c++ - Calling `shared_ptr.get()` vs copy-initialization, what's the difference? -

first version print use_count = 2;

shared_ptr<int> s{make_shared<int>(15)};   auto b = s;   cout<<s.use_count()<<endl;   auto c = s.get();   cout<<s.use_count()<<endl;   cout<<*c<<endl;   

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second version use_count = 3;

shared_ptr<int> s{make_shared<int>(15)};   auto b = s;   cout<<s.use_count()<<endl;   auto c = s;   cout<<s.use_count()<<endl;   cout<<*c<<endl;   

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question:

• why behavior different between 2 versions?

introduction

every time make copy of shared_ptr use-count increased since have additional handle underlying resource being tracked.

in order obtain value of underlying pointer, shared_ptr has member-function named get. function return address of tracked resource, not create additional shared_ptr tracking same pointer.

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potential dangers

in first snippet have 2 instances of shared_ptr refers same resource, in latter have three.

the difference being;

• auto = s create copy of s, , type of a of s (a shared_ptr)
• auto b = s.get () initialize b address of tracked resource managed s, hence raw-pointer.

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if don't know doing, calling .get () might dangerous, can seen in following snippet.

int * ptr = nullptr;  {   std::shared_ptr<int> sp { make_shared<int> (15) };   ptr = sp.get (); }  std::cout << *ptr << std::endl; // (a), dangling pointer 

inside our nested scope create shared_ptr named sp, , ask keep track of int dynamic storage duration created our call make_shared.

we obtain address of int calling sp.get (), , assign ptr.

when scope of sp ends, sp release resource managed since there's no more *shared_ptr*s refers it.

by trying print value of *ptr invoking undefined behavior in (a) since value isn't available, destroyed destruction of sp.