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std::shared_ptr<T>::operator*, std::shared_ptr<T>::operator->

From cppreference.com
< cpp‎ | memory‎ | shared ptr
 
C++
 
Memory management library
(exposition only*)
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(until C++20*)
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(until C++20*)

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std::shared_ptr
Member functions
Modifiers
Observers
shared_ptr::operator*shared_ptr::operator->
(C++17)
(until C++20*)
(C++26)
(C++26)
Non-member functions
(C++20)
(C++20)
(C++17)
(until C++20)(until C++20)(until C++20)(until C++20)(until C++20)(C++20)
functions (until C++26*)
Helper classes
(C++20)
Deduction guides(C++17)
 
T& operator*() const noexcept;
 (1) (since C++11)
T* operator->() const noexcept;
 (2) (since C++11)

Dereferences the stored pointer. The behavior is undefined if the stored pointer is null.

Contents

[edit] Parameters

(none)

[edit] Return value

1) The result of dereferencing the stored pointer, i.e., *get().
2) The stored pointer, i.e., get().

[edit] Remarks

When T is an array type or (possibly cv-qualified)(since C++17) void, it is unspecified whether function (1) is declared. If it is declared, it is unspecified what its return type is, except that the declaration (although not necessarily the definition) of the function shall be well formed. This makes it possible to instantiate std::shared_ptr<void>

When T is an array type, it is unspecified whether function (2) is declared. If it is declared, it is unspecified what its return type is, except that the declaration of the function shall be well-formed.

(since C++17)

[edit] Example

Run this code
#include <iostream>
#include <memory>
 
struct Foo
{
    Foo(int in) : a(in) {}
    void print() const
    {
        std::cout << "a = " << a << '\n';
    }
    int a;
};
 
int main()
{
    auto ptr = std::make_shared<Foo>(10);
    ptr->print();
    (*ptr).print();
}

Output: 

a = 10
a = 10

[edit] See also

 returns the stored pointer
(public member function) [edit]
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