Why smart pointer type member variable can't be initialized at the declaring place in a class?

Question

When I want to add a member variable with smart pointer type to a class, I found that it can't be initialized at the declaring place:

class Foo {
 public:
  std::shared_ptr<int> intSharedPtr = new int;  // not ok
  Foo() {}
};

But I can do this:

class Foo {
 public:
  std::shared_ptr<int> intSharedPtr;  // ok
  int* intPtr = new int; // ok
  Foo() {
    intSharedPtr.reset(new int);
  }
};

It seems that smart pointer is quite different form the normal pointer, Why this happens?

songyuanyao · Accepted Answer · 2022-04-01T10:42:29.410

std::shared_ptr can't be copy-initialized from raw pointer, the conversion constructor is marked as explicit.

class Foo {
 public:
  std::shared_ptr<int> intSharedPtr {new int};
  Foo() {}
};

Or initialize from an std::shared_ptr:

class Foo {
 public:
  std::shared_ptr<int> intSharedPtr = std::shared_ptr<int>(new int);
  Foo() {}
};

And better to use std::make_shared:

class Foo {
 public:
  std::shared_ptr<int> intSharedPtr = std::make_shared<int>();
  Foo() {}
};

1 Answers1