https://gcc.gnu.org/bugzilla/show_bug.cgi?id=113041
Bug ID: 113041
Summary: misleading diagnostic for variable of non-literal type
in constexpr function in C++20 mode
Product: gcc
Version: 14.0
Status: UNCONFIRMED
Severity: normal
Priority: P3
Component: c++
Assignee: unassigned at gcc dot gnu.org
Reporter: ppalka at gcc dot gnu.org
Target Milestone: ---
Here f<int>() is not constexpr in C++20 (but is in C++23 after P2242R3) due to
the local variable a of non-literal type A, so the testcase is invalid:
struct A { ~A(); };
void g();
template<class T>
constexpr int f() {
if (__builtin_is_constant_evaluated())
return 42;
g();
A a;
}
constexpr int n = f<int>();
But our diagnostic incorrectly blames the call to the non-constexpr g() for
f<int>() being non-constexpr, rather than the variable a:
constexpr-nonliteral.C:13:25: error: ‘constexpr int f() [with T = int]’ called
in a constant expression
13 | constexpr int n = f<int>();
| ~~~~~~^~
constexpr-nonliteral.C:6:15: note: ‘constexpr int f() [with T = int]’ is not
usable as a ‘constexpr’ function because:
6 | constexpr int f() {
| ^
constexpr-nonliteral.C:9:4: error: call to non-‘constexpr’ function ‘void g()’
9 | g();
| ~^~
constexpr-nonliteral.C:3:6: note: ‘void g()’ declared here
3 | void g();
| ^
