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Re: Newbie: how to handle different prototypes?

From: Paul Eggert
Subject: Re: Newbie: how to handle different prototypes?
Date: Thu, 23 Jun 2005 17:22:11 -0700
User-agent: Gnus/5.1006 (Gnus v5.10.6) Emacs/21.4 (gnu/linux)

Ralf Wildenhues <address@hidden> writes:

> * address@hidden wrote on Wed, Jun 22, 2005 at 10:17:02PM CEST:
>> For example, on linux the scandir prototype is:
>>        int scandir(const char *dir, struct dirent ***namelist,
>>               int(*select)(const struct dirent *),
>>               int(*compar)(const struct dirent **, const struct dirent **));
>> and on darwin it's:
>>      int
>>      scandir(const char *dirname, struct dirent ***namelist,
>>          int (*select)(struct dirent *),
>>          int (*compar)(const void *, const void *));
>> I.e. the const is missing from darwin's 3rd parameter.
> Can't you just ignore this difference in your program?  If you pass a
> select function which eats a const argument, you are within ANSI C
> bounds in either case.

Sorry, that won't work in general.

(Putting my language-lawyer hat on.)

On darwin, you'd be passing (int (*) (struct dirent *)) to (int (*)
(struct dirent const *)), and C89 and C99 both say that two function
types are compatible only if their arguments are compatible, and that
(struct dirent const *) is not compatible with (struct dirent *).

There is an exception for pointers; e.g., you are allowed to assign a
(T *) value to a (T const *) variable.  But this exception does not
apply to multiple levels (e.g., you can't assign (T **) to (T const
**).  In particular, it does not apply when assigning a (T1 (*) (T2
const *)) value to a (T1 (*) (T2 *)) variable.

Arguably this is a botch.  There are some cases that are safe that C
does not allow.  Perhaps the best known is assigning a (T **) value to
a (T const * const *) variable.  (I think C++ fixes this latter
botch.)  But the botch is part of Standard C and we have to work
around it in the portable code that we write.

(Language-lawyer hat off.)

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