pjh@mccc.UUCP (Pete Holsberg) (05/09/89)
I ran across the following code and, although I've figured out what it does,
I'm not sure how to explain the notation.
struct foo
{
char bar[20];
} list[100];
init()
{
int i;
for (i=0; i<100; ++i)
*list[i].bar = '\0';
}
Now I know that the last line points to the first character in the bar array
In each struct variable, but how? If bar = &bar[0], how so the '*' and the
'&' "get together" to make the last line equivalent to
list[i].bar[0]
??
--
Pete Holsberg UUCP: {...!rutgers!}princeton!mccc!pjh
Mercer College CompuServe: 70240,334
1200 Old Trenton Road GEnie: PJHOLSBERG
Trenton, NJ 08690 Voice: 1-609-586-4800ark@alice.UUCP (Andrew Koenig) (05/10/89)
In article <743@mccc.UUCP>, pjh@mccc.UUCP (Pete Holsberg) writes: > *list[i].bar = '\0'; > > Now I know that the last line points to the first character in the bar array > In each struct variable, but how? If bar = &bar[0], how so the '*' and the > '&' "get together" to make the last line equivalent to > list[i].bar[0] Easy ... `.' binds more tightly than any unary operator, so *list[i].bar is equivalent to *(list[i].bar) -- --Andrew Koenig ark@europa.att.com
scjones@sdrc.UUCP (Larry Jones) (05/11/89)
In article <743@mccc.UUCP>, pjh@mccc.UUCP (Pete Holsberg) writes: > struct foo > { > char bar[20]; > } list[100]; > > *list[i].bar = '\0'; > > Now I know that the last line points to the first character in the bar array > In each struct variable, but how? If bar = &bar[0], how so the '*' and the > '&' "get together" to make the last line equivalent to > list[i].bar[0] Well, the problem is that bar is NOT equivalent to &bar[0]. Rather, list[i].bar is equivalent to &list[i].bar[0], which should make it easier to see how the "*" and the "&" get together. What you have to realize is that bar DOES NOT EXIST -- since it is a member of a structure, it only exists as part of an actual instance of the structure and must be referenced as such. ---- Larry Jones UUCP: uunet!sdrc!scjones SDRC scjones@SDRC.UU.NET 2000 Eastman Dr. BIX: ltl Milford, OH 45150-2789 AT&T: (513) 576-2070 "You can't get a body like mine in a bottle -- unless you push REAL HARD." - Judy Tenuta / Dr. Pepper
mat@mole-end.UUCP (Mark A Terribile) (05/11/89)
In article <743@mccc.UUCP>, pjh@mccc.UUCP (Pete Holsberg) writes: > I ran across the following code and, although I've figured out what it does, > I'm not sure how to explain the notation. > struct foo { char bar[20]; } list[100]; > init() > { > int i; > for (i=0; i<100; ++i) > *list[i].bar = '\0'; > } > Now I know that the last line points to the first character in the bar array > In each struct variable, but how? If bar = &bar[0], how so the '*' and the > '&' "get together" to make the last line equivalent to > list[i].bar[0] It's really simple. Remember first that ``.'' and ``[]'' are what K&R-I call ``primary'' operators; they bind more tightly than anything else. list[ i ].bar is an array of char. Used as an expression, it is ``converted syntactically'' (it is taken to mean) a pointer-to-char pointing at the first element in the array. What array? The array in the struct. Ok, then why do we write *list[ i ].bar instead of list[ i ].*bar Because the other operators are primaries, and it is the *entire* expression from ``list'' to ``bar'' that is the pointer-to-char. The expression with the ``*'' in the middle isn't legal C, though it can be legal C++. -- (This man's opinions are his own.) From mole-end Mark Terribile