Differences between assignment constructor and others in C++

Question

Consider:

std::shared_ptr<Res> ptr=new Res();

The above statement doesn't work. The compiler complains there isn't any viable conversion....

While the below works

std::shared_ptr<Res> ptr{new Res()} ;

How is it possible?!

Actually, what are the main differences in the constructor{uniform, parentheses, assignments}?

Answer 1

The constructor of std::shared_ptr taking a raw pointer is marked as explicit; that's why = does not allow you to invoke this constructor.

Using std::shared_ptr<Res> ptr{new Res()}; is syntax that allows you to call the an explicit constructor to initialize the variable.

std::shared_ptr<Res> ptr=new Res();

would invoke an implicit constructor taking a pointer to Res as single parameter, but not an explicit one.

Here's a simplified example using a custom class with no assignment operators and just a single constructor:

class Test
{
public:
    Test(int i) { }

    // mark all automatically generated constructors and assignment operators as deleted
    Test(Test&&) = delete;
    Test& operator=(Test&&) = delete;
};

int main()
{
    Test t = 1;
}

Now change the constructor to

explicit Test(int i) { }

and the main function will no longer compile; the following alternatives would still be viable:

Test t(1);
Test t2{1};