Subj : Re: Old C-program
To   : borland.public.cpp.borlandcpp
From : Vladimir Grigoriev
Date : Tue Oct 11 2005 04:23 pm

Thank Ed
But another question arises.
If  benrty is equal to NULL why below the checking the following statement
is present
bentry->key = key ?
It is executed for any bentry (NULL or not NULL).
Maybe it is simply a bug?

Vladimir Grigoriev


"Ed Mulroy" <dont_email_me@bitbuc.ket> wrote in message
news:434b9d74$1@newsgroups.borland.com...
> Before the 1986 provisional C standard many compilers used C as taught in
> version 1 of the K&R book.  It used this for NULL,
>     #define NULL ((char *)0)
>
> The pre-1986 compilers normally did little type checking.  Later compilers
> are more strict.
>
> The cast in this line:
>     if ((char *) bentry == NULL )
> is used to allow comparing the pointer to NULL when the pointer is not a
> char* type but NULL is a char* type.
>
> Compilers which conform to ANSI C, C as used since 1986 use
>     #define NULL 0
> and with them it is valid for any pointer type to be compared to zero.
>
> .  Ed
>
> > Vladimir Grigoriev wrote in message
> > news:434b8037@newsgroups.borland.com...
> >
> > I get an old-style C program and cannot understand some code.
> > Here it is.
> >
> > typedef struct BlockEntry
> > {
> >      struct BlockEntry * next;
> >      char  name[4];
> >      unsigned long key;
> >      int lock;
> > } BEntry;
> >
> >
> > int SomeFunction( bentry )
> > BEntry * bentry;
> > {
> >        unsigned long key;
> >
> >        if ((char *) bentry == NULL )
> >           key = ONE_CONSTANT;
> >        else
> >           key = ANOTHER_CONSTANT;
> >
> > /* some code that I have skipped */
> >
> >       bentry->key = key;
> >
> >       return 0;
> > }
> >
> > I cannot understand the following code
> > if ((char *) bentry == NULL )
> > What did a programmer want to say?
> > Is the value of (char *)bentry is different form the value of bentry?
>
>

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