Given the following code sample:
boost::optional< int > opt;
opt = 12;
int* p( &*opt );
opt = 24;
assert( p == &*opt );
Is there any guarantee that the assert will always be valid?
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Julien
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What does the documentation say? – Bartek Banachewicz Oct 21 '15 at 16:13
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I did not find the answer in the documentation but maybe I missed it. – Julien Oct 21 '15 at 16:21
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Side note: As long as the pointer is in use the optional must not change it's internal state (optional = boost::none would be fatal). – Oct 21 '15 at 17:14
1 Answers
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yes it's a guarantee. The T of a boost::optional<T> is a logically a private member of the optional.
the code above is logically equivalent to:
bool opt_constructed = false;
int opt_i; // not constructed
new int (&opt_i)(12); opt_constructed = true; // in-place constructed
int*p = &opt_i;
opt_i = 24;
assert(p == &opt_i);
// destuctor
if (opt_constructed) {
// call opt_i's destructor if it has one
}
Richard Hodges
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`*opt` returns a reference to the internal member or throws an exception if it has not been constructed. `opt=24` assigns to the already-constructed member. – Richard Hodges Oct 21 '15 at 16:16
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Can you point the part of the documentation which would back your answer up? – Julien Oct 22 '15 at 06:55
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See the narrative on semantics. Note that an optional is semantically equivalent to T, plus being able to be uninitialised. http://www.boost.org/doc/libs/1_58_0/libs/optional/doc/html/boost_optional/tutorial/design_overview/the_semantics.html – Richard Hodges Oct 22 '15 at 08:00