This problem came up when answering this question about overload resolution with enums.
While the case for long long
was definitely a bug in MSVC2012NovCTP (according to the standard text and a test with gcc 4.7.1), I cannot figure out why the following behavior occurs:
#include <iostream>
enum charEnum : char { A = 'A' };
void fct(char) { std::cout << "fct(char)" << std::endl; }
void fct(int) { std::cout << "fct(int)" << std::endl; }
void fct(long long) { std::cout << "fct(long long)" << std::endl; }
int main()
{
fct('A');
fct(A);
}
Both MSVC2012NovCTP and gcc 4.7.1 agree on this output:
fct(char)
fct(int)
Shouldn't A
be converted from charEnum
to char
? Why is A
being converted to int
?
EDIT: clang complains that the call is ambiguous, which agrees with my interpretation below; that said, I would still find it much more intuitive if it were only considered to be the underlying type.
Two relevant standard excerpts are §7.2/9:
The value of an enumerator or an object of an unscoped enumeration type is converted to an integer by integral promotion (4.5)
And §4.5/4:
A prvalue of an unscoped enumeration type whose underlying type is ?xed (7.2) can be converted to a prvalue of its underlying type. Moreover, if integral promotion can be applied to its underlying type, a prvalue of an unscoped enumeration type whose underlying type is ?xed can also be converted to a prvalue of the promoted underlying type.
So charEnum
can either be converted to char
, or any integral promotion of char
, such as int
.
But this is vague to me because "can" doesn't quite say which will actually be chosen. If anything, this should be ambiguous with this wording because no preference is given between char
or any of its promotions. If you comment out fct(int)
, then the call is ambiguous. Why is int
special?
The only thing I can think of is that integral promotions are applied recursively, but nothing I see mandates it.
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