C strongly typed?

[Richard O'keefe]

Concerning whether C is strongly typed; I gave Pascal and Ada equivalents
of the fragment which was alleged to show C not to be strongly typed.
In article <1990Mar8.142433.13559 at ncsuvx.ncsu.edu>, jwb at cepmax.ncsu.EDU (John W. Baugh Jr.) writes:
> I think the issue here is "type equivalence,"  ...
> ... ISO/ANSI Pascal standard defines equivalence based on name.
> 
>   type
>     t1 = array[1..3] of boolean;
>     t2 = array[1..3] of boolean;
>   var
>     v1 : t1;
>     v2 : t2;
>    
> the types of v1 and v2 are distinct under name equivalence, but equal
> under structural equivalence.

Quite right.  But it isn't relevant to the example under discussion.
Remember, the C fragment was
	typedef int apple, orange; apple a; orange b; a = b;
The Pascal equivalent of such a typedef is
	type <id> = <type id>;
and that does NOT introduce a new name.  For example,
    type
	t1 = array[1..3] of boolean;
	t2 = t1;	(* NO NEW TYPE IS INTRODUCED HERE *)
    var
	v1: t1;
	v2: t2;
    begin
	v2 := v1;
is perfectly legal Pascal.  Someone else claimed that the fact that
	char c; int i; i = c;
or something like that is legal C also shows that C is not strongly
typed.  Again, Pascal and Ada are exactly the same.  The Pascal
equivalent of C's "char" is
	const minChar = {-128, or 0, or what have you};
	      maxChar = {127, or 255, or what have you};
	type cChar = minChar..maxChar;
The fact that
	var c: cChar; i: integer; begin i := c;
is legal is not normally held to show that Pascal is not strongly
typed; why should the same fact be held against C?  True, C has no
equivalent of Pascal's character type, but the absence of an equivalent
for some other language's data type is not fatal to being strongly typed.