Subj : Re: Old C-program
To   : borland.public.cpp.borlandcpp
From : Ed Mulroy
Date : Tue Oct 11 2005 08:12 am

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?

.