What is the difference between const int*, const int * const, and int const *?

Asked 2023-09-21 08:11:47 View 143,064

I always mess up how to use const int*, const int * const, and int const * correctly. Is there a set of rules defining what you can and cannot do?

I want to know all the do's and all don'ts in terms of assignments, passing to the functions, etc.

  • You can use the "Clockwise/Spiral Rule" to decipher most C and C++ declarations. - anyone
  • cdecl.org is a great website which auto-translates C declarations for you. - anyone
  • @Calmarius: start where the type-name is / should be, move right when you can, left when you must. int *(*)(char const * const). Start to the right of the parenthesized * then we have to move left: pointer. Outside the parens, we can move right: pointer to function of .... Then we have to move left: pointer to function of ... that returns pointer to int. Repeat to expand the parameter (the ...): pointer to function of (constant pointer to constant char) that returns pointer to int. What would the equivalent one-line declaration be in a easy-reading language like Pascal? - anyone
  • @MarkKCowan In Pascal it would be something like function(x:^char):^int. There function types are imply a pointer to a function so no need to specify it, and Pascal doesn't enforce const correctness. It can be read from left to right. - anyone
  • The first thing to the left of the "const" is what's constant. If "const" is the thing the farthest to the left, then the first thing to the right of it is what's constant. - anyone

Answers

Read it backwards (as driven by Clockwise/Spiral Rule):

  • int* - pointer to int
  • int const * - pointer to const int
  • int * const - const pointer to int
  • int const * const - const pointer to const int

Now the first const can be on either side of the type so:

  • const int * == int const *
  • const int * const == int const * const

If you want to go really crazy you can do things like this:

  • int ** - pointer to pointer to int
  • int ** const - a const pointer to a pointer to an int
  • int * const * - a pointer to a const pointer to an int
  • int const ** - a pointer to a pointer to a const int
  • int * const * const - a const pointer to a const pointer to an int
  • ...

If you're ever uncertain, you can use a tool like cdecl+ to convert declarations to prose automatically.

To make sure we are clear on the meaning of const:

int a = 5, b = 10, c = 15;

const int* foo;     // pointer to constant int.
foo = &a;           // assignment to where foo points to.

/* dummy statement*/
*foo = 6;           // the value of a can´t get changed through the pointer.

foo = &b;           // the pointer foo can be changed.



int *const bar = &c;  // constant pointer to int 
                      // note, you actually need to set the pointer 
                      // here because you can't change it later ;)

*bar = 16;            // the value of c can be changed through the pointer.    

/* dummy statement*/
bar = &a;             // not possible because bar is a constant pointer.           

foo is a variable pointer to a constant integer. This lets you change what you point to but not the value that you point to. Most often this is seen with C-style strings where you have a pointer to a const char. You may change which string you point to but you can't change the content of these strings. This is important when the string itself is in the data segment of a program and shouldn't be changed.

bar is a constant or fixed pointer to a value that can be changed. This is like a reference without the extra syntactic sugar. Because of this fact, usually you would use a reference where you would use a T* const pointer unless you need to allow NULL pointers.

Answered   2023-09-21 08:11:47

  • I would like to append a rule of thumb which may help you remember how to discover whether 'const' applies to pointer or to pointed data: split the statement at asterix sign, then, if the const keyword appears in the left part (like in 'const int * foo') - it belongs to pointed data, if it's in the right part ('int * const bar') - it's about the pointer. - anyone
  • @Michael: Kudos to Michael for such a simple rule for remembering/understanding const rule. - anyone
  • @Jeffrey: read it backwards works well as long as there are no parenthesis. Then, well... use typedefs - anyone
  • +1, though a better summary would be: read pointer declarations backwards, that means, close to @Michael 's statement: stop the normal left-to-right reading at the first asterisk. - anyone
  • @gedamial it does, it works fine, but you must assign it at the same time you declare it (because you can't reassign a "const pointer"). const int x = 0; const int *const px = &x; const int *const *const p = &px; works just fine. - anyone

For those who don't know about Clockwise/Spiral Rule: Start from the name of the variable, move clockwisely (in this case, move backward) to the next pointer or type. Repeat until expression ends.

Here is a demo:

pointer to int

const pointer to int const

pointer to int const

pointer to const int

const pointer to int

Answered   2023-09-21 08:11:47

  • @Rog it used to have all open access permissions... I didn't write the article and don't have access permissions myself, unfortunately. However, here is an archived version of the article that still works: archive.is/SsfMX - anyone
  • The complex example is still just right to left, but includes resolving parentheses the way one would normally. The whole clockwise spiral thing doesn't make that any easier. - anyone
  • Ultimate example: void (*signal(int, void (*fp)(int)))(int); from archive.is/SsfMX - anyone
  • Do not rely on this rule. This is not universal. There are some cases where it fails. - anyone
  • apart from the mentioned fact, that the spiral-shape is just misleading, why are is * const in the 2nd example colored in one color but with two colors in the 5th example where in the latter it is still interpreted as constant pointer? what is the meaning of the colors at all? - anyone

I think everything is answered here already, but I just want to add that you should beware of typedefs! They're NOT just text replacements.

For example:

typedef char *ASTRING;
const ASTRING astring;

The type of astring is char * const, not const char *. This is one reason I always tend to put const to the right of the type, and never at the start.

Answered   2023-09-21 08:11:47

  • And for me this is the reason to never typedef pointers. I don't see the benefit in things like typedef int* PINT (I assume its something that came from practices in C and many developers kept doing it). Great, I replaced that * with a P, it doesn't speed up typing, plus introducing the issue you mention. - anyone
  • @Mephane - I can see that. However, to me it seems kind of backwards to avoid a nice language feature in order to keep using an exceptional syntactical rule (about "const" placement), rather than avoiding using the exceptional syntactic rule so you can safely make use of this language feature. - anyone
  • @Mephane PINT is indeed a rather dumb usage of a typedef, especially cuz it makes me think that the system stores uses beer for memory. typedef s are pretty useful for dealing with pointers to functions, though. - anyone
  • @KazDragon THANKS! Without it, I would've messed up with all those typedefed PVOID, LPTSTR stuff in Win32 api! - anyone
  • @Mephane: I've had to use pSomething a couple of times when using certain legacy macros which were written to accept a type, but would break apart if the type wasn't a single alphanumeric identifier. :) - anyone

Like pretty much everyone pointed out:

What’s the difference between const X* p, X* const p and const X* const p?

You have to read pointer declarations right-to-left.

  • const X* p means "p points to an X that is const": the X object can't be changed via p.

  • X* const p means "p is a const pointer to an X that is non-const": you can't change the pointer p itself, but you can change the X object via p.

  • const X* const p means "p is a const pointer to an X that is const": you can't change the pointer p itself, nor can you change the X object via p.

Answered   2023-09-21 08:11:47

  • Don't forget that const X* p; == X const * p; as in "p points to an X that is const": the X object can't be changed via p. - anyone
  1. Constant reference:

    A reference to a variable (here int), which is constant. We pass the variable as a reference mainly, because references are smaller in size than the actual value, but there is a side effect and that is because it is like an alias to the actual variable. We may accidentally change the main variable through our full access to the alias, so we make it constant to prevent this side effect.

    int var0 = 0;
    const int &ptr1 = var0;
    ptr1 = 8; // Error
    var0 = 6; // OK
    
  2. Constant pointers

    Once a constant pointer points to a variable then it cannot point to any other variable.

    int var1 = 1;
    int var2 = 0;
    
    int *const ptr2 = &var1;
    ptr2 = &var2; // Error
    
  3. Pointer to constant

    A pointer through which one cannot change the value of a variable it points is known as a pointer to constant.

    int const * ptr3 = &var2;
    *ptr3 = 4; // Error
    
  4. Constant pointer to a constant

    A constant pointer to a constant is a pointer that can neither change the address it's pointing to and nor can it change the value kept at that address.

    int var3 = 0;
    int var4 = 0;
    const int * const ptr4 = &var3;
    *ptr4 = 1;     // Error
     ptr4 = &var4; // Error
    

Answered   2023-09-21 08:11:47

This question shows precisely why I like to do things the way I mentioned in my question is const after type id acceptable?

In short, I find the easiest way to remember the rule is that the "const" goes after the thing it applies to. So in your question, "int const *" means that the int is constant, while "int * const" would mean that the pointer is constant.

If someone decides to put it at the very front (eg: "const int *"), as a special exception in that case it applies to the thing after it.

Many people like to use that special exception because they think it looks nicer. I dislike it, because it is an exception, and thus confuses things.

Answered   2023-09-21 08:11:47

  • I'm torn on this issue. Logically it makes sense. However most c++ developers would write const T* and it has become more natural. How often do you ever use a T* const anyways, usually a reference will do just fine. I got bit by all this once when wanting a boost::shared_ptr<const T> and instead wrote const boost::shared_ptr<T>. Same issue in a slightly different context. - anyone
  • Actually, I use constant pointers more often than I use constants. Also, you have to think about how you are going to react in the presence of pointers to pointers (etc.) Admittedly those are rarer, but it would be nice to think about things in a way where you can handle these situations with applomb. - anyone
  • The one other nice advantage of placing the const on the right of the type is that now everything to the left of any const is the type of that which is const, and everything to its right is that which is actually const. Take int const * const * p; as an example. No I don't normally write like that, this is just an example. First const: type int, And the int that is const is the contents of the const pointer that is the contents of p. Second const: type is pointer to const int, const oblect is the contents of p - anyone
  • '"const" goes after the thing it applies to': This seems simply wrong: int *const p; clearly makes p const and nothing else, and there is no way to have it after the variable name, and it doesn't even conform with the English syntax which has attributes preceding the noun: It is a constant pointer. The lingua franca of computer science is English, not Latin or French. - anyone
  • @Peter-ReinstateMonica - No. int *const p; makes the pointer const rather than the int. Don't you think that if the statement was "simply wrong", somebody else in the last thirteen years (on a question with over ⅔ of a million views) would have pointed it out by now? - anyone

The general rule is that the const keyword applies to what precedes it immediately. Exception, a starting const applies to what follows.

  • const int* is the same as int const* and means "pointer to constant int".
  • const int* const is the same as int const* const and means "constant pointer to constant int".

Edit: For the Dos and Don'ts, if this answer isn't enough, could you be more precise about what you want?

Answered   2023-09-21 08:11:47

Simple Use of const.

The simplest use is to declare a named constant. To do this, one declares a constant as if it was a variable but add const before it. One has to initialize it immediately in the constructor because, of course, one cannot set the value later as that would be altering it. For example:

const int Constant1=96; 

will create an integer constant, unimaginatively called Constant1, with the value 96.

Such constants are useful for parameters which are used in the program but are do not need to be changed after the program is compiled. It has an advantage for programmers over the C preprocessor #define command in that it is understood & used by the compiler itself, not just substituted into the program text by the preprocessor before reaching the main compiler, so error messages are much more helpful.

It also works with pointers but one has to be careful where const to determine whether the pointer or what it points to is constant or both. For example:

const int * Constant2 

declares that Constant2 is variable pointer to a constant integer and:

int const * Constant2

is an alternative syntax which does the same, whereas

int * const Constant3

declares that Constant3 is constant pointer to a variable integer and

int const * const Constant4

declares that Constant4 is constant pointer to a constant integer. Basically ‘const’ applies to whatever is on its immediate left (other than if there is nothing there in which case it applies to whatever is its immediate right).

ref: http://duramecho.com/ComputerInformation/WhyHowCppConst.html

Answered   2023-09-21 08:11:47

It's simple but tricky. Please note that we can apply the const qualifier to any data type (int, char, float, etc.).

Let's see the below examples.


const int *p ==> *p is read-only [p is a pointer to a constant integer]

int const *p ==> *p is read-only [p is a pointer to a constant integer]


int *p const ==> Wrong Statement. Compiler throws a syntax error.

int *const p ==> p is read-only [p is a constant pointer to an integer]. As pointer p here is read-only, the declaration and definition should be in same place.


const int *p const ==> Wrong Statement. Compiler throws a syntax error.

const int const *p ==> *p is read-only

const int *const p ==> *p and p are read-only [p is a constant pointer to a constant integer]. As pointer p here is read-only, the declaration and definition should be in same place.


int const *p const ==> Wrong Statement. Compiler throws a syntax error.

int const int *p ==> Wrong Statement. Compiler throws a syntax error.

int const const *p ==> *p is read-only and is equivalent to int const *p

int const *const p ==> *p and p are read-only [p is a constant pointer to a constant integer]. As pointer p here is read-only, the declaration and definition should be in same place.

Answered   2023-09-21 08:11:47

I had the same doubt as you until I came across this book by the C++ Guru Scott Meyers. Refer the third Item in this book where he talks in details about using const.

Just follow this advice

  1. If the word const appears to the left of the asterisk, what's pointed to is constant
  2. If the word const appears to the right of the asterisk, the pointer itself is constant
  3. If const appears on both sides, both are constant

Answered   2023-09-21 08:11:47

For me, the position of const i.e. whether it appears to the LEFT or RIGHT or on both LEFT and RIGHT relative to the * helps me figure out the actual meaning.

  1. A const to the LEFT of * indicates that the object pointed by the pointer is a const object.

  2. A const to the RIGHT of * indicates that the pointer is a const pointer.

The following table is taken from Stanford CS106L Standard C++ Programming Laboratory Course Reader.

enter image description here

Answered   2023-09-21 08:11:47

  • What is the differences between "reassign" and "modify" for a pointer exactly? - anyone
  • @BattleTested_закалённыйвбою Reassign means the address pointer is pointing to can be changed which is not true ('No' as in table) for constant pointers, and if the pointer address can be changed, but the value it points to is always constant that means we can't modify the pointee. - anyone

The C and C++ declaration syntax has repeatedly been described as a failed experiment, by the original designers.

Instead, let's name the type “pointer to Type”; I’ll call it Ptr_:

template< class Type >
using Ptr_ = Type*;

Now Ptr_<char> is a pointer to char.

Ptr_<const char> is a pointer to const char.

And const Ptr_<const char> is a const pointer to const char.

Answered   2023-09-21 08:11:47

  • do you have a quote for the first sentence? - anyone
  • @sp2danny: Googling “C syntax failed experiment” only coughs up a number of interviews with Bjarne Stroustrup where he expresses his opinion in that direction, e.g. “I consider the C declarator syntax an experiment that failed” in the Slashdot interview. So I have no reference for the claim about the viewpoints of the original designers of C. I guess it can be found by a sufficiently strong research effort, or maybe disproved simply by asking them, but I think it's better the way it is now. with that part of the claim, still undecided and likely true:) - anyone
  • "The C and C++ declaration syntax has repeatedly been described as a failed experiment, by the original designers." wrong for C please change your sentence about C or provide some quotes. - anyone
  • @Stargateur: Apparently you have read the preceding comments and found something you could leverage for pedantry. Good luck with your life. Anyway, old-timers like me remember a lot that we can't prove without engaging in very time-consuming research. You could just take my word. - anyone
  • @Stargateur "Sethi (...) 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." is from DMR. Of course DMR did not invent const and volatile keywords, they came from C++ / X3J11, as evidenced on that page. - anyone

To remember in easy way :

If const is before * then value is constant.

If const is after * then address is constant.

if const are available both before and after * then both value and address are constant.

e.g.

  1. int * const var; //here address is constant.

  2. int const * var; //here value is constant.

  3. int const * const var; // both value and address are constant.

Answered   2023-09-21 08:11:47

There are many other subtle points surrounding const correctness in C++. I suppose the question here has simply been about C, but I'll give some related examples since the tag is C++ :

  • You often pass large arguments like strings as TYPE const & which prevents the object from being either modified or copied. Example :

    TYPE& TYPE::operator=(const TYPE &rhs) { ... return *this; }

    But TYPE & const is meaningless because references are always const.

  • You should always label class methods that do not modify the class as const, otherwise you cannot call the method from a TYPE const & reference. Example :

    bool TYPE::operator==(const TYPE &rhs) const { ... }

  • There are common situations where both the return value and the method should be const. Example :

    const TYPE TYPE::operator+(const TYPE &rhs) const { ... }

    In fact, const methods must not return internal class data as a reference-to-non-const.

  • As a result, one must often create both a const and a non-const method using const overloading. For example, if you define T const& operator[] (unsigned i) const;, then you'll probably also want the non-const version given by :

    inline T& operator[] (unsigned i) { return const_cast<char&>( static_cast<const TYPE&>(*this)[](i) ); }

Afaik, there are no const functions in C, non-member functions cannot themselves be const in C++, const methods might have side effects, and the compiler cannot use const functions to avoid duplicate function calls. In fact, even a simple int const & reference might witness the value to which it refers be changed elsewhere.

Answered   2023-09-21 08:11:47

The const with the int on either sides will make pointer to constant int:

const int *ptr=&i;

or:

int const *ptr=&i;

const after * will make constant pointer to int:

int *const ptr=&i;

In this case all of these are pointer to constant integer, but none of these are constant pointer:

 const int *ptr1=&i, *ptr2=&j;

In this case all are pointer to constant integer and ptr2 is constant pointer to constant integer. But ptr1 is not constant pointer:

int const *ptr1=&i, *const ptr2=&j;

Answered   2023-09-21 08:11:47

  • if const is to the left of *, it refers to the value (it doesn't matter whether it's const int or int const)
  • if const is to the right of *, it refers to the pointer itself
  • it can be both at the same time

An important point: const int *p does not mean the value you are referring to is constant!!. It means that you can't change it through that pointer (meaning, you can't assign $*p = ...`). The value itself may be changed in other ways. Eg

int x = 5;
const int *p = &x;
x = 6; //legal
printf("%d", *p) // prints 6
*p = 7; //error 

This is meant to be used mostly in function signatures, to guarantee that the function can't accidentally change the arguments passed.

Answered   2023-09-21 08:11:47

I drew an image below to explain this, maybe helpful.

int const v and const int v are identical.

enter image description here

Answered   2023-09-21 08:11:47

This mostly addresses the second line: best practices, assignments, function parameters etc.

General practice. Try to make everything const that you can. Or to put that another way, make everything const to begin with, and then remove exactly the minimum set of consts necessary to allow the program to function. This will be a big help in attaining const-correctness, and will help ensure that subtle bugs don't get introduced when people try and assign into things they're not supposed to modify.

Avoid const_cast<> like the plague. There are one or two legitimate use cases for it, but they are very few and far between. If you're trying to change a const object, you'll do a lot better to find whoever declared it const in the first pace and talk the matter over with them to reach a consensus as to what should happen.

Which leads very neatly into assignments. You can assign into something only if it is non-const. If you want to assign into something that is const, see above. Remember that in the declarations int const *foo; and int * const bar; different things are const - other answers here have covered that issue admirably, so I won't go into it.

Function parameters:

Pass by value: e.g. void func(int param) you don't care one way or the other at the calling site. The argument can be made that there are use cases for declaring the function as void func(int const param) but that has no effect on the caller, only on the function itself, in that whatever value is passed cannot be changed by the function during the call.

Pass by reference: e.g. void func(int &param) Now it does make a difference. As just declared func is allowed to change param, and any calling site should be ready to deal with the consequences. Changing the declaration to void func(int const &param) changes the contract, and guarantees that func can now not change param, meaning what is passed in is what will come back out. As other have noted this is very useful for cheaply passing a large object that you don't want to change. Passing a reference is a lot cheaper than passing a large object by value.

Pass by pointer: e.g. void func(int *param) and void func(int const *param) These two are pretty much synonymous with their reference counterparts, with the caveat that the called function now needs to check for nullptr unless some other contractual guarantee assures func that it will never receive a nullptr in param.

Opinion piece on that topic. Proving correctness in a case like this is hellishly difficult, it's just too damn easy to make a mistake. So don't take chances, and always check pointer parameters for nullptr. You will save yourself pain and suffering and hard to find bugs in the long term. And as for the cost of the check, it's dirt cheap, and in cases where the static analysis built into the compiler can manage it, the optimizer will elide it anyway. Turn on Link Time Code Generation for MSVC, or WOPR (I think) for GCC, and you'll get it program wide, i.e. even in function calls that cross a source code module boundary.

At the end of the day all of the above makes a very solid case to always prefer references to pointers. They're just safer all round.

Answered   2023-09-21 08:11:47

Just for the sake of completeness for C following the others explanations, not sure for C++.

  • pp - pointer to pointer
  • p - pointer
  • data - the thing pointed, in examples x
  • bold - read-only variable

Pointer

  • p data - int *p;
  • p data - int const *p;
  • p data - int * const p;
  • p data - int const * const p;

Pointer to pointer

  1. pp p data - int **pp;
  2. pp p data - int ** const pp;
  3. pp p data - int * const *pp;
  4. pp p data - int const **pp;
  5. pp p data - int * const * const pp;
  6. pp p data - int const ** const pp;
  7. pp p data - int const * const *pp;
  8. pp p data - int const * const * const pp;
// Example 1
int x;
x = 10;
int *p = NULL;
p = &x;
int **pp = NULL;
pp = &p;
printf("%d\n", **pp);

// Example 2
int x;
x = 10;
int *p = NULL;
p = &x;
int ** const pp = &p; // Definition must happen during declaration
printf("%d\n", **pp);

// Example 3
int x;
x = 10;
int * const p = &x; // Definition must happen during declaration
int * const *pp = NULL;
pp = &p;
printf("%d\n", **pp);

// Example 4
int const x = 10; // Definition must happen during declaration
int const * p = NULL;
p = &x;
int const **pp = NULL;
pp = &p;
printf("%d\n", **pp);

// Example 5
int x;
x = 10;
int * const p = &x; // Definition must happen during declaration
int * const * const pp = &p; // Definition must happen during declaration
printf("%d\n", **pp);

// Example 6
int const x = 10; // Definition must happen during declaration
int const *p = NULL;
p = &x;
int const ** const pp = &p; // Definition must happen during declaration
printf("%d\n", **pp);

// Example 7
int const x = 10; // Definition must happen during declaration
int const * const p = &x; // Definition must happen during declaration
int const * const *pp = NULL;
pp = &p;
printf("%d\n", **pp);

// Example 8
int const x = 10; // Definition must happen during declaration
int const * const p = &x; // Definition must happen during declaration
int const * const * const pp = &p; // Definition must happen during declaration
printf("%d\n", **pp);

N-levels of Dereference

Just keep going, but may the humanity excommunicate you.

int x = 10;
int *p = &x;
int **pp = &p;
int ***ppp = &pp;
int ****pppp = &ppp;

printf("%d \n", ****pppp);

Answered   2023-09-21 08:11:47

  1. const int* - pointer to constant int object.

You can change the value of the pointer; you can not change the value of the int object, the pointer points to.


  1. const int * const - constant pointer to constant int object.

You can not change the value of the pointer nor the value of the int object the pointer points to.


  1. int const * - pointer to constant int object.

This statement is equivalent to 1. const int* - You can change the value of the pointer but you can not change the value of the int object, the pointer points to.


Actually, there is a 4th option:

  1. int * const - constant pointer to int object.

You can change the value of the object the pointer points to but you can not change the value of the pointer itself. The pointer will always point to the same int object but this value of this int object can be changed.


If you want to determine a certain type of C or C++ construct you can use the Clockwise/Spiral Rule made by David Anderson; but not to confuse with Anderson`s Rule made by Ross J. Anderson, which is something quite distinct.

Answered   2023-09-21 08:11:47

Nobody has mentioned the system underlying declarations which Kernighan and Ritchie pointed out in their C book:

Declarations mimic expressions.

I'll repeat this because it so essential and gives a clear strategy to parse even the most complicated declarations:

Declarations mimic expressions.

The declarations contain the same operators as expressions the declared identifier can appear in later, with the same priority they have in expressions. This is why the "clockwise spiral rule" is wrong: The evaluation order is strictly determined by the operator precedences, with complete disregard for left, right or rotational directions.

Here are a few Examples, in order of increasing complexity:

  • int i;: When i is used as-is, it is an expression of type int. Therefore, i is an int.
  • int *p;: When p is dereferenced with *, the expression is of type int. Therefore, p is a pointer to int.
  • const int *p;: When p is dereferenced with *, the expression is of type const int. Therefore, p is a pointer to const int.
  • int *const p;: p is const. If this constant expression is dereferenced with *, the expression is of type int. Therefore, p is a const pointer to int.
  • const int *const p;: p is const. If this constant expression is dereferenced with *, the expression is of type const int. Therefore, p is a const pointer to const int.

So far we didn't have any issues with operator precedence yet: We simply evaluated right-to-left. This changes when we have fun with arrays of pointers and pointers to arrays. You may want to have a cheat sheet open.

  • int a[3];: When we apply the array indexing operator to a, the result is an int. Therefore, a is an array of int.
  • int *a[3];: Here the indexing operator has higher precedence, so we apply it first: When we apply the array indexing operator to a, the result is an int *. Therefore, a is an array of pointers to int. This is not uncommon.
  • int (*a)[3];: Here the operator precedence is overridden by round parentheses, exactly as in any expression. Consequently, we dereference first. We know now that a is a pointer to some type. *a, the dereferenced pointer, is an expression of that type. When we apply the array indexing operator to *a, we obtain a plain int, which means that *a is an array of three ints, and a is a pointer to that array. This is fairly uncommon outside of C++ templates, which is why the operator precedences are not catering to this case. Note how the use of such a pointer is the model for its declaration: int i = (*a)[1];. The parentheses are mandatory to dereference first.
  • int (*a)[3][2];: There is nothing preventing anybody from having pointers to multi-dimensional arrays, one case where circular spiral clockwise advice becomes obvious nonsense.

A thing that sometimes comes up in real life are function pointers. We need parentheses there as well because the function call operator (operator()() in C++, simple syntax rule in C) has higher priority than the dereferencing operator*(), again because it's more common to have functions returning pointers than pointers to functions:

  • int *f();: Function call first, so f is a function. The call must be dereferenced to result in an int, so the return value is a pointer to int. Usage: int i = *f();.

  • int (*fp)();: Parentheses change order of operator application. Because we must dereference first we know that fp is a pointer to something. Because we can apply the function call operator to *fp we know (in C) that fp is a pointer to a function; in C++ we only know that it is something for which operator()() is defined. Since the call takes no parameters and returns an int, fp is in C++ a pointer to a function with that signature. (In C, an empty parameter list indicates that nothing is known about the parameters, but future C specifications may forbid that obsolete use.)

  • int *(*fp)();: Of course we can return pointers to int from a function pointed to.

  • int (*(*fp)())[3];: Dereference first, hence a pointer; apply function call operator next, hence a pointer to function; dereference the return value again, hence a pointer to a function returning a pointer; apply the indexing operator to that: pointer to function returning pointer to array. The result is an int, hence pointer to function returning pointer to array of ints.-

    All parentheses are necessary: As discussed, we must prioritize dereferencing of the function pointer with (*fp) before anything else happens. Obviously, we need the function call; and since the function returns a pointer to an array (not to its first element!), we must dereference that as well before we can index it. I admit that I wrote a test program to check this because I wasn't sure, even with this fool-proof method ;-). Here it is:

#include <iostream>
using namespace std;

int (*f())[3]
{
  static int arr[3] = {1,2,3};
  return &arr;
}

int (*(*fp)())[3] = &f;

int main()
{
  for(int i=0; i<3; i++)
  {
    cout << (*(*fp)())[i] << endl;
  }
}

Note how beautifully the declaration mimics the expression!

Answered   2023-09-21 08:11:47

simple mnemonic:

type pointer <- * -> pointee name


I like to think of int *i as declaring "the dereference of i is int"; in this sense, const int *i means "the deref of i is const int", while int *const i means "deref of const i is int".

(the one danger of thinking like this is it may lead to favoring int const *i style of declaration, which people might hate/disallow)

Answered   2023-09-21 08:11:47

Lot of people answered correctly I will just organize well here and put some Extra info which is missing in given Answers.

Const is keyword in C language also known as qualifier. Const can applied to the declaration of any variable to specify that it's value will not changed

const int a=3,b;

a=4;  // give error
b=5;  // give error as b is also const int 

you have to intialize while declaring itself as no way to assign
it afterwards.

How to read ?

just read from right to left every statement works smoothly

3 main things

type a.    p is ptr to const int

type b.    p is const ptr to int 
 
type c.    p is const ptr to const int

[Error]

if * comes before int 

two types

1. const int *

2. const const int *

we look first

Major type 1. const int*

ways to arrange 3 things at 3 places 3!=6

i. * at start

*const int p      [Error]
*int const p      [Error]

ii. const at start

const int *p      type a. p is ptr to const int 
const *int p      [Error]

iii. int at start

int const *p      type a. 
int * const p     type b. p is const ptr to int

Major type 2. const const int*

ways to arrange 4 things at 4 places in which 2 are alike 4!/2!=12

i. * at start

* int const const p     [Error]
* const int const p     [Error]
* const const int p     [Error]
 

ii. int at start

int const const *p      type a. p is ptr to const int
int const * const p     type c. p is const ptr to const int
int * const const p     type b. p is const ptr to int

iii. const at start

const const int *p     type a.
const const * int p    [Error]

const int const *p      type a.
const int * const p     type c.

const * int const p    [Error]
const * const int p    [Error]

squeezing all in one

type a. p is ptr to const int (5)

const int *p
int const *p

int const const *p
const const int  *p
const int  const *p

type b. p is const ptr to int (2)

int * const p
int * const const p;

type c. p is const ptr to const int (2)

int const * const p
const int * const p

just little calculation

1. const int * p        total arrangemets (6)   [Errors] (3)
2. const const int * p  total arrangemets (12)  [Errors] (6)

little Extra

int const * p,p2 ;

here p is ptr to const int  (type a.) 
but p2 is just const int please note that it is not ptr

int * const p,p2 ;

similarly 
here p is const ptr to int  (type b.)   
but p2 is just int not even cost int

int const * const p,p2 ;

here p is const ptr to const int  (type c.)
but p2 is just const int. 

Finished

Answered   2023-09-21 08:11:47