🚀 go-pugleaf

In article mco@netlab.cs.rpi.edu, bstowers@pobox.com (Brad Stowers) writes:
>#include <stdio.h>
>
>typedef enum { mdType1, mdType2, mdType3 } myType;
>
>void FooBar(int x,
>            int* y,
>            myType type = mdType1)
>{
>  printf("x=%i\ny=%p\ntype=%type\n", x, y, type);
>}
>
>main()
>{
>  int x = 10;
>  int* y = &x;
>  FooBar(x, y, mdType2);
>  FooBar(x, mdType1);
>}

>  Is it just me, or should the last line
>of the main() function ( FooBar(x, mdType1); ) have caused a type mismatch
>error? 

The code is valid (although incorrect) as it stands.

> I think I know why it didn't:  mdType1 evaluates to an integer with
>a value of 0, which is a valid value to pass as a pointer.

That's about right. In more detail: Parameter y needs an int*, but the
actual parameter is a constant of enumerated type. In an expression, an
enumerator gets promoted to type int (or to a wider type if necessary),
so at this point we have an int constant with value zero. The definition
of "null pointer constant" is an integer constant-expression with value zero,
so mdType1 qualifies for conversion to a null int*.

>Obviously, the work-around is simple:
>
>  typedef enum { mdType1 = 1, mdType2, mdType3 } myType;
>
>While this will solve the problem, I really think that it is the compiler's
>job to see that mdType1 does not evaluate to the proper type no matter what
>it's internal representation might be.

It isn't the internal representation, but the zero value. This is
yet another C compatibility hack that tends to subvert the type system.

---
Steve Clamage, stephen.clamage@eng.sun.com




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