What is "template using owner = T;"?

Question

Below is excerpted from gsl.h of Microsoft's gsl library (https://github.com/microsoft/gsl):

namespace gsl
{
    //
    // GSL.owner: ownership pointers 
    //
    using std::unique_ptr;
    using std::shared_ptr;

    template<class T>
    using owner = T;
    ...
};

I cannot understand what the following alias template means:

template<class T>
using owner = T;

Any explanations?

Possible duplicate of [What is the difference between 'typedef' and 'using' in C++11?](http://stackoverflow.com/questions/10747810/what-is-the-difference-between-typedef-and-using-in-c11) — Qix - MONICA WAS MISTREATED, Jul 19 '16 at 04:01

score 7 · Accepted Answer · answered Jul 19 '16 at 04:00

7

It means that for every T, owner<T> is an alias for T.

answered Jul 19 '16 at 04:00

Brian Bi

111,498
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Isn't it that owner is an alias of T. Not owner – West Jul 19 '16 at 04:09
3

@West, No, `owner` is a template. `T` is a type. – chris Jul 19 '16 at 04:11
Yup I was so wrong on this. For everyone like me this compiles. static_assert(is_same>(),"hello"); – West Jul 19 '16 at 04:24

score 2 · Answer 2 · answered Jul 13 '17 at 08:30

2

It can be used as annotation to show which pointers are 'owner' ie:

Example of non owning raw pointer

template<typename T>
class X2 {
    // ...
public:
    owner<T*> p;  // OK: p is owning
    T* q;         // OK: q is not owning
};

answered Jul 13 '17 at 08:30

arenard

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12

What is "template using owner = T;"?

2 Answers2

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