What should "*nft = *(struct need_to_free *) ntf->addr;" do?

[Dan Merget]

> {
>	struct need_to_free {
>		struct need_to_free	*addr;
>		int			nbytes;
>		} *ntf = &kmem_info.need_to_free_list;
>
>	*ntf = *(struct need_to_free *) ntf->addr;
>    }

I don't claim to be any C guru (I'm still trying to decipher that ANSI
paragraph you quoted).  However, it seems to me that your code simply copies
the first sizeof(struct need_to_free) bytes from the second object in the
linked list to the first object.  In that case, I think that your code would
give you the following linked list:

	        +-------------+        +------------+        +-------------+
		|             |        |            |        |             |
	ntf ->	|  addr p2 ----------> |   addr p3 ------->  |    addr p4  |
		|             |        |            |        |             |
		+-------------+        +------------+        +-------------+
		|             |        |            |        |             |
		|  nbytes n1  |        |  nbytes n3 |        |  nbytes n4  |
		|             |        |            |        |             |
		+-------------+        +------------+        +-------------+

To get the result you're looking for, the last line of code should be either:

	ntf = ntf->addr;

or:

	*(need_to_free_list_type *) ntf = *(need_to_free_list_type *) ntf->addr;

where "need_to_free_list_type" is the type of kmem_info.need_to_free_list.

On the other hand, maybe I simply need to sign up for "Oversimplifiers
Anonymous". :-)

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    Dan Merget
    danm at hpnmdla.HP.COM