an lvalue then the behaviour is defined.
I don't believe that it is illegal, but I do believe that the behaviour of &array[5] is undefined.
5.2.1 [expr.sub] E1[E2] is identical (by definition) to *((E1)+(E2))
5.3.1 [expr.unary.op] unary * operator ... the result is an lvalue referring to the object or function to which the expression points.
At this point you have undefined behaviour because the expression ((E1)+(E2)) didn't actually point to an object and the standard does say what the result should be unless it does.
1.3.12 [defns.undefined] Undefined behaviour may also be expected when this International Standard omits the description of any explicit definition of behaviour.
As noted elsewhere, array + 5 and &array[0] + 5 are valid and well defined ways of obtaining a pointer one beyond the end of array.
In addition to the above answers, I'll point out operator& can be overridden for classes. So even if it was valid for PODs, it probably isn't a good idea to do for an object you know isn't valid (much like overriding operator&() in the first place).
This is legal:
int array[5];
int *array_begin = &array[0];
int *array_end = &array[5];
Section 5.2.1 Subscripting The expression E1[E2] is identical (by definition) to *((E1)+(E2))
So by this we can say that array_end is equivalent too:
int *array_end = &(*((array) + 5)); // or &(*(array + 5))
Section 5.3.1.1 Unary operator '*': The unary * operator performs indirection: the expression to which it is applied shall be a pointer to an object type, or
a pointer to a function type and the result is an lvalue referring to the object or function to which the expression points.
If the type of the expression is “pointer to T,” the type of the result is “T.” [ Note: a pointer to an incomplete type (other
than cv void) can be dereferenced. The lvalue thus obtained can be used in limited ways (to initialize a reference, for
example); this lvalue must not be converted to an rvalue, see 4.1. — end note ]
The important part of the above:
'the result is an lvalue referring to the object or function'.
The unary operator '*' is returning a lvalue referring to the int (no de-refeference). The unary operator '&' then gets the address of the lvalue.
As long as there is no de-referencing of an out of bounds pointer then the operation is fully covered by the standard and all behavior is defined. So by my reading the above is completely legal.
The fact that a lot of the STL algorithms depend on the behavior being well defined, is a sort of hint that the standards committee has already though of this and I am sure there is a something that covers this explicitly.
The comment section below presents two arguments:
(please read: but it is long and both of us end up trollish)
Argument 1
this is illegal because of section 5.7 paragraph 5
When an expression that has integral 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 integral 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.
And though the section is relevant; it does not show undefined behavior. All the elements in the array we are talking about are either within the array or one past the end (which is well defined by the above paragraph).
Argument 2:
The second argument presented below is: * is the de-reference operator.
And though this is a common term used to describe the '*' operator; this term is deliberately avoided in the standard as the term 'de-reference' is not well defined in terms of the language and what that means to the underlying hardware.
Though accessing the memory one beyond the end of the array is definitely undefined behavior. I am not convinced the unary * operator accesses the memory (reads/writes to memory) in this context (not in a way the standard defines). In this context (as defined by the standard (see 5.3.1.1)) the unary * operator returns a lvalue referring to the object. In my understanding of the language this is not access to the underlying memory. The result of this expression is then immediately used by the unary & operator operator that returns the address of the object referred to by the lvalue referring to the object.
Many other references to Wikipedia and non canonical sources are presented. All of which I find irrelevant. C++ is defined by the standard.
Conclusion:
I am wiling to concede there are many parts of the standard that I may have not considered and may prove my above arguments wrong. NON are provided below. If you show me a standard reference that shows this is UB. I will
Leave the answer.
Put in all caps this is stupid and I am wrong for all to read.
This is not an argument:
Not everything in the entire world is defined by the C++ standard. Open your mind.
Working draft (n2798):
"The result of the unary & operator is
a pointer to its operand. The operand
shall be an lvalue or a qualified-id.
In the first case, if the type of the
expression is “T,” the type of the
result is “pointer to T.”" (p. 103)
array[5] is not a qualified-id as best I can tell (the list is on p. 87); the closest would seem to be identifier, but while array is an identifier array[5] is not. It is not an lvalue because "An lvalue refers to an object or function. " (p. 76). array[5] is obviously not a function, and is not guaranteed to refer to a valid object (because array + 5 is after the last allocated array element).
Obviously, it may work in certain cases, but it's not valid C++ or safe.
Note: It is legal to add to get one past the array (p. 113):
"if the expression P [a pointer]
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"
But it is not legal to do so using &.
Even if it is legal, why depart from convention? array + 5 is shorter anyway, and in my opinion, more readable.
Edit: If you want it to by symmetric you can write
int* array_begin = array;
int* array_end = array + 5;
It should be undefined behaviour, for the following reasons:
Trying to access out-of-bounds elements results in undefined behaviour. Hence the standard does not forbid an implementation throwing an exception in that case (i.e. an implementation checking bounds before an element is accessed). If & (array[size]) were defined to be begin (array) + size, an implementation throwing an exception in case of out-of-bound access would not conform to the standard anymore.
It's impossible to make this yield end (array) if array is not an array but rather an arbitrary collection type.
C++ standard, 5.19, paragraph 4:
An address constant expression is a pointer to an lvalue....The pointer shall be created explicitly, using the unary & operator...or using an expression of array (4.2)...type. The subscripting operator []...can be used in the creation of an address constant expression, but the value of an object shall not be accessed by the use of these operators. If the subscripting operator is used, one of its operands shall be an integral constant expression.
Looks to me like &array[5] is legal C++, being an address constant expression.
If your example is NOT a general case but a specific one, then it is allowed. You can legally, AFAIK, move one past the allocated block of memory.
It does not work for a generic case though i.e where you are trying to access elements farther by 1 from the end of an array.
Just searched C-Faq : link text
It is perfectly legal.
The vector<> template class from the stl does exactly this when you call myVec.end(): it gets you a pointer (here as an iterator) which points one element past the end of the array.