C++ function definition and variable declaration mismatch?

Question

Consider this very simple code:

#include 

class Foo
{
public:
    Foo() {};
};

class Bar
{
public:
    Bar( const std::shared_ptr&         


        

Answer 1

This issue is about C++'s most vexing parse. The statement:

Bar bar( std::shared_ptr<Foo>( foo ) );

declares a function called bar that returns Bar and takes an argument called foo of type std::shared_ptr<Foo>.

The innermost parenthesis have no effect. It is as if you would have written the following:

Bar bar( std::shared_ptr<Foo> foo);

---

Assuming C++11 (since you are already using std::shared_ptr) you could use the brace syntax instead of parenthesis:

Bar bar(std::shared_ptr<Foo>{foo});

This would actually construct an object bar of type Bar, since the statement above can't be interpreted as a declaration because of the braces.