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Internet Message Format | 1996-08-06 | 4.4 KB

Path: chronicle.mti.sgi.com!austern
From: [e-mail] (J. Kanze)
Newsgroups: comp.std.c++
Subject: Re: Enumerated type converted to pointer? Legal?
Date: 22 Mar 1996 09:28:31 PST
Organization: GABI Software, Sarl.
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In article <4hobji$[e-mail]> [e-mail] (Brad
Stowers) writes:

    [I've stripped all but comp.std.c++ out of the newsgroups line.  In
    general, it is very awkward for the moderators if you cross-post to
    several moderated groups, and comp.std.c++ seemed the most
    appropriate.]

|> #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);
                                           ^

Just a warning: this is undefined behavior, although it will work on a
lot of machines.  You must cast any parameter corresponding to a `%p' to
void* (if it isn't void* already).

Also: what is %type?

(The above sound like two very good reasons for switching to iostream to
me.)

|> }

|> main()
|> {
|>   int x = 10;
|>   int* y = &x;
|>   FooBar(x, y, mdType2);
|>   FooBar(x, mdType1);
|> }

|> ------------------EOF------------------------------

|> Compile and run the above program.  Is it just me, or should the last line
|> of the main() function ( FooBar(x, mdType1); ) have caused a type mismatch
|> error?

Definitly.

|> 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.  In support of
|> this theory, the line:

|>   FooBar(x, mdType2);

|> will not compile, but rather generates the expected type mismatch error.
|> 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.

Correct.  There is nothing in the standard that allows converting an
enum to a pointer, irrespective of the numerical value of the enum.

|> I have verified that this happens on both Borland and Symantec compilers.
|> Can anyone verify it on Watcom and Microsoft for me?

No.  On a Sun, however, g++ (2.7.2) gives the correct error message.
Sun CC (4.1) has the same error as Borland and Symantec.

The error may be historically conditioned.  As I interpret the C
standard, an enumerated constant has type int, and a constant integral
expression which evalutates to 0 is a null pointer.  Even in the ARM,
however, ``An enumeration is a distinct integral type.''

(In fact, the question is less clear that the above would make it
appear.  While the C standard definitly says that an enumerator has type
int, it also implies that the enum statement itself defines a new
distinct type, ``compatible with an integer type.''  And while in the
absense of a definite statement to the contrary in the ARM, and the
insistence that the enumeration is a distinct type, I think that the
intent is clear that the enumerator should have the same type as the
enumeration, rather than int, there are no explicit words to this
effect.)

This is all passe, however.  From the Jan., 1996 draft standard: ``The
type of an enumerator is its enumeration.''  You cannot get any clearer
than that.
--
James Kanze           ([phone]          email: [e-mail]
GABI Software, Sarl., 8 rue des Francs Bourgeois, 67000 Strasbourg, France
Conseils en informatique industrielle --
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