CODEWITHSUNDEEP
c++boostshared-ptr
OopsUser
OopsUser

Reputation: 4784

Using default parameters when using boost shared_ptr

I have boost shared_ptr as a parameter to function, i would like this parameter to have some default value.

void foo(boost::shared_ptr<myclass> ptr = nullptr);

because ptr is not a pointer, but a class..

so what can i do ?

i found a similar question : boost::shared_ptr and nullptr in default template function argument

But the solution there is just to switch to std::shared_ptr, but i can't do it

Thanks

Upvotes: 0

Views: 4087

Answers (4)

Igor R.
Igor R.

Reputation: 15075

It's perfectly ok to initialize boost::shared_ptr with nullptr. It has a special constructor for this:

#if !defined( BOOST_NO_CXX11_NULLPTR )
shared_ptr( boost::detail::sp_nullptr_t ) BOOST_NOEXCEPT : px( 0 ), pn() // never throws
{
}
#endif

(boost::detail::sp_nullptr_t resolves to nullptr_t in a portable manner)

Upvotes: 2

Jonathan Wakely
Jonathan Wakely

Reputation: 171283

Since you're using nullptr you must be using C++11, so you should be able to do this:

void foo(boost::shared_ptr<myclass> ptr = {});

That makes the default argument a default-constructed shared_ptr, which is the same as one initialized with nullptr. It requires your compiler to support uniform initialization syntax, instead of requiring your Boost version to support constructing shared_ptr from nullptr_t.

If your compiler can't handle that then use ForEveR's answer.

Upvotes: 2

Balog Pal
Balog Pal

Reputation: 17163

I think the problem is not "being class" but that the involved constructor is explicit. And your form would require implicit conversion. You can work it around using an explicit form: boost::shared_ptr<myclass>(nullptr) or boost::shared_ptr<myclass>(), whichever is supported for your version.

Upvotes: 1

ForEveR
ForEveR

Reputation: 55887

Why not simply?

void foo(boost::shared_ptr<myclass> ptr = boost::shared_ptr<myclass>());

Upvotes: 1

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