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typedef and #ifdef

P: n/a
How can you check to see if a type is built-in or typedef'ed? For
example, an 8-bit integer could be u_int8_t, uint8_t, or unsigned
__int8, depending on the OS/compiler. I'd like to write code like:

#if defined(u_int8_t)
typedef u_int8_t BYTE;
#elif defined (uint8_t)
typedef uint8_t BYTE;
#elif defined (__int8)
typedef unsigned __int8 BYTE
#endif

However, this does not work, because built-in and typedef'ed types are
not "defined" according to #ifdef. So two questions: Is there a
work-around? And why doesn't C++ allow the above code?
Jul 19 '05 #1
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P: n/a
On 11 Aug 2003 10:26:51 -0700, bo**@stanford.edu (F. Edward Boas) wrote:
How can you check to see if a type is built-in or typedef'ed?


You can check whether a type is built-in by checking for all
built-in types (this list is not necessarily limited to the
C types); use template classes for this.

You cannot check whether a C++ type is a typedef, because a
C++ typedef just introduces an alternate _name_ for a type.

A typedef behaves differently in the language D.
Jul 19 '05 #2

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