Why isn't the boost::shared_ptr -> operator declared inline?

Question

Since boost::shared_ptr could be called very frequently and simply returns a pointer, isn't the -> operator a good candidate for being inlined ?

T * operator-> () const // never throws
{
    BOOST_ASSERT(px != 0);
    return px;
}

Would a good compiler automatically inline this anyway?

Should I lose any sleep over this? :-)

Answer 1

Functions defined (ie with a body) inside a class are implicitly candidates for inlining. There is no need to use the inline keyword in these cases, and it is unusual to do so.

Answer 2

Would a good compiler automatically inline this anyway?

Quite probably, yes, it would.

Should I lose any sleep over this?

Better not. If you want to be super-sure (or you are super-curious), check the assembly that's going out from your compiler.

Answer 3

Please note that shared_ptr is a class template , so its member functions are actually function templates .

Since they are not export ed, they must not only be declared , but also defined in all translation units where they are used, just like a function defined with the inline storage specifier.

In a way, template also means inline .