https://gcc.gnu.org/bugzilla/show_bug.cgi?id=110828
Bug ID: 110828
Summary: union constexpr dtor not constexpr when used in member
array
Product: gcc
Version: 13.1.0
Status: UNCONFIRMED
Severity: normal
Priority: P3
Component: c++
Assignee: unassigned at gcc dot gnu.org
Reporter: gccbugbjorn at fahller dot se
Target Milestone: ---
The following code does not compile with g++ 13.1 or g++ trunk
82c2a34b2f2c1a06eff672eba2e447b53f35d7b0
union type {
constexpr type(){}
constexpr ~type() {}
int t;
};
struct S
{
constexpr S() = default;
constexpr bool f() const { return true;}
private:
type v[1];
};
static_assert(S{}.f());
The error message is:
source.cpp:16:20: error: non-constant condition for static assertion
16 | static_assert(S{}.f());
| ~~~~~^~
source.cpp:16:23: in 'constexpr' expansion of '(&<anonymous>)->S::~S()'
source.cpp:7:8: error: '(((type*)(&<anonymous>.S::v)) != 0)' is not a constant
expression
7 | struct S
| ^
Two simple changes can each make it compile.
1. Comment out the constexpr destructor for 'type'
2. Remove [1] from the declaration of member variable S::v such that it is not
an array.
Link to compiler explorer: https://godbolt.org/z/q3KcKrPh1
