boost optional and user-defined conversion - c++

Question

I am unable to write a correct user defined conversion for a type Item. This is what I've tried:
#include <iostream>
#include <boost/optional.hpp>
struct A
{
int x;
};
struct Item
{
boost::optional<int> x_;
Item(){}
Item(const A& s)
: x_(s.x)
{
}
operator boost::optional<A>() const {
boost::optional<A> s;
if (x_) {
s->x = *x_;
}
return s;
}
};
std::vector<A> getA(const std::vector<Item> &items) {
std::vector<A> a;
for (const auto &i : items) {
if (i.x_) {
a.push_back(*static_cast<boost::optional<A>>(i)); // <- this line causes error
}
}
return a;
}
That is how I use it:
int main() {
A a;
a.x = 3;
Item i(a);
auto v = getA({i});
return 0;
}
g++ -std=c++11 says:
In file included from /usr/include/boost/optional.hpp:15:0,
from test.cpp:2:
/usr/include/boost/optional/optional.hpp: In instantiation of ‘void boost::optional_detail::optional_base<T>::construct(const Expr&, const void*) [with Expr = Item; T = A]’:
/usr/include/boost/optional/optional.hpp:262:25: required from ‘boost::optional_detail::optional_base<T>::optional_base(const Expr&, const Expr*) [with Expr = Item; T = A]’
/usr/include/boost/optional/optional.hpp:559:78: required from ‘boost::optional<T>::optional(const Expr&) [with Expr = Item; T = A]’
test.cpp:30:55: required from here
/usr/include/boost/optional/optional.hpp:392:8: error: no matching function for call to ‘A::A(const Item&)’
new (m_storage.address()) internal_type(expr) ;
^
/usr/include/boost/optional/optional.hpp:392:8: note: candidates are:
test.cpp:3:8: note: A::A()
struct A
^
test.cpp:3:8: note: candidate expects 0 arguments, 1 provided
test.cpp:3:8: note: constexpr A::A(const A&)
test.cpp:3:8: note: no known conversion for argument 1 from ‘const Item’ to ‘const A&’
test.cpp:3:8: note: constexpr A::A(A&&)
test.cpp:3:8: note: no known conversion for argument 1 from ‘const Item’ to ‘A&&’
Why does it try to find A struct constructor instead of use user defined conversion operator?
You may point me directly to any position of the user-defined conversion page because I am unable to find any reason for this. For example,
User-defined conversion function is invoked on the second stage of the implicit conversion, which consists of zero or one converting constructor or zero or one user-defined conversion function.
in my opinion directly says that if no conversion constructor is defined then user-defined conversion function will be used. Am I wrong? And if yes, how can I implement user-defined conversion then without defining conversion cunstructor in struct A ?

Answer 1

You have two issues with your code. Your optional operator never initializes the boost::optional. If you don't do that, accessing members is undefined behavior. What you have to do is:
operator boost::optional<A>() const {
boost::optional<A> s;
if (x_) {
s = A{*x_};
}
return s;
}
The second issue is when you do:
static_cast<boost::optional<A>>(i);
That is equivalent to:
boost::optional<A> __tmp(i);
But it turns out that boost::optional has an explicit template constructor. That will be preferred to your conversion function. The error you're seeing is the compiling going down the path of this factory constructor, where Item is not such a factory.
You could simply use boost::optional<A> directly:
std::vector<A> getA(const std::vector<Item> &items) {
std::vector<A> a;
for (boost::optional<A> opt : items) {
if (opt) {
a.push_back(*opt);
}
}
return a;
}
Or, since the constructor template is explicit, you could use the conversion operator in a non-explicit context:
boost::optional<A> opt = i;
a.push_back(*opt);
This has the added benefit of also being easier to read.