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2948. unique_ptr does not define operator<< for stream output

Section: 20.3.1 [unique.ptr] Status: C++20 Submitter: Peter Dimov Opened: 2017-03-19 Last modified: 2021-02-25

Priority: 0

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Discussion:

shared_ptr does define operator<<, and unique_ptr should too, for consistency and usability reasons.

[2017-07 Toronto Wed Issue Prioritization]

Priority 0; move to Ready

Proposed resolution:

This wording is relative to N4659.

1. Change 20.2.2 [memory.syn], header <memory> synopsis, as indicated:

   namespace std {
     […]
     
     // 20.3.1 [unique.ptr], class template unique_ptr
     […]
     template <class T, class D>
     bool operator>=(nullptr_t, const unique_ptr<T, D>& y);
     
     template<class E, class T, class Y, class D>
     basic_ostream<E, T>& operator<< (basic_ostream<E, T>& os, const unique_ptr<Y, D>& p);
     […]
   }
   

2. Change 20.3.1 [unique.ptr], class template unique_ptr synopsis, as indicated:

   namespace std {
     […]
     template <class T, class D>
     bool operator>=(nullptr_t, const unique_ptr<T, D>& y);
     
     template<class E, class T, class Y, class D>
     basic_ostream<E, T>& operator<< (basic_ostream<E, T>& os, const unique_ptr<Y, D>& p);
   }
   

3. Add a new subclause following subclause 20.3.1.6 [unique.ptr.special] as indicated:

   23.11.1.?? unique_ptr I/O [unique.ptr.io]

   template<class E, class T, class Y, class D>
     basic_ostream<E, T>& operator<< (basic_ostream<E, T>& os, const unique_ptr<Y, D>& p);
   

   -?- Effects: Equivalent to os << p.get();

   -?- Returns: os.

   -?- Remarks: This function shall not participate in overload resolution unless os << p.get() is a valid expression.