https://gcc.gnu.org/bugzilla/show_bug.cgi?id=120710
Bug ID: 120710
Summary: C23 enum member does not have during processing the
type indicated by C23 6.7.3.4:12
Product: gcc
Version: 15.1.0
Status: UNCONFIRMED
Severity: normal
Priority: P3
Component: c
Assignee: unassigned at gcc dot gnu.org
Reporter: pascal_cuoq at hotmail dot com
Target Milestone: ---
Testcase:
#include <stdio.h>
enum e {
A = 0,
B = 0x80000000,
C = 2
};
#define ty(e) _Generic(e, int:"int", unsigned:"unsigned", long:"long", unsigned
long:"unsigned long")
enum e1 {
G = 0,
H = -0x80000001L,
I,
J = _Generic(I, int:147, long:1046)
};
int main(void) {
printf ("A B C %s %s %s\n\n", ty(A), ty(B), ty(C));
printf ("G H I J %s %s %s %s\n", ty(G), ty(H), ty(I), ty(J));
printf("I %d", (int)J);
}
Compiled with GCC 15.1 for x86-64, the program outputs:
A B C unsigned unsigned unsigned
G H I J long long long long
I 147
Compiler Explorer link: https://gcc.godbolt.org/z/1dY7n7Kco
The surprising behavior is in the last output line that shows that I had type
int during the processing of enum e1. https://cigix.me/c23#6.7.3.4 says:
During the processing of each enumeration constant in the enumerator list, the
type of the enumeration constant shall be:
…
- the type of the value from the previous enumeration constant with one added
to it. If such an integer constant expression would overflow or wraparound the
value of the previous enumeration constant from the addition of one, [this is
not the case in the example above]
