std::shared_ptr::operator*, std::shared_ptr::operator->
From cppreference.com
< cpp | memory | shared ptr
| T& operator*() const; |
(1) | (since C++11) |
| T* operator->() const; |
(2) | (since C++11) |
Dereferences pointer to the managed object.
Contents |
[edit] Parameters
(none)
[edit] Return value
1) reference to the managed object.
2) pointer to the managed object.
[edit] Exceptions
noexcept specification:
noexcept
[edit] Notes
When element_type is 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 is guaranteed to be legal. This makes it possible to instantiate std::shared_ptr<void>.
[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 a pointer to the managed object (public member function) |
