将const和decltype与指针变量一起使用

Question

The following code allows me to change the value at *p2 even though p2 is declared with const.

int *p1;

const decltype(p1) p2 = new int(100);
*p2 = 200;
cout << "*p2: " << *p2 << endl; // Outputs *p2: 200

However, if I use "int *" instead of "decltype(p1)", then the compiler flags an error.

const int * p2 = new int(100);
*p2 = 200;
cout << "*p2: " << *p2 << endl;

error: assignment of read-only location ‘* p2’
  *p2 = 200;
      ^

I am using g++ (Ubuntu 4.8.2-19ubuntu1) 4.8.2.

Does decltype ignore const specifier when applied on pointer variable?

Answer 1

const decltype(p1) p2 means int* const p2 .

This means that you cannot change p2 , but you may change the things being pointed to.

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const T always applies const to the so-called "top level" of T . When T is a composite type (that is, a type built up from the basic set of types) then you will have to jump through hoops if you want to apply const to the lower levels.

When T is int * , adding top level const would give int * const . The * demarcates a level; to get at the stuff below the * you have to manually remove the * , apply const , then put the * back.

A possible solution is:

const std::remove_pointer<decltype(p1)>::type *p2 = new int(100);

Answer 2

std::pointer_traits would come in handy here. std::pointer_traits::element_type and std::pointer_traits::rebind allow you to write a generic expression which will work well for any pointer-like type:

using element_type = std::pointer_traits<decltype(p1)>::element_type;
using pointer_like_to_const = std::pointer_traits<decltype(p1)>::rebind<std::add_const_t<element_type>>;
pointer_like_to_const p2 = new int(100);

Note that this code would work even if p1 was shared_ptr<int> or unique_ptr<int> .