9.15 — Pointers and const

By Alex on July 16th, 2007 | last modified by nascardriver on January 30th, 2021

Pointing to const variables

So far, all of the pointers you’ve seen are non-const pointers to non-const values:

1 2 3

int value{ 5 }; int* ptr{ &value }; *ptr = 6; // change value to 6

However, what happens if value is const?

1 2 3

const int value{ 5 }; // value is const int* ptr{ &value }; // compile error: cannot convert const int* to int* *ptr = 6; // change value to 6

The above snippet won’t compile -- we can’t set a non-const pointer to a const variable. This makes sense: a const variable is one whose value can not be changed. Hypothetically, if we could set a non-const pointer to a const value, then we would be able perform indirection through the non-const pointer and change the value. That would violate the intention of const.

Pointer to const value

A pointer to a const value is a (non-const) pointer that points to a constant value.

To declare a pointer to a const value, use the const keyword before the data type:

1 2 3

const int value{ 5 }; const int* ptr{ &value }; // this is okay, ptr is a non-const pointer that is pointing to a "const int" *ptr = 6; // not allowed, we can't change a const value

In the above example, ptr points to a const int.

So far, so good, right? Now consider the following example:

1 2	int value{ 5 }; // value is not constant const int* ptr{ &value }; // this is still okay

A pointer to a constant variable can point to a non-constant variable (such as variable value in the example above). Think of it this way: a pointer to a constant variable treats the variable as constant when it is accessed through the pointer, regardless of whether the variable was initially defined as const or not.

Thus, the following is okay:

1 2 3

int value{ 5 }; const int* ptr{ &value }; // ptr points to a "const int" value = 6; // the value is non-const when accessed through a non-const identifier

But the following is not:

1 2 3

int value{ 5 }; const int* ptr{ &value }; // ptr points to a "const int" *ptr = 6; // ptr treats its value as const, so changing the value through ptr is not legal

Because a pointer to a const value is not const itself (it just points to a const value), the pointer can be redirected to point at other values:

1 2 3 4 5

int value1{ 5 }; const int *ptr{ &value1 }; // ptr points to a const int   int value2{ 6 }; ptr = &value2; // okay, ptr now points at some other const int

Const pointers

We can also make a pointer itself constant. A const pointer is a pointer whose value can not be changed after initialization

To declare a const pointer, use the const keyword between the asterisk and the pointer name:

1 2	int value{ 5 }; int* const ptr{ &value };

Just like a normal const variable, a const pointer must be initialized to a value upon declaration. This means a const pointer will always point to the same address. In the above case, ptr will always point to the address of value (until ptr goes out of scope and is destroyed).

1 2 3 4 5

int value1{ 5 }; int value2{ 6 };   int* const ptr{ &value1 }; // okay, the const pointer is initialized to the address of value1 ptr = &value2; // not okay, once initialized, a const pointer can not be changed.

However, because the value being pointed to is still non-const, it is possible to change the value being pointed to indirection through the const pointer:

1 2 3

int value{ 5 }; int* const ptr{ &value }; // ptr will always point to value *ptr = 6; // allowed, since ptr points to a non-const int

Const pointer to a const value

Finally, it is possible to declare a const pointer to a const value by using the const keyword both before the type and before the variable name:

1 2	int value{ 5 }; const int* const ptr{ &value };

A const pointer to a const value can not be set to point to another address, nor can the value it is pointing to be changed through the pointer.

Recapping

To summarize, you only need to remember 4 rules, and they are pretty logical:

• A non-const pointer can be redirected to point to other addresses.
• A const pointer always points to the same address, and this address can not be changed.

• A pointer to a non-const value can change the value it is pointing to. These can not point to a const value.
• A pointer to a const value treats the value as const (even if it is not), and thus can not change the value it is pointing to.

Keeping the declaration syntax straight can be challenging. Just remember that the type of value the pointer points to is always on the far left:

1 2 3 4

int value{ 5 }; const int* ptr1{ &value }; // ptr1 points to a "const int", so this is a pointer to a const value. int* const ptr2{ &value }; // ptr2 points to an "int", so this is a const pointer to a non-const value. const int* const ptr3{ &value }; // ptr3 points to a "const int", so this is a const pointer to a const value.

Conclusion

Pointers to const values are primarily used in function parameters (for example, when passing an array to a function) to help ensure the function doesn’t inadvertently change the passed in argument. We will discuss this further in the section on functions.

9.16 -- Reference variables

Index

9.14 -- Dynamically allocating arrays

C++ Tutorial |  Print This Post

116 comments to 9.15 — Pointers and const

• kio

  March 10, 2021 at 12:46 pm · Reply

  Summary,

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

  int main() {       int nonConstVariable{ 5 };     int nonConstVarible2{ 6 };     const int constVariable{ 5 };       //// A non-const pointer can be redirected to point to other addresses.     int* pointer{ &nonConstVariable };     pointer = nullptr;       pointer = &nonConstVarible2;     pointer = nullptr;       //// A const pointer always points to the same address, and this address can not be changed.     int* const constPointer{ &nonConstVariable };     // constPointer = nullptr; // can't point to diffrente address even nullptr?     *constPointer = 10;        // works, because it can change the value at the given address but not the address itself     delete constPointer;       //// A pointer to a non-const value can change the value it is pointing to. These can not point to a const value.     pointer = &nonConstVarible2;     // pointer = &constVariable; // can't point the non-const pointer to the const value?       //// A pointer to a const value treats the value as const (even if it is not), and thus can not change the value it is pointing to.     const int* const constPointer2{ &constVariable };     // *constPointer2 = 15; // l-value must be modifiable value (constVariable isn't modifible l-value) it's constant          delete pointer;     }

• Waldo Lemmer

  January 29, 2021 at 1:04 pm · Reply

  1. The snippets in this lesson use copy initialization instead of uniform initialization, I don't know if it's intentional.

  2. Section "Const pointers", snippet 2, line 4:

  1	int * const ptr = &value1; // okay, the const pointer is initialized to the address of value1

  This is the only place where you have a space after the asterisk, so I assume it's a typo.

  This lesson is otherwise written well :)

  • nascardriver

    January 30, 2021 at 5:42 am · Reply

    Thank you once again :)

• Andrew Johnson

  November 23, 2020 at 7:20 am · Reply

  Following on from a comment on 3rd Nov 2020 by Hersel99 - this code works but I'm interested to know if it's problematic in that a constant value IS changed once defined:

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

  #include <iostream>   int main() {     int a {3};     const int b {5};       int* pToA {&a};     int* pToB {&a}; // initially pointing at A but this will change       std::cout << "Before change:" << '\n';     std::cout << "A: " << pToA << " " << *pToA << '\n';        // 3?     std::cout << "B: " << &b << " " << *(&b) << '\n';    // 5       --pToB; //Trick to acces address of constant int `b`     *pToB = 6; // This should change const int `b` to 5 (Mua, ha, ha!)       std::cout << "After change:" << '\n';     std::cout << "A: " << pToA << " " << *pToA << '\n';        // 3?     std::cout << "B: " << &b << " " << *(&b) << '\n';    // 5 }   Outputs: Before change: A: 0xffffcbdc 3 B: 0xffffcbd8 5 After change: A: 0xffffcbdc 3 B: 0xffffcbd8 6

  Also, can I add that these lessons are excellent but I find your use of the asterisk (*) problematic in that I have to think a lot harder about how its being used.  What I mean is that you use it in the following ways:

  1 2 3 4 5 6

  int * anIntPtr {&value}            // to define a pointer to an integer variable int *anIntPtr {&value}             // to define a pointer to an integer variable int *const anIntPtr {&value}    // to define a pointer to a constant integer variable int * const anIntPtr {&value}    // to define a pointer to a constant integer variable int* aFunction() {..}             // to define a function that returns an integer pointer *anInt = 4                         // to change a value at a specific address

  As I understand it so far, the first 5 statements are all creating a pointer to an integer; the last is actually changing a value at an address pointed at by an integer pointer.
  Unless it is common practice, it would be a lot less confusing (by which I mean cognitively disrupting) if you could define pointer types as you would other types:

  1 2 3 4

  int* anIntPtr {&value}            // to define a pointer to an integer variable int* const anIntPtr {&value}    // to define a pointer to a constant integer variable int* aFunction() {..}             // to define a function that returns an integer pointer *anInt = 4                         // to change a value at a specific address

  You also get rid of two variations (albeit probably typed with spacing by mistake)!  

  Keep up the great work.

  • nascardriver

    November 24, 2020 at 7:17 am · Reply

    1 2	int * anIntPtr int * const anIntPtr

    Should never appear. The other ones are very slowly being changed to `int*`.

    Line 14 invokes undefined behavior. Pointer arithmetic is only allowed on pointers that point to arrays and only in the array's range.

  • Sahil

    January 31, 2021 at 6:40 am · Reply

    How can you be sure that line 14 will give you access to b every time when you run your program? b could be stored anywhere on the stack and you might as well end up corrupting some other data when you assign a different value to pToB. This would fall under undefined behavior, unless I'm missing something

• Alek

  May 27, 2020 at 11:42 pm · Reply

  hello I have a question although I'm 99% percent sure I cannnot be satisfied with it.why did they make these variables this way xD too hard to remember this can be so dangerous when one has around 1k line of code in ahead so it's normal to forget and mistakenly do such thing as below :

  1 2 3

  const int *b; int const *b; /*if we put astrisk symbol here instead of before const it will be equivalent to "const int *b" and it wouldn't treat it as a const pointer right ?*/

• z0dr

  February 20, 2020 at 12:06 pm · Reply

  in section pointing to const variable
  int *ptr = &value; // compile error: cannot convert const int* to int*

  your explaination : "This makes sense: a const variable is one whose value can not be changed"

  This compiler error is type convertion error but you explain this as error thrown due to "const cannot be changed". Did I miss something there?

  • nascardriver

    February 22, 2020 at 12:47 am · Reply

    The error is the conversion, because `const int*` can't be cast to `int*`.
    The reason for this conversion to be illegal is that a `const int*` can't be changed, but an `int*` can.

    • z0dr

      February 23, 2020 at 6:25 pm · Reply

      Thanks, that is what i am saying in my comment but this error explained as :
      "This makes sense: a const variable is one whose value can not be changed"

• Ged

  November 24, 2019 at 6:54 am · Reply

  I commented the line which works and which doesn't. Just wanted to ask. When we initialize a variable with some data type we use braces. But when we want to assign to it a new value or in this case a new address, we cannot use braces again?

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

  #include <iostream>   int main() {     int value1{ 5 };     const int* ptr{ &value1 };     std::cout << ptr << '\n';     std::cout << *ptr << '\n';     std::cout << "--------------------------\n";     int value2{ 6 };     ptr{ &value2 }; // This doesn't work         ptr = &value2; // This works     std::cout << ptr << '\n';     std::cout << *ptr << '\n';     return 0; }

  • nascardriver

    November 24, 2019 at 6:59 am · Reply

    Correct. We use braces for initialization (Or parentheses or equals, but braces provide higher type-safety). Initialization can only occur once. You can assign as many times as you like, with an equals-sign.

• hero

  November 16, 2019 at 5:55 am · Reply

  do you also have something about "this->"?

  • Alex

    November 16, 2019 at 6:00 pm · Reply

    Yes, https://www.learncpp.com/cpp-tutorial/8-8-the-hidden-this-pointer/

• hersel99

  November 3, 2019 at 9:40 am · Reply

  Hi! I was trying to see what happen if I access a const int with a pointer to a non-const int.
  And I end up with a great idea using increment or decrement on my pointer (I know it's dangerous but I'm curious and want to know) as you can see bellow.
  However things doesn't go as I expected it.

  1 2 3 4 5 6 7 8 9 10

  int a {3}; const int b {5}; int *p {&a}; --p; //Trick to acces address of constant int `b`   *p = 6; // This should change const int `b` to 5 (Mua, ha, ha!)   std::cout << &b << " " << *(&b) << '\n';    // 5 std::cout << p << " " << *p << '\n';        // 6? std::cout << (p == &b);                     // true

  This was compiled in codeblocks, gcc 6.3, debian.
  So why my pointer `p` points to a value 6? Where does this "magic" number comes from?

  Also I do have a gravatar image but seems like it isn't working right now.

  • nascardriver

    November 4, 2019 at 3:35 am · Reply

    Hi!

    > I do have a gravatar image but seems like it isn't working right now
    Your email is correct, you should have a cat avatar. If your avatar still doesn't show up the next time you post a comment, mention it again.

    > Where does this "magic" number comes from?
    You're setting it in line 6.

    Line 6 causes undefined behavior. For one, because `p` is an invalid pointer, then again because you're trying to modify a `const` variable.
    Your compiler probably replaced `*(&b)` in line 8 with 5. It's allowed to do that, because `b` is const, so its value can never change anyway. Line 9 then accesses `b` location in memory, which you changed to 6.

• Andy

  October 29, 2019 at 12:24 am · Reply

  Hi,
  A very good tutorial, with good advice on best practice.
  One thing I was wondering about with respect to const pointers, would it be worthwhile recommending that pointers to dynamically allocated objects should be const?
  This would prevent one of the memory leak errors where the pointer is assigned to a different address, eg:

  1 2 3

  int value = 5; const int *ptr = new int; // allocate memory ptr = &value; // compile error, old address not lost, memory leak does not occur

  But a disadvantage of this is that would not allow the pointer to be assigned to nullptr after the object has been deleted

  • nascardriver

    October 29, 2019 at 2:58 am · Reply

    Hi

    Your `ptr` points to a const `int`, but the pointer itself is non-const.

    1 2 3 4 5 6 7

    const int* ptr{ new int{} }; *ptr = 3; // Illegal ptr = nullptr; // Legal   int* const ptr{ new int{} }; *ptr = 3; // Legal ptr = nullptr; // Illegal

    I wouldn't say this is too useful. There'll still be pointers that you can't initialize with a new expression, so they can't be const.
    Fortunately this isn't too big of a problem, because C++ has smart pointers (Covered later). With them you don't have to worry about leaks.

    • Andy

      October 30, 2019 at 1:36 am · Reply

      Thanks for correcting my mistake, I realised it after going through a later chapter.
      And thanks for the explanation, I will look forward to smart pointers...

• hellmet

  October 21, 2019 at 9:17 am · Reply

  In recap, I think it would be better if it was a table, like so

  1 2 3 4 5 6 7 8 9 10 11

  Legend..   p is a pointer   v is value   CC = 'can change', abbreviated for space constraints :P     Replace 'int' with any other type required           p, v types        CC p?  CC v?    CC v with p?    Declaration of p nonconst p to nonconst v  Yes    Yes    Yes             int *p nonconst p to    const v  Yes    No     No              const int *p    const p to nonconst v  No     Yes    Yes             int *const p    const p to    const v  No     No     No              const int *const p

  Hope that's correct!

• Behroo

  February 28, 2019 at 10:40 am · Reply

  Hi!
  I've tried this code :

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37

  #include "pch.h" #include <iostream>   void showValue(int *ptr) {     std::cout << ptr << " Is a pointer that's pointing "         << "to a non-const value " << *ptr << '\n'; }   void showValue(const int *ptr) {     std::cout << ptr << " Is a pointer that's poiting "         << "to a const Or non-const value " << *ptr << '\n'; }   void showValue(const int *const ptr) {     std::cout << ptr << " Is a const pointer that's poiting "         << "to a const Or non-const value " << *ptr << '\n'; }   int main() {     int value1{ 10 };     const int value2{ 12 };     int *ptr1{ &value1 };     const int *ptr2{ &value1 };     const int *ptr3{ &value2 };     const int *const ptr4{ &value1 };     const int *const ptr5{ &value2 };       showValue(ptr1);     showValue(ptr2);     showValue(ptr3);     showValue(ptr4);     showValue(ptr5); }

  I was expecting to run and each function call finds its path but I got these errors :

  Error    C2264    'showValue': error in function definition or declaration; function not called    33    
  Error    C2264    'showValue': error in function definition or declaration; function not called    32    
  Error    C2264    'showValue': error in function definition or declaration; function not called    31    
  Error    C2264    'showValue': error in function definition or declaration; function not called    30    
  Error    C2084    function 'void showValue(const int *)' already has a body    15    

  can you explain me why did that happen ?

  • nascardriver

    March 2, 2019 at 4:33 am · Reply

    Hi Behroo!

    The last @showValue is ambiguous. Here's a smaller example that's hopefully easier to understand:

    1 2 3 4 5 6 7

    void fn(int i) {}   void fn(const int i) {}   fn(3); // Which @fn is supposed to be called?

    • Shrushti Vora

      March 29, 2019 at 3:35 am · Reply

      Hi,
      My question is just an extension to the question asked by Behroo. I fixed the part of the code that was giving the errors.

      1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37

      #include <iostream>   void showValue(int *ptr) {     std::cout << ptr << " is a pointer that's pointing to a non-const value " << *ptr << '\n'; }   //! Part of the original code which is giving errors //! void showValue(const int *ptr) //! { //!     std::cout << ptr << " is a pointer that's pointing to a const or non-const value " << *ptr << '\n'; //! }   void showValue(const int *const ptr) {     std::cout << ptr << " is a const pointer that's pointing to a const or non-const value "               << *ptr << '\n'; }   int main() {     int value1{10};     const int value2{12};     int *ptr1{&value1};     const int *ptr2{&value1};     const int *ptr3{&value2};     const int *const ptr4{&value1};     const int *const ptr5{&value2};       showValue(ptr1);     showValue(ptr2);     showValue(ptr3);     showValue(ptr4);     showValue(ptr5);       return 0; }

      My question is that when the function showValue is called  for

      1 2	showValue(ptr2); showValue(ptr3);

      I would expect the compiler would either complaint since none of the above-defined functions have matching arguments. But instead, this is what I got as output on my machine:

      0x7ffeea785728 is a pointer that's pointing to a non-const value 10
      0x7ffeea785728 is a const pointer that's pointing to a const or non-const value 10
      0x7ffeea785724 is a const pointer that's pointing to a const or non-const value 12
      0x7ffeea785728 is a const pointer that's pointing to a const or non-const value 10
      0x7ffeea785724 is a const pointer that's pointing to a const or non-const value 12

      Can you explain why it called the second function since it only takes a const pointer to a const int?

      • nascardriver

        March 29, 2019 at 3:55 am · Reply

        const parameters can be initialized with const and non-const types. The const that's not part of the pointer-type declaration only affects the contents of the function, but not the call.

        1	void fn(const int i);

        can be called with a const or non-const int. The const only means that @i cannot be modified inside of @fn.

        The same goes for pointers

        1	void fn(const int *const p);

        The first const means that the pointer cannot be used to modify the pointed-to object. This is a restriction the caller has to follow.
        The second const only means that @p cannot be modified inside of @fn. This doesn't affect the caller, since they won't see modifications to @p anyway.

• Ian

  February 1, 2019 at 10:30 am · Reply

  Hi,

  const int value = 5;
  const int *ptr = &value; // this is okay, ptr is a non-const pointer that is pointing to a "const int"
  *ptr = 6; // not allowed, we can't change a const value

  Should line two here not be "this is okay, ptr is a const pointer...

  Just trying to correct for future generations...

  • nascardriver

    February 2, 2019 at 4:13 am · Reply

    Hi Ian!

    Nope, it's correct as it is.
    The pointer itself is not const.

    1 2	// This is allowed, as is any other reassignment. ptr = nullptr;

    const pointers use this syntax

    1 2 3 4 5 6 7 8

    // Assuming @value is non-const int *const ptr{ &value };   // Not allowed ptr = nullptr;   // Allowed *ptr = 6;

• Raindrop7

  January 26, 2019 at 6:00 pm · Reply

  good explanation. would you add multiple indirection pointers?  I mean pointers to pointers.

  • Alex

    January 27, 2019 at 5:03 pm · Reply

    Those are covered later in this chapter.

• Nguyen

  October 16, 2018 at 9:23 pm · Reply

  Hi,

  1 2 3

  int value = 5; // value is not constant const int *ptr = &value; // this is still okay

  A pointer(ptr) to a constant variable(???) can point to a non-constant variable(such as variable value in the example above).

  I am confused.  What is the constant variable?  

  Thanks, Have a great day

  • nascardriver

    October 17, 2018 at 6:22 am · Reply

    @ptr is non-const
    @value is non-const
    But @ptr treats @value as a const int.

    1 2 3

    value = 9; // Legal *ptr = 9; // Illegal ptr = nullptr; // Legal

    • Nguyen

      January 8, 2019 at 11:44 pm · Reply

      But the following is not:

      1 2 3 4

      int value = 5; const int *ptr = &value; // ptr points to a "const int" *ptr = 6; // ptr treats its value as const, so changing the value through ptr is not legal

      Sorry, the comments in line 4 confuse me.  
      To me, the variable value is different from &value.  So, I am not sure if "its value" should be understood as &value or the variable value in "ptr treats its value as const".  Another thing is that ptr is not the same as *ptr, so "changing the value through ptr is not legal" confuses me.  It would make more sense to me if it was said "change the value through *ptr instead of ptr is not legal".

      • Piyush

        January 19, 2019 at 2:18 am · Reply

        1 2 3 4 5 6 7 8 9 10 11

        int value{5}; const int *ptr =&value; //This line means that when ptr will be dereferenced the value obtained will be treated as constant it is true if the value being pointed is const or not. //So you can do this. value=6; //But not this. *ptr=6; //Why because as soon as ptr is dereferenced the compiler treats it as a constant hence it can't be changed. //So basically you can't change the value of the variable "value" in the above program using ptr . like this. *ptr=SomeValue; // :( //But you can change it using value itself. value=someValue; // :)

• Bouke285

  February 22, 2018 at 10:10 am · Reply

  This isn't directly related to this lesson, but more general. I've heard the practice, "Declaration is initialization." Which I believe means any time you declare a member, you should initialize it to some value, even if not necessary.

  Is this a good practice, to initialize all variables to 0 or similar when you don't have an actual value to initialize to, even when it will be re-assigned inside a loop a couple of lines down the execution path (as an example)?

  I just notice that I've been having some trouble forgetting to initialize values to 0 when it is required, such as with a sum variable. Not all compilers will crash, but some will since this is undefined.

  Thank you!

  • nascardriver

    February 23, 2018 at 4:07 am · Reply

    Hi Bouke!

    Always initialize your variables! It appears you already figured why.

  • Alex

    February 26, 2018 at 11:35 am · Reply

    It's a good rule of thumb. It may be slightly inefficient in certain cases, but better inefficient than wrong. :)

• Sameer Verma

  January 15, 2018 at 11:45 pm · Reply

  Const pointer to a const value

  Finally, it is possible to declare a const pointer to a const value by using the const keyword both before the type and before the variable name:

  1 2	int value = 5;     //Shouldn't it be "const int value = 5" to declare the value as constant? const int *const ptr = &value;

  A const pointer to a const value can not be set to point to another address, nor can the value it is pointing to be changed through the pointer.

  • Alex

    January 17, 2018 at 6:08 pm · Reply

    It's not necessary. A pointer to a const value can point to a non-const value -- that value is considered const when accessed through the pointer.

    It's the same thing with references. A const reference can reference a non-const value.

• aditya rawat

  October 21, 2017 at 5:31 am · Reply

  hey Alex,
  pointers point to the address of a variable they are pointed to, so any change in the pointers will lead to a change in the original variable.

  now consider this program

  #include<iostream>
  using namespace std;
  int main()
  {
      int a=22;//a is assigned a value 22
      const int *ptr=&a;//ok, const ptr points to the address of a.
      cout<<*ptr*2<<endl;//gives 44
      cout<<a;// but why does the value of original variable does not change.
  }

  i know the question may be silly, but i didn't seem to to understand the  function of pointer in this particular case

  • Alex

    October 21, 2017 at 12:47 pm · Reply

    Be careful when you say "any change in the pointers", because you can change what address a pointer points to, and you can dereference the pointer to change the value it points to.

    In your example, you set ptr to point to variable a's address. Then you access a's value through *ptr, and multiply that value by 2, and print it. This doesn't affect a's value -- it's just a result that's calculated and thrown away.

    If you'd done this instead:

    1	*ptr = *ptr * 2; // assuming ptr wasn't const

    That would actually change a's value (through the pointer)

• Lamont Peterson

  September 8, 2017 at 6:43 am · Reply

  Alex,

  If I'm understanding the intent of this example snippet properly:

  Because a pointer to a const value is not const itself (it just points to a const value), the pointer can be redirected to point at other values:

  1 2 3 4 5

  int value1 = 5; const int *ptr = &value1; // ptr points to a const int int value2 = 6; ptr = &value2; // okay, ptr now points at some other const int

  ... then, shouldn't both of those variable declarations (for value1 & value2) be "const int"?

  • Alex

    September 9, 2017 at 8:14 pm · Reply

    They can be, but it's not necessary. A pointer to a const value can point at either a const value or a non-const value. In either case, the value will be treated as const when accessed through the pointer.

    • Lamont Peterson

      September 11, 2017 at 12:32 pm · Reply

      Alex,

      Good point.  Definitely worth noting.

      I think that while in the context of the lesson, the way it read for that example lead me to believe that the goal was for the pointer to be pointing to const int values (as in, it should be used only with type "const int" variables) but the pointer itself is NOT const (as in, the pointer can be changed to point somewhere else).  Thus, the ambiguity which led to my question.  Perhaps I was taking you too literally?

      Perhaps both points ought be mentioned in this lesson?

• haihui li

  August 31, 2017 at 7:25 pm · Reply

  Hi,Alex
  For beginners, it is difficult to distinguish between variables that are pointer types or normal variables.

  • Alex

    August 31, 2017 at 10:34 pm · Reply

    Do you think so? Variables that are pointer types are declared with an asterisk, while normal variables are not...

    • haihui li

      September 1, 2017 at 1:11 am · Reply

      thanks for your reply.If you add CONST, there will be chemical reactions between them。just like（int  *const ptr = ...）The asterisk is far from ptr，but it is pointer type.Although I understand now,This problem bothering me for a long time。So，I think the way of thinking is more important than its grammar.Thanks again.

• Finn St John

  August 28, 2017 at 10:33 am · Reply

  Hi Alex,

  you say: A pointer to a const value is a (non-const) pointer that points to a constant value.

  You clearly say that the pointer SHOULD NOT BE constant but in you very first example after that, the pointer you initialize IS a constant.

  If this a mistake or am i missing something?

  • Alex

    August 28, 2017 at 5:15 pm · Reply

    In that example, ptr is not a const pointer -- it's a non-const pointer pointing to a "const int".

• tf

  July 7, 2017 at 3:45 pm · Reply

  Hello,
  I was wondering how pointers interact with constexpr (variables getting evaluated at compile-time):

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

  int main() {     constexpr int value1{ 1 };     constexpr int value2{ 2 };     constexpr int value3{ 3 };     constexpr int value4{ 4 };     constexpr int value5{ 5 };          int *ptr1 = &value1;                    //not okay, since we point with a non const pointer at a const value.     const int *ptr2 = &value2;                //okay, since we point at a constexpr variable(?)     constexpr int *ptr3 = &value3;                //not okay, since it can't convert "const int *" -> "int *const "     const int *const ptr4 = &value4;            //okay, pointing a constant pointer at a constant virable     constexpr const int *ptr5 = &value5;            //okay, evaluates similar to ptr4?         return 0; }

  VS17 output:

  1 2 3 4 5 6 7

  1>------ Build started: Project: learncpp, Configuration: Debug Win32 ------ 1>learncpp.cpp 1>learncpp.cpp(9): error C2440: 'initializing': cannot convert from 'const int *' to 'int *' 1>learncpp.cpp(9): note: Conversion loses qualifiers 1>learncpp.cpp(11): error C2440: 'initializing': cannot convert from 'const int *' to 'int *const ' 1>learncpp.cpp(11): note: Conversion loses qualifiers ========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========

  I'm confused about ptr3 and ptr5. From my understanding ptr3 should not work, because the memory for value3 gets assigned at runtime and not at compile time, meaning that pointers can't be constexpr, since the compiler can't predict what memory address the OS will assign. Yet it only complains about being able to convert to a *const pointer (since it seems to be missing const). Isn't const implied by constexpr?

  VS2017 seems to treat "constexpr const int *ptr" exactly the same to "const int const* ptr"?  

  When I run the same stuff in Clang 4.0 (clang++ prog.cc -Wall -Wextra -std=gnu++1z)

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

  Start prog.cc:9:7: error: cannot initialize a variable of type 'int *' with an rvalue of type 'const int *'         int *ptr1 = &value1;                                                ^      ~~~~~~~ prog.cc:11:17: error: cannot initialize a variable of type 'int *const' with an rvalue of type 'const int *'         constexpr int *ptr3 = &value3;                                          ^      ~~~~~~~ prog.cc:13:23: error: constexpr variable 'ptr5' must be initialized by a constant expression         constexpr const int *ptr5 = &value5;                                  ^      ~~~~~~~ prog.cc:13:23: note: pointer to 'value5' is not a constant expression prog.cc:7:16: note: declared here         constexpr int value5{ 5 };                       ^ 3 errors generated. 1 Finish

  It complains about "constexpr const int *ptr5 = &value5;" not being a constant expression, which makes sense to me since the compiler does not know what the memory address will be at compile time.

  Is this just a quirk in VS2017 or is it undefined behavior?

  • Alex

    July 7, 2017 at 11:33 pm · Reply

    I think this is a quirk of VS2017. Since the address of a variable isn't known at compile time, I wouldn't expect you to be able to set a constexpr pointer to a constexpr value. But this is pushing my knowledge of the subject, so it's very possible I could be wrong.

• justAnotherConfusedStudent

  June 28, 2017 at 3:22 pm · Reply

  So I'm confused. When you initialize a pointer (not a const pointer) and have it point to a const int, can the pointer be redirected to point at other, non-const values?

  • Alex

    June 29, 2017 at 4:02 pm · Reply

    You can't make a non-const pointer point at a const int.

    • Rex Lucas

      November 30, 2017 at 6:20 am · Reply

      Alex
      In the light of your fundamental statement at the top of this page:
      "A pointer to a const value is a (non-const) pointer that points to a constant value."

      I guess I am missing some subtle point in your reply to "justAnotherConfusedStudent". Are you able to elaborate, please?

      • Alex

        December 2, 2017 at 5:27 pm · Reply

        I think I misread his question.

        A non-const pointer to a const value can be redirected.
        A const pointer to a non-const value can not be redirected.

        Basically, the constness of the pointer determines whether the pointer can have the address it holds changed, just like the constness of a variable determines whether that variables value can be changed.

• jenifer

  June 7, 2017 at 8:58 am · Reply

  how can i deal with pointers to static int?

  • Alex

    June 7, 2017 at 2:19 pm · Reply

    A pointer to a static int works the same way as a pointer to a non-static int.

    • jenifer

      June 7, 2017 at 9:10 pm · Reply

      What about the life time of the pointer to static int? Is it same as the static int, if so i can use it to return by address in a function.

      • Alex

        June 8, 2017 at 11:16 am · Reply

        The lifetime of a static int is until the end of the program, so yes, you could return a pointer to a local static variable and be fine.

        That said, I can't think of a good use case for actually doing this -- if the caller needs access to the static int's value, why wouldn't you return by value instead?

• Mehul

  April 26, 2017 at 2:34 am · Reply

  Hi Alex

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

  I have eritten below code and i am really confused how it is working       int k;     const int j = 0;       int *p = (int *)&j;          *p = 25;       cout << &j << endl;     cout << p << endl;       cout << j << endl;     cout << *p << endl;          cin >> k;

  O/P
  0x7ffff9bf6f80
  0x7ffff9bf6f80
  0
  25

  Queries
  Here pointer is pointing to a constant variable, And still modifying the values
  And strange thing is j remains 0 and *p is only updated to 25

  Can you please help me on understanding this

  • Alex

    April 26, 2017 at 12:19 pm · Reply

    You're casting away a const, which you should never do unless you're sure the original value was non-const. This will result in undefined behavior, which is what you're seeing.

    • Mehul

      April 29, 2017 at 5:27 am · Reply

      Thanks Alex