??? 04/30/07 14:06 Read: times |
#138289 - C99 says Responding to: ???'s previous message |
C99 6.7.2.2 said:
Each enumerated type shall be compatible with char, a signed integer type, or an unsigned integer type. The choice of type is implementation-defined, but shall be capable of representing the values of all the members of the enumeration. What would you expect from a language which does not know what is an integer? JW (the C-hater) |
Topic | Author | Date |
Quick C enumerated types question | 01/01/70 00:00 | |
Not to worry i've found out anyway. | 01/01/70 00:00 | |
What's a "quick C" | 01/01/70 00:00 | |
The answer was..... | 01/01/70 00:00 | |
not really | 01/01/70 00:00 | |
as has been discussed before | 01/01/70 00:00 | |
of course not | 01/01/70 00:00 | |
Not true | 01/01/70 00:00 | |
C99 says | 01/01/70 00:00 | |
more details | 01/01/70 00:00 | |
OK | 01/01/70 00:00 | |
not any more | 01/01/70 00:00 | |
In C99 | 01/01/70 00:00 | |
Enumeration type vs. constant | 01/01/70 00:00 | |
Thanks for that everyone | 01/01/70 00:00 |