when operating smart pointers, something confused me.

Hers is the error message

no known conversion from 'std::shared_ptr<int>' to 'int *' for 1st argument

and here's the code I ran

#include <memory>
#include <vector>

class test {
public:
  int x_;
  test(int *x) {}
};

int main(int argc, char *argv[]) {

  auto x = std::make_shared<int>(5);
  std::shared_ptr<test> t = std::make_shared<test>(x);
}

I think this error came from the different types of pointers. And compilation can succeed when changing

std::shared_ptr<test> t = std::make_shared<test>(x);

into

std::shared_ptr<test> t = std::make_shared<test>(&(*x));

But this operation &(*), in my opinion, looks weird. I'm not sure that is a common usage when people program.

Is there any suggestion for this question? Thanks.

CodePudding user response：

Use x.get().

It's not implicitly convertible because that would make it too easy to make mistakes, like creating another shared_ptr from the same raw pointer without proper sharing.

CodePudding user response：

You are trying to pass std::shared_ptr<int> to a constructor accepting int *. There is no implicit conversion for it.

You have two choices -

1. change to std::shared_ptr<test> t = std::make_shared<test>(x.get());
2. or change the constructor to test(std::shared_ptr<int> x) {}