Declaration

I was reading page 131 in K&R C 2nd ed. book.
>
In page 63,
>
"The commas that seperate function arguments, variables in declarations, etc
are not comma operators, and do not guarantee left to right evaluation."
>
and bottom of page 131 there's example of
struct rect r, *rp = &r;
also in page 216, near bottom,
int i, *pi, *const cpi = &i;
>
my question is if comma in declaration does not gurantee left to right
evaluation, how did in examples latter valuables declared and intialized to
value of previous valuables on same line?
>
couln't the *rp declared before r is declared?

>I was reading page 131 in K&R C 2nd ed. book.
>
>In page 63,
>
>"The commas that seperate function arguments, variables in declarations, etc
>are not comma operators, and do not guarantee left to right evaluation."
>
>and bottom of page 131 there's example of
struct rect r, *rp = &r;
>also in page 216, near bottom,
int i, *pi, *const cpi = &i;
>
>my question is if comma in declaration does not gurantee left to right
>evaluation, how did in examples latter valuables declared and intialized to
>value of previous valuables on same line?
>
>couln't the *rp declared before r is declared?
>

On Sun, 31 Dec 2006 14:14:26 -0500, "marko" <marko@validnow.com>
wrote:
>>
Even though the comma is not a comma operator, section 6.2.1-4 states
that the scope of an identifier is determined by the placement of its
declaration. 6.2.1-7 states that the scope begins just after the
completion of its declarator. The comma marks the end of the
declaration of r and the start of rp. r is therefore in scope at the
time rp is declared and the compiler is not allowed reverse the order.

Since there is no sequence point between the declarators though, the
following would be undefined:
>
int i = 1;
int x = i++, y = i++;

Robert Gamble wrote:>
N1124, 6.8 Statements and blocks
...
4 ... Each of the following is a full expression: an initializer; ...
The end of a full expression is a sequence point.