1

What is the difference between the following two pointer definitions 

int i = 0;
const int *p = &i;
constexpr int *cp = &i;
asked Feb 9, 2015 at 22:16
12
  • 2
    That one compiles, whilst the other doesn't. (ideone.com/NIO41u) Commented Feb 9, 2015 at 22:18
  • 3
    @OliverCharlesworth You're the one who put it inside a function body, causing the error. That's not in the question. Commented Feb 9, 2015 at 22:21
  • @hvd: That's a fair point. Commented Feb 9, 2015 at 22:21
  • 2
    @0x499602D2 according to c++ primer constexpr *p is a constant pointer to int and not a pointer to a const int Commented Feb 9, 2015 at 22:30
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    Can't a very complete answer be found in this SO question : stackoverflow.com/questions/14116003/… Commented Feb 9, 2015 at 22:40

1 Answer 1

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const int *p = &i; means:

  • p is non-const: it can be reassigned to point to a different int
  • i cannot be modified through p without use of a cast

constexpr int *cp = &i; means:

  • cp is const: it cannot be reassigned
  • i can be modified through p

In both cases, p is an address constant if and only if i has static storage duration. However, adding constexpr will cause a compilation error if applied to something that is not an address constant.

To put that another way: constexpr int *cp = &i; and int *const cp = &i; are very similar; the only difference is that the first one will fail to compile if cp would not be an address constant.

answered Feb 9, 2015 at 23:56
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