Enforce strong type checking in C (type strictness for typedefs)

quinmars picture quinmars · Dec 18, 2008 · Viewed 15.7k times · Source

Is there a way to enforce explicit cast for typedefs of the same type? I've to deal with utf8 and sometimes I get confused with the indices for the character count and the byte count. So it be nice to have some typedefs:

typedef unsigned int char_idx_t;
typedef unsigned int byte_idx_t;

With the addition that you need an explicit cast between them:

char_idx_t a = 0;
byte_idx_t b;

b = a; // compile warning
b = (byte_idx_t) a; // ok

I know that such a feature doesn't exist in C, but maybe you know a trick or a compiler extension (preferable gcc) that does that.


EDIT I still don't really like the Hungarian notation in general. I couldn't use it for this problem because of project coding conventions, but I used it now in another similar case, where also the types are the same and the meanings are very similar. And I have to admit: it helps. I never would go and declare every integer with a starting "i", but as in Joel's example for overlapping types, it can be life saving.

Answer

Tim Lesher picture Tim Lesher · Dec 18, 2008

For "handle" types (opaque pointers), Microsoft uses the trick of declaring structures and then typedef'ing a pointer to the structure:

#define DECLARE_HANDLE(name) struct name##__ { int unused; }; \
                             typedef struct name##__ *name

Then instead of

typedef void* FOOHANDLE;
typedef void* BARHANDLE;

They do:

DECLARE_HANDLE(FOOHANDLE);
DECLARE_HANDLE(BARHANDLE);

So now, this works:

FOOHANDLE make_foo();
BARHANDLE make_bar();
void do_bar(BARHANDLE);

FOOHANDLE foo = make_foo();  /* ok */
BARHANDLE bar = foo;         /* won't work! */
do_bar(foo);                 /* won't work! */