Since the constructor of std::shared_ptr
is marked as explicit one, so expressions like auto p = std::make_shared<int>(1); p = new int(6);
is wrong.
My question is why does std::make_shared<int>(1); p = nullptr;
compile?
Here is the aforementioned code snippet:
#include <memory>
#include <iostream>
int main()
{
auto p = std::make_shared<int>(1);
//p = new int(6);
p = nullptr;
if(!p)
{
std::cout << "not accessable any more" << std::endl;
}
p.reset();
}
Such code is seen at std::shared_ptr: reset() vs. assignment
CodePudding user response:
The raw pointer constructor is explicit
to prevent you accidentally taking ownership of a pointer. As there is no concern taking ownership of nullptr
the constructor taking std::nullptr_t
is not marked explicit.
Note that this only applies to nullptr
these other assignments of a null pointer might not work (depending on your compiler):
auto p = std::make_shared<int>(1);
p = NULL;
int * a = nullptr;
p = a;