I came across this construction inside a function (e is a parameter passed to the function):
short (*tt)[][2] = (short (*)[][2])(heater_ttbl_map[e]);
and its use (where i is a counter inside a for loop):
(*tt)[i][0]
I think I got the first part of the assignment:
short (*tt)[][2]
for what I understand tt is declared as a pointer to an array of arrays of shorts.
The second part is confusing me though, looks like some sort of cast but I'm not sure I understand what it does, expecially this: (*). How does it work?
heater_ttbl_map is declared like this (where pointer1 and pointer2 are both bidimensional arrays of shorts):
static void *heater_ttbl_map[2] = {(void*)pointer1, (void*)pointer2};
as for its use I understand that what is pointed at by tt is dereferenced (and its the content of the third index of the i index of the array, which is a short) but why writing it like this:
(*tt)[i][0]
and not like this:
*tt[i][0]
is it because tt is not an array itself but a pointer to an array?
Due to operator precedence ([] has precedence over * operator), there is difference in two statements -
(*tt)[i][0]
In this you access the element at index [i][0] of array to which pointer tt points to .
Whereas, in this -
*tt[i][0]
First the element at index [i][0](may be 2-d array of pointers) is accessed and then dereferenced.
Using them interchangeably can cause access or dereferencing unauthorized memory location and lead to undefined behaviour.
As ameyCU explained, the [] subscript operator has higher precedence than the unary * operator, so the expression *a[i] will be parsed as *(a[i]); IOW, you're indexing into a and dereferencing the result.
This works if a is an array of T (or a pointer to T; more on that below). However, if a is a pointer to an array of T, that won't do what you want. This is probably best explained visually.
Assume the declarations:
int arr[3] = { 0, 1, 2 };
int (*parr)[3] = &arr; // type of &arr is int (*)[3], not int **
Here's what things look like in memory (sort of; addresses are pulled out of thin air):Address Item Memory cell
------- ---- -----------
+---+
0x8000 arr: | 0 | <--------+
+---+ |
0x8004 | 1 | |
+---+ |
0x8008 | 2 | |
+---+ |
... |
+---+ |
0x8080 parr: | | ----------+
+---+
...
So you see the array arr with its three elements, and the pointer parr pointing to arr. We want to access the second element of arr (value 1 at address 0x8004) through the pointer parr. What happens if we write *parr[1]?
First of all, remember that the expression a[i] is defined as *(a + i); that is, given a pointer value a1, offset i elements (not bytes) from a and dereference the result. But what does it mean to offset i elements from a?
Pointer arithmetic is based on the size of the pointed-to type; if p is a pointer to T, then p+1 will give me the location of the next object of type T. So, if p points to an int object at address 0x1000, then p+1 will give me the address of the int object following p - 0x1000 + sizeof (int).
So, if we write parr[1], what does that give us? Since parr points to a 3-element array if int, parr + 1 will give us the address of the next 3-element array of int - 0x8000 + sizeof (int [3]), or 0x800c (assuming 4-byte int type).
Remember from above that [] has higher precedence than unary *, so the expression *parr[1] will be parsed as *(parr[1]), which evaluates to *(0x800c).
That's not what we want. To access arr[1] through parr, we must make sure parr has been dereferenced before the subscript operation is applied by explicitly grouping the * operator with parentheses: (*parr)[1]. *parr evaluates to 0x8000 which has type "3-element array of int"; we then access the second element of that array (0x8000 + sizeof (int), or 0x8004) to get the desired value.
Now, let's look at something - if a[i] is equivalent to *(a+i), then it follows that a[0] is equivalent to *a. That means we can write (*parr)[1] as (parr[0])[1], or just parr[0][1]. Now, you don't want to do that for this case since parr is just a pointer to a 1D array, not a 2D array. But this is how 2D array indexing works. Given a declaration like T a[M][N];, the expression a will "decay" to type T (*)[N] in most circumstances. If I wrote something like
int arr[3][2] = {{1,2},{3,4},{5,6}};
int (*parr)[2] = arr; // don't need the & this time, since arr "decays" to type
// int (*)[2]
then to access an element of arr through parr, all I need to do is write parr[i][j]; parr[i] implicitly dereferences the parr pointer.
This is where things get confusing; arrays are not pointers, and they don't store any pointers internally. Instead, of an array expression is not the operand of the sizeof or unary * operators, its type is converted from "N-element array of T" to "pointer to T", and the value of the expression is the address of the first element of the array. This is why you can use the [] operator on both array and pointer objects.
This is also why we used the & operator to get the address of arr in our code snippet; if it's not the operand of the `&` operator, the expression "decays" from type "3-element array of int" to "pointer to int"
void DoSomeThing(CHAR parm[])
{
}
int main()
{
DoSomeThing(NULL);
}
Is the passing NULL for array parameter allowed in C/C++?
What you cannot do is pass an array by value. The signature of the function
void f( char array[] );
is transformed into:
void f( char *array );
and you can always pass NULL as argument to a function taking a pointer.
You can however pass an array by reference in C++, and in that case you will not be able to pass a NULL pointer (or any other pointer):
void g( char (&array)[ 10 ] );
Note that the size of the array is part of the type, and thus part of the signature, which means that g will only accept lvalue-expressions of type array of 10 characters
Short answer is yes, you can pass NULL in this instance (at least for C, and I think the same is true for C++).
There are two reasons for this. First, in the context of a function parameter declaration, the declarations T a[] and T a[N] are synonymous with T *a; IOW, despite the array notation, a is declared as a pointer to T, rather than an array of T. From the C language standard:
6.7.5.3 Function declarators (including prototypes)
...
7 A declaration of a parameter as ‘‘array of type’’ shall be adjusted to ‘‘qualified pointer to
type’’, where the type qualifiers (if any) are those specified within the [ and ] of the
array type derivation. If the keyword static also appears within the [ and ] of the
array type derivation, then for each call to the function, the value of the corresponding
actual argument shall provide access to the first element of an array with at least as many
elements as specified by the size expression.
The second reason is that when an expression of array type appears in most contexts (such as in a function call), the type of that expression is implicitly converted ("decays") to a pointer type, so what actually gets passed to the function is a pointer value, not an array:
6.3.2.1 Lvalues, arrays, and function designators
...
3 Except when it is the operand of the sizeof operator or the unary & operator, or is a
string literal used to initialize an array, an expression that has type ‘‘array of type’’ is
converted to an expression with type ‘‘pointer to type’’ that points to the initial element of
the array object and is not an lvalue. If the array object has register storage class, the
behavior is undefined.
First off, what you're passing to the function is a pointer and not an array. The array notation used for the parm parameter is just syntactic sugar for CHAR *parm.
So yes, you can pass a NULL pointer.
The only issue would be if DoSomeThing were overloaded, in which case the compiler might not be able to distinguish which function to call. But casting to the appropriate type would theoretically fix that.
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.