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I'm just starting to learn about C++, so forgive me if this is obvious. What is the difference between the below pointers?

const int i = 42;
constexpr const int *p = &i;
const int const* p2 = &i;

To me, it seems like they are both constant pointers to a constant int. Is there a difference? When would I use each type of declaration?

Thanks :)

japata
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    Note: `constexpr` is `c++11` and is unsupported syntax in older versions. Also its meaning in static and function call contexts has changed in `c++17`. For more details see: http://en.cppreference.com/w/cpp/language/constexpr – Jason R. Mick Mar 08 '17 at 18:17

2 Answers2

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  1)Meaning of constexpr int *p = &i;
    p is const: it cannot be reassigned
    i can be modified through p

  2)Meaning of const int *p2 = &i; 
    p2 is non-const: it can be reassigned to point to a different int
    i cannot be modified through p2 without use of a cast
Deepak Tatyaji Ahire
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This code is invalid C++ and in most cases won't even compile. You were probably interested in this article: What is the difference between const int*, const int * const, and int const *?

All three declarations

const int const* p2 = &i;
const int * p2 = &i;
int const* p2 = &i;

are just different ways to write the same thing: a (non-const pointer) to (const int). Most compilers will balk when it's declared like in your code through, because it's not a standard C++ and a likely error.

You will be able to write

p2 = p3; // assign value of (pointer p3) to (pointer p2)

because (pointer) is not const, but not 

*p2 = *p3; // assign (object) pointed by (pointer p3) to (object) pointed by (pointer p2)

because (object) that is pointed by a pointer is const (const int in this case)

But if you will write

int * const p2 = &i;

then it'll be a (const pointer) pointing to (int) object. Then things will be completely opposite:

p2 = p3; // don't work
*p2 = *p3; // work!

You can also write const int * const to prohibit both options or don't use const to allow both.

Constexpr is a very different keyword. Rather than say that it's simply a (const object), it says that it's a (const object with value known at compile time). Most "const objects" in C++ are not 'constants' in mathematical sense since they could be different in different runs of same program. They are created at runtime and we just tell compiler that we're not going to change them later. On the other hands, 'constexpr' means that it's a 'real' constant that is defined once and for all. It's always the same and thereby it gets hardcoded in compiled binary code. Since you are beginner, think about it as a fancy C++ version of #define. In your case you tried to declare a pointer to const int to be constexpr, but it won't happen since this pointer is going to be different in different runs of your program and cannot be determined at compile time.

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Sergei Ozerov
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  • Thanks for your help. Did something change about `constexpr` from C++11 to C++14? I'm reading **C++ Primer** for C++11, and one of the examples straight of the book is: `constexpr const int *p = &i;` But when I try to compile it in a C++14 project, it crashes with `'& i' is not a constant expression`. – japata Mar 08 '17 at 20:55
  • No, as far as I know `constexpr` hasn't changed and it won't make any sense to mark a pointer as constexpr, except for few rare cases where it's a pointer to static object or function pointer and thereby it can be determined at compile-time. I don't have this book on hands, but most likely you've changed an example from the book in a subtle way (i.e. changed i from static variable to local one). – Sergei Ozerov Mar 09 '17 at 08:56