I've recently decided that I just have to finally learn C/C++, and there is one thing I do not really understand about pointers or more precisely, their definition.
How about these examples:
int* test;
int *test;
int * test;
int* test,test2;
int *test,test2;
int * test,test2;
Now, to my understanding, the first three cases are all doing the same: Test is not an int, but a pointer to one.
The second set of examples is a bit more tricky. In case 4, both test and test2 will be pointers to an int, whereas in case 5, only test is a pointer, whereas test2 is a "real" int. What about case 6? Same as case 5?
4, 5, and 6 are the same thing, only test is a pointer. If you want two pointers, you should use:
int *test, *test2;
Or, even better (to make everything clear):
int* test;
int* test2;
White space around asterisks have no significance. All three mean the same thing:
int* test;
int *test;
int * test;
The "int *var1, var2" is an evil syntax that is just meant to confuse people and should be avoided. It expands to:
int *var1;
int var2;
Many coding guidelines recommend that you only declare one variable per line. This avoids any confusion of the sort you had before asking this question. Most C++ programmers I've worked with seem to stick to this.
A bit of an aside I know, but something I found useful is to read declarations backwards.
int* test; // test is a pointer to an int
This starts to work very well, especially when you start declaring const pointers and it gets tricky to know whether it's the pointer that's const, or whether its the thing the pointer is pointing at that is const.
int* const test; // test is a const pointer to an int
int const * test; // test is a pointer to a const int ... but many people write this as
const int * test; // test is a pointer to an int that's const
Use the "Clockwise Spiral Rule" to help parse C/C++ declarations;
There are three simple steps to follow:
Starting with the unknown element, move in a spiral/clockwise
direction; when encountering the following elements replace them with
the corresponding english statements:
[X] or []: Array X size of... or Array undefined size of...
(type1, type2): function passing type1 and type2 returning...
*: pointer(s) to...
Keep doing this in a spiral/clockwise direction until all tokens have been covered.
Always resolve anything in parenthesis first!
Also, declarations should be in separate statements when possible (which is true the vast majority of times).
There are three pieces to this puzzle.
The first piece is that whitespace in C and C++ is normally not significant beyond separating adjacent tokens that are otherwise indistinguishable.
During the preprocessing stage, the source text is broken up into a sequence of tokens - identifiers, punctuators, numeric literals, string literals, etc. That sequence of tokens is later analyzed for syntax and meaning. The tokenizer is "greedy" and will build the longest valid token that's possible. If you write something like
inttest;
the tokenizer only sees two tokens - the identifier inttest followed by the punctuator ;. It doesn't recognize int as a separate keyword at this stage (that happens later in the process). So, for the line to be read as a declaration of an integer named test, we have to use whitespace to separate the identifier tokens:
int test;
The * character is not part of any identifier; it's a separate token (punctuator) on its own. So if you write
int*test;
the compiler sees 4 separate tokens - int, *, test, and ;. Thus, whitespace is not significant in pointer declarations, and all of
int *test;
int* test;
int*test;
int * test;
are interpreted the same way.
The second piece to the puzzle is how declarations actually work in C and C++1. Declarations are broken up into two main pieces - a sequence of declaration specifiers (storage class specifiers, type specifiers, type qualifiers, etc.) followed by a comma-separated list of (possibly initialized) declarators. In the declaration
unsigned long int a[10]={0}, *p=NULL, f(void);
the declaration specifiers are unsigned long int and the declarators are a[10]={0}, *p=NULL, and f(void). The declarator introduces the name of the thing being declared (a, p, and f) along with information about that thing's array-ness, pointer-ness, and function-ness. A declarator may also have an associated initializer.
The type of a is "10-element array of unsigned long int". That type is fully specified by the combination of the declaration specifiers and the declarator, and the initial value is specified with the initializer ={0}. Similarly, the type of p is "pointer to unsigned long int", and again that type is specified by the combination of the declaration specifiers and the declarator, and is initialized to NULL. And the type of f is "function returning unsigned long int" by the same reasoning.
This is key - there is no "pointer-to" type specifier, just like there is no "array-of" type specifier, just like there is no "function-returning" type specifier. We can't declare an array as
int[10] a;
because the operand of the [] operator is a, not int. Similarly, in the declaration
int* p;
the operand of * is p, not int. But because the indirection operator is unary and whitespace is not significant, the compiler won't complain if we write it this way. However, it is always interpreted as int (*p);.
Therefore, if you write
int* p, q;
the operand of * is p, so it will be interpreted as
int (*p), q;
Thus, all of
int *test1, test2;
int* test1, test2;
int * test1, test2;
do the same thing - in all three cases, test1 is the operand of * and thus has type "pointer to int", while test2 has type int.
Declarators can get arbitrarily complex. You can have arrays of pointers:
T *a[N];
you can have pointers to arrays:
T (*a)[N];
you can have functions returning pointers:
T *f(void);
you can have pointers to functions:
T (*f)(void);
you can have arrays of pointers to functions:
T (*a[N])(void);
you can have functions returning pointers to arrays:
T (*f(void))[N];
you can have functions returning pointers to arrays of pointers to functions returning pointers to T:
T *(*(*f(void))[N])(void); // yes, it's eye-stabby. Welcome to C and C++.
and then you have signal:
void (*signal(int, void (*)(int)))(int);
which reads as
signal -- signal
signal( ) -- is a function taking
signal( ) -- unnamed parameter
signal(int ) -- is an int
signal(int, ) -- unnamed parameter
signal(int, (*) ) -- is a pointer to
signal(int, (*)( )) -- a function taking
signal(int, (*)( )) -- unnamed parameter
signal(int, (*)(int)) -- is an int
signal(int, void (*)(int)) -- returning void
(*signal(int, void (*)(int))) -- returning a pointer to
(*signal(int, void (*)(int)))( ) -- a function taking
(*signal(int, void (*)(int)))( ) -- unnamed parameter
(*signal(int, void (*)(int)))(int) -- is an int
void (*signal(int, void (*)(int)))(int); -- returning void
and this just barely scratches the surface of what's possible. But notice that array-ness, pointer-ness, and function-ness are always part of the declarator, not the type specifier.
One thing to watch out for - const can modify both the pointer type and the pointed-to type:
const int *p;
int const *p;
Both of the above declare p as a pointer to a const int object. You can write a new value to p setting it to point to a different object:
const int x = 1;
const int y = 2;
const int *p = &x;
p = &y;
but you cannot write to the pointed-to object:
*p = 3; // constraint violation, the pointed-to object is const
However,
int * const p;
declares p as a const pointer to a non-const int; you can write to the thing p points to
int x = 1;
int y = 2;
int * const p = &x;
*p = 3;
but you can't set p to point to a different object:
p = &y; // constraint violation, p is const
Which brings us to the third piece of the puzzle - why declarations are structured this way.
The intent is that the structure of a declaration should closely mirror the structure of an expression in the code ("declaration mimics use"). For example, let's suppose we have an array of pointers to int named ap, and we want to access the int value pointed to by the i'th element. We would access that value as follows:
printf( "%d", *ap[i] );
The expression *ap[i] has type int; thus, the declaration of ap is written as
int *ap[N]; // ap is an array of pointer to int, fully specified by the combination
// of the type specifier and declarator
The declarator *ap[N] has the same structure as the expression *ap[i]. The operators * and [] behave the same way in a declaration that they do in an expression - [] has higher precedence than unary *, so the operand of * is ap[N] (it's parsed as *(ap[N])).
As another example, suppose we have a pointer to an array of int named pa and we want to access the value of the i'th element. We'd write that as
printf( "%d", (*pa)[i] );
The type of the expression (*pa)[i] is int, so the declaration is written as
int (*pa)[N];
Again, the same rules of precedence and associativity apply. In this case, we don't want to dereference the i'th element of pa, we want to access the i'th element of what pa points to, so we have to explicitly group the * operator with pa.
The *, [] and () operators are all part of the expression in the code, so they are all part of the declarator in the declaration. The declarator tells you how to use the object in an expression. If you have a declaration like int *p;, that tells you that the expression *p in your code will yield an int value. By extension, it tells you that the expression p yields a value of type "pointer to int", or int *.
So, what about things like cast and sizeof expressions, where we use things like (int *) or sizeof (int [10]) or things like that? How do I read something like
void foo( int *, int (*)[10] );
There's no declarator, aren't the * and [] operators modifying the type directly?
Well, no - there is still a declarator, just with an empty identifier (known as an abstract declarator). If we represent an empty identifier with the symbol λ, then we can read those things as (int *λ), sizeof (int λ[10]), and
void foo( int *λ, int (*λ)[10] );
and they behave exactly like any other declaration. int *[10] represents an array of 10 pointers, while int (*)[10] represents a pointer to an array.
And now the opinionated portion of this answer. I am not fond of the C++ convention of declaring simple pointers as
T* p;
and consider it bad practice for the following reasons:
It's not consistent with the syntax;
It introduces confusion (as evidenced by this question, all the duplicates to this question, questions about the meaning of T* p, q;, all the duplicates to those questions, etc.);
It's not internally consistent - declaring an array of pointers as T* a[N] is asymmetrical with use (unless you're in the habit of writing * a[i]);
It cannot be applied to pointer-to-array or pointer-to-function types (unless you create a typedef just so you can apply the T* p convention cleanly, which...no);
The reason for doing so - "it emphasizes the pointer-ness of the object" - is spurious. It cannot be applied to array or function types, and I would think those qualities are just as important to emphasize.
In the end, it just indicates confused thinking about how the two languages' type systems work.
There are good reasons to declare items separately; working around a bad practice (T* p, q;) isn't one of them. If you write your declarators correctly (T *p, q;) you are less likely to cause confusion.
I consider it akin to deliberately writing all your simple for loops as
i = 0;
for( ; i < N; )
{
...
i++;
}
Syntactically valid, but confusing, and the intent is likely to be misinterpreted. However, the T* p; convention is entrenched in the C++ community, and I use it in my own C++ code because consistency across the code base is a good thing, but it makes me itch every time I do it.
I will be using C terminology - the C++ terminology is a little different, but the concepts are largely the same.
As others mentioned, 4, 5, and 6 are the same. Often, people use these examples to make the argument that the * belongs with the variable instead of the type. While it's an issue of style, there is some debate as to whether you should think of and write it this way:
int* x; // "x is a pointer to int"
or this way:
int *x; // "*x is an int"
FWIW I'm in the first camp, but the reason others make the argument for the second form is that it (mostly) solves this particular problem:
int* x,y; // "x is a pointer to int, y is an int"
which is potentially misleading; instead you would write either
int *x,y; // it's a little clearer what is going on here
or if you really want two pointers,
int *x, *y; // two pointers
Personally, I say keep it to one variable per line, then it doesn't matter which style you prefer.
#include <type_traits>
std::add_pointer<int>::type test, test2;
In 4, 5 and 6, test is always a pointer and test2 is not a pointer. White space is (almost) never significant in C++.
The rationale in C is that you declare the variables the way you use them. For example
char *a[100];
says that *a[42] will be a char. And a[42] a char pointer. And thus a is an array of char pointers.
This because the original compiler writers wanted to use the same parser for expressions and declarations. (Not a very sensible reason for a langage design choice)
I would say that the initial convention was to put the star on the pointer name side (right side of the declaration
in the c programming language by Dennis M. Ritchie the stars are on the right side of the declaration.
by looking at the linux source code at https://github.com/torvalds/linux/blob/master/init/main.c
we can see that the star is also on the right side.
You can follow the same rules, but it's not a big deal if you put stars on the type side.
Remember that consistency is important, so always but the star on the same side regardless of which side you have choose.
In my opinion, the answer is BOTH, depending on the situation.
Generally, IMO, it is better to put the asterisk next to the pointer name, rather than the type. Compare e.g.:
int *pointer1, *pointer2; // Fully consistent, two pointers
int* pointer1, pointer2; // Inconsistent -- because only the first one is a pointer, the second one is an int variable
// The second case is unexpected, and thus prone to errors
Why is the second case inconsistent? Because e.g. int x,y; declares two variables of the same type but the type is mentioned only once in the declaration. This creates a precedent and expected behavior. And int* pointer1, pointer2; is inconsistent with that because it declares pointer1 as a pointer, but pointer2 is an integer variable. Clearly prone to errors and, thus, should be avoided (by putting the asterisk next to the pointer name, rather than the type).
However, there are some exceptions where you might not be able to put the asterisk next to an object name (and where it matters where you put it) without getting undesired outcome — for example:
MyClass *volatile MyObjName
void test (const char *const p) // const value pointed to by a const pointer
Finally, in some cases, it might be arguably clearer to put the asterisk next to the type name, e.g.:
void* ClassName::getItemPtr () {return &item;} // Clear at first sight
The pointer is a modifier to the type. It's best to read them right to left in order to better understand how the asterisk modifies the type. 'int *' can be read as "pointer to int'. In multiple declarations you must specify that each variable is a pointer or it will be created as a standard variable.
1,2 and 3) Test is of type (int *). Whitespace doesn't matter.
4,5 and 6) Test is of type (int *). Test2 is of type int. Again whitespace is inconsequential.
I have always preferred to declare pointers like this:
int* i;
I read this to say "i is of type int-pointer". You can get away with this interpretation if you only declare one variable per declaration.
It is an uncomfortable truth, however, that this reading is wrong. The C Programming Language, 2nd Ed. (p. 94) explains the opposite paradigm, which is the one used in the C standards:
The declaration of the pointer ip,
int *ip;
is intended as a mnemonic; it says that the expression *ip is an
int. The syntax of the declaration for a variable mimics the syntax
of expressions in which the variable might appear. This reasoning
applies to function declarations as well. For example,
double *dp, atof(char *);
says that in an expression *dp and atof(s) have values of type
double, and that the argument of atof is a pointer to char.
So, by the reasoning of the C language, when you declare
int* test, test2;
you are not declaring two variables of type int*, you are introducing two expressions that evaluate to an int type, with no attachment to the allocation of an int in memory.
A compiler is perfectly happy to accept the following:
int *ip, i;
i = *ip;
because in the C paradigm, the compiler is only expected to keep track of the type of *ip and i. The programmer is expected to keep track of the meaning of *ip and i. In this case, ip is uninitialized, so it is the programmer's responsibility to point it at something meaningful before dereferencing it.
A good rule of thumb, a lot of people seem to grasp these concepts by: In C++ a lot of semantic meaning is derived by the left-binding of keywords or identifiers.
Take for example:
int const bla;
The const applies to the "int" word. The same is with pointers' asterisks, they apply to the keyword left of them. And the actual variable name? Yup, that's declared by what's left of it.
Related
when I do this (in my class)
public:
Entity()
{
re_sprite_eyes = new sf::Sprite();
re_sprite_hair = new sf::Sprite();
re_sprite_body = new sf::Sprite();
}
private:
sf::Sprite* re_sprite_hair;
sf::Sprite* re_sprite_body;
sf::Sprite* re_sprite_eyes;
Everything works fine. However, if I change the declarations to this:
private:
sf::Sprite* re_sprite_hair, re_sprite_body, re_sprite_eyes;
I get this compiler error:
error: no match for 'operator=' in '((Entity*)this)->Entity::re_sprite_eyes = (operator new(272u), (<statement>, ((sf::Sprite*)<anonymous>)))
And then it says candidates for re_sprite_eyes are sf::Sprite objects and/or references.
Why does this not work? Aren't the declarations the same?
sf::Sprite* re_sprite_hair, re_sprite_body, re_sprite_eyes;
Does not declare 3 pointers - it is one pointer and 2 objects.
sf::Sprite* unfortunately does not apply to all the variables declared following it, just the first. It is equivalent to
sf::Sprite* re_sprite_hair;
sf::Sprite re_sprite_body;
sf::Sprite re_sprite_eyes;
You want to do:
sf::Sprite *re_sprite_hair, *re_sprite_body, *re_sprite_eyes;
You need to put one star for each variable. In such cases I prefer to keep the star on the variable's side, rather than the type, to make exactly this situation clear.
In both C and C++, the * binds to the declarator, not the type specifier. In both languages, declarations are based on the types of expressions, not objects.
For example, suppose you have a pointer to an int named p, and you want to access the int value that p points to; you do so by dereferencing the pointer with the unary * operator, like so:
x = *p;
The type of the expression *p is int; thus, the declaration of p is
int *p;
This is true no matter how many pointers you declare within the same declaration statement; if q and r also need to be declared as pointers, then they also need to have the unary * as part of the declarator:
int *p, *q, *r;
because the expressions *q and *r have type int. It's an accident of C and C++ syntax that you can write T *p, T* p, or T * p; all of those declarations will be interpreted as T (*p).
This is why I'm not fond of the C++ style of declaring pointer and reference types as
T* p;
T& r;
because it implies an incorrect view of how C and C++ declaration syntax works, leading to the exact kind of confusion that you just experienced. However, I've written enough C++ to realize that there are times when that style does make the intent of the code clearer, especially when defining container types.
But it's still wrong.
This is a (two years late) response to Lightness Races in Orbit (and anyone else who objects to my labeling the T* p convention as "wrong")...
First of all, you have the legion of questions just like this one that arise specifically from the use of the T* p convention, and how it doesn't work like people expect. How many questions on this site are on the order of "why doesn't T* p, q declare both p and q as pointers?"
It introduces confusion - that by itself should be enough to discourage its use.
But beyond that, it's inconsistent. You can't separate array-ness or function-ness from the declarator, why should you separate pointer-ness from it?
"Well, that's because [] and () are postfix operators, while * is unary". Yes, it is, so why aren't you associating the operator with its operand? In the declaration T* p, T is not the operand of *, so why are we writing the declaration as though it is?
If a is "an array of pointers", why should we write T* a[N]? If f is "a function returning a pointer", why should we write T* f()? The declarator system makes more sense and is internally consistent if you write those declarations as T *a[N] and T *f(). That should be obvious from the fact that I can use T as a stand-in for any type (indeed, for any sequence of declaration specifiers).
And then you have pointers to arrays and pointers to functions, where the * must be explicitly bound to the declarator1:
T (*a)[N];
T (*f)();
Yes, pointer-ness is an important property of the thing you're declaring, but so are array-ness and function-ness, and emphasizing one over the other creates more problems than it solves. Again, as this question shows, the T* p convention introduces confusion.
Because * is unary and a separate token on its own you can write T* p, T *p, T*p, and T * p and they'll all be accepted by the compiler, but they will all be interpreted as T (*p). More importantly, T* p, q, r will be interpreted as T (*p), q, r. That interpretation is more obvious if you write T *p, q, r. Yeah, yeah, yeah, "declare only one thing per line and it won't be a problem." You know how else to not make it a problem? Write your declarators properly. The declarator system itself will make more sense and you will be less likely to make mistake.
We're not arguing over an "antique oddity" of the language, it's a fundamental component of the language grammar and its philosophy. Pointer-ness is a property of the declarator, just like array-ness and function-ness, and pretending it's somehow not just leads to confusion and makes both C and C++ harder to understand than they need to be.
I would argue that making the dereference operator unary as opposed to postfix was a mistake2, but that's how it worked in B, and Ritchie wanted to keep as much of B as possible. I will also argue that Bjarne's promotion of the T* p convention is a mistake.
At this point in the discussion, somebody will suggest using a typedef liketypedef T arrtype[N];
arrtype* p;
which just totally misses the point and earns the suggester a beating with the first edition of "C: The Complete Reference" because it's big and heavy and no good for anything else.
Writing T a*[N]*() as opposed to T (*(*a)[N])() is definitely less eye-stabby and scans much more easily.
In C++11 you have a nice little workaround, which might be better than shifting spaces back and forth:
template<typename T> using type=T;
template<typename T> using func=T*;
// I don't like this style, but type<int*> i, j; works ok
type<int*> i = new int{3},
j = new int{4};
// But this one, imho, is much more readable than int(*f)(int, int) = ...
func<int(int, int)> f = [](int x, int y){return x + y;},
g = [](int x, int y){return x - y;};
Another thing that may call your attention is the line:
int * p1, * p2;
This declares the two pointers used in the previous example. But notice that there is an asterisk (*) for each pointer, in order for both to have type int* (pointer to int). This is required due to the precedence rules. Note that if, instead, the code was:
int * p1, p2;
p1 would indeed be of type int*, but p2 would be of type int. Spaces do not matter at all for this purpose. But anyway, simply remembering to put one asterisk per pointer is enough for most pointer users interested in declaring multiple pointers per statement. Or even better: use a different statemet for each variable.
From http://www.cplusplus.com/doc/tutorial/pointers/
The asterisk binds to the pointer-variable name. The way to remember this is to notice that in C/C++, declarations mimic usage.
The pointers might be used like this:
sf::Sprite *re_sprite_body;
// ...
sf::Sprite sprite_bod = *re_sprite_body;
Similarly,
char *foo[3];
// ...
char fooch = *foo[1];
In both cases, there is an underlying type-specifier, and the operator or operators required to "get to" an object of that type in an expression.
I'm not sure if this is a proper programming question, but it's something that has always bothered me, and I wonder if I'm the only one.
When initially learning C++, I understood the concept of references, but pointers had me confused. Why, you ask? Because of how you declare a pointer.
Consider the following:
void foo(int* bar)
{
}
int main()
{
int x = 5;
int* y = NULL;
y = &x;
*y = 15;
foo(y);
}
The function foo(int*) takes an int pointer as parameter. Since I've declared y as int pointer, I can pass y to foo, but when first learning C++ I associated the * symbol with dereferencing, as such I figured a dereferenced int needed to be passed. I would try to pass *y into foo, which obviously doesn't work.
Wouldn't it have been easier to have a separate operator for declaring a pointer? (or for dereferencing). For example:
void test(int# x)
{
}
In The Development of the C Language, Dennis Ritchie explains his reasoning thusly:
The second innovation that most clearly distinguishes C from its
predecessors is this fuller type structure and especially its
expression in the syntax of declarations... given an object of any
type, it should be possible to describe a new object that gathers
several into an array, yields it from a function, or is a pointer to
it.... [This] led to a
declaration syntax for names mirroring that of the expression syntax
in which the names typically appear. Thus,
int i, *pi, **ppi; declare an integer, a pointer to an integer, a
pointer to a pointer to an integer. The syntax of these declarations
reflects the observation that i, *pi, and **ppi all yield an int type
when used in an expression.
Similarly, int f(), *f(), (*f)(); declare
a function returning an integer, a function returning a pointer to an
integer, a pointer to a function returning an integer. int *api[10],
(*pai)[10]; declare an array of pointers to integers, and a pointer to
an array of integers.
In all these cases the declaration of a
variable resembles its usage in an expression whose type is the one
named at the head of the declaration.
An accident of syntax contributed to the perceived complexity of the
language. The indirection operator, spelled * in C, is syntactically a
unary prefix operator, just as in BCPL and B. This works well in
simple expressions, but in more complex cases, parentheses are
required to direct the parsing. For example, to distinguish
indirection through the value returned by a function from calling a
function designated by a pointer, one writes *fp() and (*pf)()
respectively. The style used in expressions carries through to
declarations, so the names might be declared
int *fp(); int (*pf)();
In more ornate but still realistic cases,
things become worse: int *(*pfp)(); is a pointer to a function
returning a pointer to an integer.
There are two effects occurring.
Most important, C has a relatively rich set of ways of describing
types (compared, say, with Pascal). Declarations in languages as
expressive as C—Algol 68, for example—describe objects equally hard to
understand, simply because the objects themselves are complex. A
second effect owes to details of the syntax. Declarations in C must be
read in an `inside-out' style that many find difficult to grasp.
Sethi [Sethi 81] observed that many of the nested
declarations and expressions would become simpler if the indirection
operator had been taken as a postfix operator instead of prefix, but
by then it was too late to change.
The reason is clearer if you write it like this:
int x, *y;
That is, both x and *y are ints. Thus y is an int *.
That is a language decision that predates C++, as C++ inherited it from C. I once heard that the motivation was that the declaration and the use would be equivalent, that is, given a declaration int *p; the expression *p is of type int in the same way that with int i; the expression i is of type int.
Because the committee, and those that developed C++ in the decades before its standardisation, decided that * should retain its original three meanings:
A pointer type
The dereference operator
Multiplication
You're right to suggest that the multiple meanings of * (and, similarly, &) are confusing. I've been of the opinion for some years that it they are a significant barrier to understanding for language newcomers.
Why not choose another symbol for C++?
Backwards-compatibility is the root cause... best to re-use existing symbols in a new context than to break C programs by translating previously-not-operators into new meanings.
Why not choose another symbol for C?
It's impossible to know for sure, but there are several arguments that can be — and have been — made. Foremost is the idea that:
when [an] identifier appears in an expression of the same form as the declarator, it yields an object of the specified type. {K&R, p216}
This is also why C programmers tend to[citation needed] prefer aligning their asterisks to the right rather than to the left, i.e.:
int *ptr1; // roughly C-style
int* ptr2; // roughly C++-style
though both varieties are found in programs of both languages, varyingly.
Page 65 of Expert C Programming: Deep C Secrets includes the following: And then, there is the C philosophy that the declaration of an object should look like its use.
Page 216 of The C Programming Language, 2nd edition (aka K&R) includes: A declarator is read as an assertion that when its identifier appears in an expression of the same form as the declarator, it yields an object of the specified type.
I prefer the way van der Linden puts it.
Haha, I feel your pain, I had the exact same problem.
I thought a pointer should be declared as &int because it makes sense that a pointer is an address of something.
After a while I thought for myself, every type in C has to be read backwards, like
int * const a
is for me
a constant something, when dereferenced equals an int.
Something that has to be dereferenced, has to be a pointer.
Is there a "good" way to write "pointer to something" in C/C++ ?
I use to write void foo( char *str ); But sometimes I find it quite illogical because the type of str is "pointer to char", then it should more logical to attach the * to the type name.
Is there a rule to write pointers ?
char*str;
char* str;
char *str;
char * str;
There is no strict rule, but bear in mind that the * attaches to the variable, so:
char *str1, *str2; // str1 and str2 are pointers
char* str1, str2; // str1 is a pointer, str2 is a char
Some people like to do char * str1 as well, but it's up to you or your company's coding standard.
The common C convention is to write T *p, whereas the common C++ convention is to write T* p. Both parse as T (*p); the * is part of the declarator, not the type specifier. It's purely an accident of pointer declaration syntax that you can write it either way.
C (and by extension, C++) declaration syntax is expression-centric; IOW, the form of a declaration should match the form of an expression of the same type in the code.
For example, suppose we had a pointer to int, and we wanted to access that integer value. To do so, we dereference the pointer with the * indirection operator, like so:
x = *p;
The type of the expression *p is int; thus, it follows that the declaration of p should be
int *p
The int-ness of p is provided by the type specifier int, but the pointer-ness of p is provided by the declarator *p.
As a slightly more complicated example, suppose we had a pointer to an array of float, and wanted to access the floating point value at the i'th element of the array through the pointer. We dereference the array pointer and subscript the result:
f = (*ap)[i];
The type of the expression (*ap)[i] is float, so it follows that the declaration of the array pointer is
float (*ap)[N];
The float-ness of ap is provided by the type specifier float, but the pointer-ness and array-ness are provided by the declarator (*ap)[N]. Note that in this case the * must explicitly be bound to the identifer; [] has a higher precedence than unary * in both expression and declaration syntax, so float* ap[N] would be parsed as float *(ap[N]), or "array of pointers to float", rather than "pointer to array of float". I suppose you could write that as
float(* ap)[N];
but I'm not sure what the point would be; it doesn't make the type of ap any clearer.
Even better, how about a pointer to a function that returns a pointer to an array of pointer to int:
int *(*(*f)())[N];
Again, at least two of the * operators must explicitly be bound in the declarator; binding the last * to the type specifier, as in
int* (*(*f)())[N];
just indicates confused thinking IMO.
Even though I use it in my own C++ code, and even though I understand why it became popular, the problem I have with the reasoning behind the T* p convention is that it just doesn't apply outside of the simplest of pointer declarations, and it reinforces a simplistic-to-the-point-of-being-wrong view of C and C++ declaration syntax. Yes, the type of p is "pointer to T", but that doesn't change the fact that as far as the language grammar is concerned * binds to the declarator, not the type specifier.
For another case, if the type of a is "N-element array of T", we don't write
T[N] a;
Obviously, the grammar doesn't allow it. Again, the argument just doesn't apply in this case.
EDIT
As Steve points out in the comments, you can use typedefs to hide some of the complexity. For example, you could rewrite
int *(*(*f)())[N];
as something like
typedef int *iptrarr[N]; // iptrarr is an array of pointer to int
typedef iptrarr *arrptrfunc(); // arrptrfunc is a function returning
// a pointer to iptrarr
arrptrfunc *f; // f is a pointer to arrptrfunc
Now you can cleanly apply the T* p convention, declaring f as arrptrfunc* f. I personally am not fond of doing things this way, since it's not necessarily clear from the typedef how f is supposed to be used in an expression, or how to use an object of type arrptrfunc. The non-typedef'd version may be ugly and difficult to read, but at least it tells you everything you need to know up front; you don't have to go digging through all the typedefs.
The "good way" depends on
internal coding standards in your project
your personal preferences
(probably) in that order.
There is no right or wrong in this. The important thing is to pick one coding standard and stick to it.
That being said, I personally believe that the * belongs with the type and not the variable name, as the type is "pointer to char". The variable name is not a pointer.
I think this is going to be heavily influenced by the general pattern in how one declares the variables.
For example, I have a tendency to declare only one variable per line. This way, I can add a comment reminding me how the variable is to be used.
However, there are times, when it is practical to declare several variables of the same type on one line. Under such circumstances, my personal coding rule is to never, NEVER, EVER declare pointers on the same line as non-pointers. I find that mixing them can be a source of errors, so I try to make it easier to see "wrongness" by avoiding mixing.
As long as I follow the first guideline, I find that it does not matter overly much how I declare the pointers so long as I am consistent.
However, if I use the second guideline and declare several pointers on the same line, I find the following style to be most beneficial and clear (of course others may disagree) ...
char *ptr1, *ptr2, *ptr3;
By having no space between the * and the pointer name, it becomes easier to spot whether I have violated the second guideline.
Now, if I wanted to be consistent between my two personal style guidelines, when declaring only one pointer on a line, I would use ...
char *ptr;
Anyway, that's my rationale for part of why I do what I do. Hope this helps.
I'm relatively new to C++ (about one year of experience, on and off). I'm curious about what led to the decision of type * name as the syntax for defining pointers. It seems to me that the syntax should be type & name as the & symbol is used everywhere else in code to refer to the variable's memory address. So, to use the traditional example of int pointers:
int a = 1;
int * b = &a;
would become
int a = 1;
int & b = &a
I'm sure there's some reason for this that I'm just not seeing, and I'd love to hear some input from C++ veterans.
Thanks,
-S
C++ adopts the C syntax. As revealed in "The Development of the C Language" (by Dennis Ritchie) C uses * for pointers in type declarations because it was decided that type syntax should follow use.
For each object of [a compound type], there was already a way to mention the underlying object: index the array, call the function, use the indirection operator [*] on the pointer. Analogical reasoning led to a declaration syntax for names mirroring that of the expression syntax in which the names typically appear. Thus,
int i, *pi, **ppi;
declare an integer, a pointer to an integer, a pointer to a pointer to an integer. The syntax of these declarations reflects the observation that i, *pi, and **ppi all yield an int type when used in an expression.
Here's a more complex example:
int *(*foo)[4][];
This declaration means an expression *(*foo)[4][0] has type int, and from that (and that [] has higher precedence than unary *) you can decode the type: foo is a pointer to an array of size 4 of array of pointers to ints.
This syntax was adopted in C++ for compatibility with C. Also, don't forget that C++ has a use for & in declarations.
int & b = a;
The above line means a reference variable refering to another variable of type int. The difference between a reference and pointer roughly is that references are initialized only, and you can not change where they point, and finally they are always dereferenced automatically.
int x = 5, y = 10;
int& r = x;
int sum = r + y; // you do not need to say '*r' automatically dereferenced.
r = y; // WRONG, 'r' can only have one thing pointing at during its life, only at its infancy ;)
I think that Dennis Ritchie answered this in The Development of the C Language:
For each object of such a composed
type, there was already a way to
mention the underlying object: index
the array, call the function, use the
indirection operator on the pointer.
Analogical reasoning led to a
declaration syntax for names mirroring
that of the expression syntax in which
the names typically appear. Thus,
int i, *pi, **ppi;
declare an integer, a pointer to an
integer, a pointer to a pointer to an
integer. The syntax of these
declarations reflects the observation
that i, *pi, and **ppi all yield an
int type when used in an expression.
Similarly,
int f(), *f(), (*f)();
declare a function returning an
integer, a function returning a
pointer to an integer, a pointer to a
function returning an integer;
int *api[10], (*pai)[10];
declare an array of pointers to
integers, and a pointer to an array of
integers. In all these cases the
declaration of a variable resembles
its usage in an expression whose type
is the one named at the head of the
declaration.
So we use type * var to declare a pointer because this allows the declaration to mirror the usage (dereferencing) of the pointer.
In this article, Ritchie also recounts that in "NB", an extended version of the "B" programming language, he used int pointer[] to declare a pointer to an int, as opposed to int array[10] to declare an array of ints.
If you are a visual thinker, it may help to imagine the asterisk as a black hole leading to the data value. Hence, it is a pointer.
The ampersand is the opposite end of the hole, think of it as an unraveled asterisk or a spaceship wobbling about in an erratic course as the pilot gets over the transition coming out of the black hole.
I remember being very confused by C++ overloading the meaning of the ampersand, to give us references. In their desperate attempt to avoid using any more characters, which was justified by the international audience using C and known issues with keyboard limitations, they added a major source of confusion.
One thing that may help in C++ is to think of references as pre-prepared dereferenced pointers. Rather than using &someVariable when you pass in an argument, you've already used the trailing ampersand when you defined someVariable. Then again, that might just confuse you further!
One of my pet hates, which I was unhappy to see promulgated in Apple's Objective-C samples, is the layout style int *someIntPointer instead of int* someIntPointer
IMHO, keeping the asterisk with the variable is an old-fashioned C approach emphasizing the mechanics of how you define the variable, over its data type.
The data type of someIntPointer is literally a pointer to an integer and the declaration should reflect that. This does lead to the requirement that you declare one variable per line, to avoid subtle bugs such as:
int* a, b; // b is a straight int, was that our intention?
int *a, *b; // old-style C declaring two pointers
int* a;
int* b; // b is another pointer to an int
Whilst people argue that the ability to declare mixed pointers and values on the same line, intentionally, is a powerful feature, I've seen it lead to subtle bugs and confusion.
Your second example is not valid C code, only C++ code. The difference is that one is a pointer, whereas the other is a reference.
On the right-hand side the '&' always means address-of. In a definition it indicates that the variable is a reference.
On the right-hand side the '*' always means value-at-address. In a definition it indicates that the variable is a pointer.
References and pointers are similar, but not the same. This article addresses the differences.
Instead of reading int* b as "b is a pointer to int", read it as int *b: "*b is an int". Then, you have & as an anti-*: *b is an int. The address of *b is &*b, or just b.
I think the answer may well be "because that's the way K&R did it."
K&R are the ones who decided what the C syntax for declaring pointers was.
It's not int & x; instead of int * x; because that's the way the language was defined by the guys who made it up -- K&R.
I've recently decided that I just have to finally learn C/C++, and there is one thing I do not really understand about pointers or more precisely, their definition.
How about these examples:
int* test;
int *test;
int * test;
int* test,test2;
int *test,test2;
int * test,test2;
Now, to my understanding, the first three cases are all doing the same: Test is not an int, but a pointer to one.
The second set of examples is a bit more tricky. In case 4, both test and test2 will be pointers to an int, whereas in case 5, only test is a pointer, whereas test2 is a "real" int. What about case 6? Same as case 5?
4, 5, and 6 are the same thing, only test is a pointer. If you want two pointers, you should use:
int *test, *test2;
Or, even better (to make everything clear):
int* test;
int* test2;
White space around asterisks have no significance. All three mean the same thing:
int* test;
int *test;
int * test;
The "int *var1, var2" is an evil syntax that is just meant to confuse people and should be avoided. It expands to:
int *var1;
int var2;
Many coding guidelines recommend that you only declare one variable per line. This avoids any confusion of the sort you had before asking this question. Most C++ programmers I've worked with seem to stick to this.
A bit of an aside I know, but something I found useful is to read declarations backwards.
int* test; // test is a pointer to an int
This starts to work very well, especially when you start declaring const pointers and it gets tricky to know whether it's the pointer that's const, or whether its the thing the pointer is pointing at that is const.
int* const test; // test is a const pointer to an int
int const * test; // test is a pointer to a const int ... but many people write this as
const int * test; // test is a pointer to an int that's const
Use the "Clockwise Spiral Rule" to help parse C/C++ declarations;
There are three simple steps to follow:
Starting with the unknown element, move in a spiral/clockwise
direction; when encountering the following elements replace them with
the corresponding english statements:
[X] or []: Array X size of... or Array undefined size of...
(type1, type2): function passing type1 and type2 returning...
*: pointer(s) to...
Keep doing this in a spiral/clockwise direction until all tokens have been covered.
Always resolve anything in parenthesis first!
Also, declarations should be in separate statements when possible (which is true the vast majority of times).
There are three pieces to this puzzle.
The first piece is that whitespace in C and C++ is normally not significant beyond separating adjacent tokens that are otherwise indistinguishable.
During the preprocessing stage, the source text is broken up into a sequence of tokens - identifiers, punctuators, numeric literals, string literals, etc. That sequence of tokens is later analyzed for syntax and meaning. The tokenizer is "greedy" and will build the longest valid token that's possible. If you write something like
inttest;
the tokenizer only sees two tokens - the identifier inttest followed by the punctuator ;. It doesn't recognize int as a separate keyword at this stage (that happens later in the process). So, for the line to be read as a declaration of an integer named test, we have to use whitespace to separate the identifier tokens:
int test;
The * character is not part of any identifier; it's a separate token (punctuator) on its own. So if you write
int*test;
the compiler sees 4 separate tokens - int, *, test, and ;. Thus, whitespace is not significant in pointer declarations, and all of
int *test;
int* test;
int*test;
int * test;
are interpreted the same way.
The second piece to the puzzle is how declarations actually work in C and C++1. Declarations are broken up into two main pieces - a sequence of declaration specifiers (storage class specifiers, type specifiers, type qualifiers, etc.) followed by a comma-separated list of (possibly initialized) declarators. In the declaration
unsigned long int a[10]={0}, *p=NULL, f(void);
the declaration specifiers are unsigned long int and the declarators are a[10]={0}, *p=NULL, and f(void). The declarator introduces the name of the thing being declared (a, p, and f) along with information about that thing's array-ness, pointer-ness, and function-ness. A declarator may also have an associated initializer.
The type of a is "10-element array of unsigned long int". That type is fully specified by the combination of the declaration specifiers and the declarator, and the initial value is specified with the initializer ={0}. Similarly, the type of p is "pointer to unsigned long int", and again that type is specified by the combination of the declaration specifiers and the declarator, and is initialized to NULL. And the type of f is "function returning unsigned long int" by the same reasoning.
This is key - there is no "pointer-to" type specifier, just like there is no "array-of" type specifier, just like there is no "function-returning" type specifier. We can't declare an array as
int[10] a;
because the operand of the [] operator is a, not int. Similarly, in the declaration
int* p;
the operand of * is p, not int. But because the indirection operator is unary and whitespace is not significant, the compiler won't complain if we write it this way. However, it is always interpreted as int (*p);.
Therefore, if you write
int* p, q;
the operand of * is p, so it will be interpreted as
int (*p), q;
Thus, all of
int *test1, test2;
int* test1, test2;
int * test1, test2;
do the same thing - in all three cases, test1 is the operand of * and thus has type "pointer to int", while test2 has type int.
Declarators can get arbitrarily complex. You can have arrays of pointers:
T *a[N];
you can have pointers to arrays:
T (*a)[N];
you can have functions returning pointers:
T *f(void);
you can have pointers to functions:
T (*f)(void);
you can have arrays of pointers to functions:
T (*a[N])(void);
you can have functions returning pointers to arrays:
T (*f(void))[N];
you can have functions returning pointers to arrays of pointers to functions returning pointers to T:
T *(*(*f(void))[N])(void); // yes, it's eye-stabby. Welcome to C and C++.
and then you have signal:
void (*signal(int, void (*)(int)))(int);
which reads as
signal -- signal
signal( ) -- is a function taking
signal( ) -- unnamed parameter
signal(int ) -- is an int
signal(int, ) -- unnamed parameter
signal(int, (*) ) -- is a pointer to
signal(int, (*)( )) -- a function taking
signal(int, (*)( )) -- unnamed parameter
signal(int, (*)(int)) -- is an int
signal(int, void (*)(int)) -- returning void
(*signal(int, void (*)(int))) -- returning a pointer to
(*signal(int, void (*)(int)))( ) -- a function taking
(*signal(int, void (*)(int)))( ) -- unnamed parameter
(*signal(int, void (*)(int)))(int) -- is an int
void (*signal(int, void (*)(int)))(int); -- returning void
and this just barely scratches the surface of what's possible. But notice that array-ness, pointer-ness, and function-ness are always part of the declarator, not the type specifier.
One thing to watch out for - const can modify both the pointer type and the pointed-to type:
const int *p;
int const *p;
Both of the above declare p as a pointer to a const int object. You can write a new value to p setting it to point to a different object:
const int x = 1;
const int y = 2;
const int *p = &x;
p = &y;
but you cannot write to the pointed-to object:
*p = 3; // constraint violation, the pointed-to object is const
However,
int * const p;
declares p as a const pointer to a non-const int; you can write to the thing p points to
int x = 1;
int y = 2;
int * const p = &x;
*p = 3;
but you can't set p to point to a different object:
p = &y; // constraint violation, p is const
Which brings us to the third piece of the puzzle - why declarations are structured this way.
The intent is that the structure of a declaration should closely mirror the structure of an expression in the code ("declaration mimics use"). For example, let's suppose we have an array of pointers to int named ap, and we want to access the int value pointed to by the i'th element. We would access that value as follows:
printf( "%d", *ap[i] );
The expression *ap[i] has type int; thus, the declaration of ap is written as
int *ap[N]; // ap is an array of pointer to int, fully specified by the combination
// of the type specifier and declarator
The declarator *ap[N] has the same structure as the expression *ap[i]. The operators * and [] behave the same way in a declaration that they do in an expression - [] has higher precedence than unary *, so the operand of * is ap[N] (it's parsed as *(ap[N])).
As another example, suppose we have a pointer to an array of int named pa and we want to access the value of the i'th element. We'd write that as
printf( "%d", (*pa)[i] );
The type of the expression (*pa)[i] is int, so the declaration is written as
int (*pa)[N];
Again, the same rules of precedence and associativity apply. In this case, we don't want to dereference the i'th element of pa, we want to access the i'th element of what pa points to, so we have to explicitly group the * operator with pa.
The *, [] and () operators are all part of the expression in the code, so they are all part of the declarator in the declaration. The declarator tells you how to use the object in an expression. If you have a declaration like int *p;, that tells you that the expression *p in your code will yield an int value. By extension, it tells you that the expression p yields a value of type "pointer to int", or int *.
So, what about things like cast and sizeof expressions, where we use things like (int *) or sizeof (int [10]) or things like that? How do I read something like
void foo( int *, int (*)[10] );
There's no declarator, aren't the * and [] operators modifying the type directly?
Well, no - there is still a declarator, just with an empty identifier (known as an abstract declarator). If we represent an empty identifier with the symbol λ, then we can read those things as (int *λ), sizeof (int λ[10]), and
void foo( int *λ, int (*λ)[10] );
and they behave exactly like any other declaration. int *[10] represents an array of 10 pointers, while int (*)[10] represents a pointer to an array.
And now the opinionated portion of this answer. I am not fond of the C++ convention of declaring simple pointers as
T* p;
and consider it bad practice for the following reasons:
It's not consistent with the syntax;
It introduces confusion (as evidenced by this question, all the duplicates to this question, questions about the meaning of T* p, q;, all the duplicates to those questions, etc.);
It's not internally consistent - declaring an array of pointers as T* a[N] is asymmetrical with use (unless you're in the habit of writing * a[i]);
It cannot be applied to pointer-to-array or pointer-to-function types (unless you create a typedef just so you can apply the T* p convention cleanly, which...no);
The reason for doing so - "it emphasizes the pointer-ness of the object" - is spurious. It cannot be applied to array or function types, and I would think those qualities are just as important to emphasize.
In the end, it just indicates confused thinking about how the two languages' type systems work.
There are good reasons to declare items separately; working around a bad practice (T* p, q;) isn't one of them. If you write your declarators correctly (T *p, q;) you are less likely to cause confusion.
I consider it akin to deliberately writing all your simple for loops as
i = 0;
for( ; i < N; )
{
...
i++;
}
Syntactically valid, but confusing, and the intent is likely to be misinterpreted. However, the T* p; convention is entrenched in the C++ community, and I use it in my own C++ code because consistency across the code base is a good thing, but it makes me itch every time I do it.
I will be using C terminology - the C++ terminology is a little different, but the concepts are largely the same.
As others mentioned, 4, 5, and 6 are the same. Often, people use these examples to make the argument that the * belongs with the variable instead of the type. While it's an issue of style, there is some debate as to whether you should think of and write it this way:
int* x; // "x is a pointer to int"
or this way:
int *x; // "*x is an int"
FWIW I'm in the first camp, but the reason others make the argument for the second form is that it (mostly) solves this particular problem:
int* x,y; // "x is a pointer to int, y is an int"
which is potentially misleading; instead you would write either
int *x,y; // it's a little clearer what is going on here
or if you really want two pointers,
int *x, *y; // two pointers
Personally, I say keep it to one variable per line, then it doesn't matter which style you prefer.
#include <type_traits>
std::add_pointer<int>::type test, test2;
In 4, 5 and 6, test is always a pointer and test2 is not a pointer. White space is (almost) never significant in C++.
The rationale in C is that you declare the variables the way you use them. For example
char *a[100];
says that *a[42] will be a char. And a[42] a char pointer. And thus a is an array of char pointers.
This because the original compiler writers wanted to use the same parser for expressions and declarations. (Not a very sensible reason for a langage design choice)
I would say that the initial convention was to put the star on the pointer name side (right side of the declaration
in the c programming language by Dennis M. Ritchie the stars are on the right side of the declaration.
by looking at the linux source code at https://github.com/torvalds/linux/blob/master/init/main.c
we can see that the star is also on the right side.
You can follow the same rules, but it's not a big deal if you put stars on the type side.
Remember that consistency is important, so always but the star on the same side regardless of which side you have choose.
In my opinion, the answer is BOTH, depending on the situation.
Generally, IMO, it is better to put the asterisk next to the pointer name, rather than the type. Compare e.g.:
int *pointer1, *pointer2; // Fully consistent, two pointers
int* pointer1, pointer2; // Inconsistent -- because only the first one is a pointer, the second one is an int variable
// The second case is unexpected, and thus prone to errors
Why is the second case inconsistent? Because e.g. int x,y; declares two variables of the same type but the type is mentioned only once in the declaration. This creates a precedent and expected behavior. And int* pointer1, pointer2; is inconsistent with that because it declares pointer1 as a pointer, but pointer2 is an integer variable. Clearly prone to errors and, thus, should be avoided (by putting the asterisk next to the pointer name, rather than the type).
However, there are some exceptions where you might not be able to put the asterisk next to an object name (and where it matters where you put it) without getting undesired outcome — for example:
MyClass *volatile MyObjName
void test (const char *const p) // const value pointed to by a const pointer
Finally, in some cases, it might be arguably clearer to put the asterisk next to the type name, e.g.:
void* ClassName::getItemPtr () {return &item;} // Clear at first sight
The pointer is a modifier to the type. It's best to read them right to left in order to better understand how the asterisk modifies the type. 'int *' can be read as "pointer to int'. In multiple declarations you must specify that each variable is a pointer or it will be created as a standard variable.
1,2 and 3) Test is of type (int *). Whitespace doesn't matter.
4,5 and 6) Test is of type (int *). Test2 is of type int. Again whitespace is inconsequential.
I have always preferred to declare pointers like this:
int* i;
I read this to say "i is of type int-pointer". You can get away with this interpretation if you only declare one variable per declaration.
It is an uncomfortable truth, however, that this reading is wrong. The C Programming Language, 2nd Ed. (p. 94) explains the opposite paradigm, which is the one used in the C standards:
The declaration of the pointer ip,
int *ip;
is intended as a mnemonic; it says that the expression *ip is an
int. The syntax of the declaration for a variable mimics the syntax
of expressions in which the variable might appear. This reasoning
applies to function declarations as well. For example,
double *dp, atof(char *);
says that in an expression *dp and atof(s) have values of type
double, and that the argument of atof is a pointer to char.
So, by the reasoning of the C language, when you declare
int* test, test2;
you are not declaring two variables of type int*, you are introducing two expressions that evaluate to an int type, with no attachment to the allocation of an int in memory.
A compiler is perfectly happy to accept the following:
int *ip, i;
i = *ip;
because in the C paradigm, the compiler is only expected to keep track of the type of *ip and i. The programmer is expected to keep track of the meaning of *ip and i. In this case, ip is uninitialized, so it is the programmer's responsibility to point it at something meaningful before dereferencing it.
A good rule of thumb, a lot of people seem to grasp these concepts by: In C++ a lot of semantic meaning is derived by the left-binding of keywords or identifiers.
Take for example:
int const bla;
The const applies to the "int" word. The same is with pointers' asterisks, they apply to the keyword left of them. And the actual variable name? Yup, that's declared by what's left of it.