问题
My understanding is that C++ enumerations are converted to integral types according to Integral Promotion. And during Integral Promotion, we should try converting a value to int first and if the value cannot be represented by an int, unsigned int should be used:
C++03 conv.prom:
2) .... An rvalue of an enumeration type (7.2) can be converted to an rvalue of the first of the following types that can represent all the values of the enumeration (i.e. the values in the range bmin to bmax as described in 7.2: int, unsigned int, long, or unsigned long.
But my testing in VC++2010 shows contradicting result:
enum {A1=60, A2=61,A3=63,A4=64,A5=66,A6=0xffffFFF1};
const bool b1 = A6 < A1;
//result b1=true;
A6 is 0xffffFFF1 which should be an unsigned value (decimal 4294967281). And, since this value cannot be represented by an int, it should be converted to unsigned int. But clearly, when used in the comparison, A6 was converted to -15. Why is it so?
回答1:
It is a bug of MS VC++. I described it here
Though the description is written in Russian you can translate it into English using for example google service translate.
Also there is a defect relative to enumerations in the C++ Standard itself. I also described it here
And one more reference relative to a discussion of enumerations (at this time in English) can be accessed here
来源：https://stackoverflow.com/questions/24802322/sign-of-c-enum-type-incorrect-after-converting-to-integral-type