complier problem

[neha]

Hi

My question is : Would the expression *p++=c be disallowed by the
complier.
The answer given in the book is: No.Because here even though the vlue
of p is acced twice it is used to modify two different objects p and
*p.
so can anyone explain this answer to me detail

[Chris Torek]

>Chris Torek wrote:

>>[the compiler can produce varying machine code for the C source:]
save = p++;

In article <r9************ *************** ***@comcast.com >,
Ark <ak*****@macroe xpressions.comw rote:

>Looks like a storm in a teacup. ['save' doesn't have to exist at all,
depending on the instruction set,

The intent of this later example -- which I admit is not clear if
you follow the whole thread and think that it is *not* a separate
example -- is that the "save = p++;" is a complete line of C source
code. Hence "save" *does* exist, right there in the source code. :-)

>and *p++=c; may execute in a single instruction]

On some architectures (try it on a SPARC or MIPS!); but this is a
different line of source code anyway. :-) And of course, while a
number of architectures have a single instruction that implements
"*p++ = expr" and "*--p = expr", many of those do not have a single
instruction that implements "*++p = expr" or "*p-- = expr". (But
some do.)

>p++ has a value (old p) and a side effect (p = old p + 1).
*p++ dereferences the value of p++ (old p).

(Correct -- but can you say that in no fewer than 1000 words? :-) )

>I'd care about implementation only when shopping for a compiler (or
squeezing out that last optimization)

Indeed, this is an appropriate attitude.
--
In-Real-Life: Chris Torek, Wind River Systems
Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) [phone]
email: forget about it http://web.torek.net/torek/index.html
Reading email is like searching for food in the garbage, thanks to spammers.

[Dann Corbit]

"neha" <ne*********@gm ail.comwrote in message
news:11******** **************@ i42g2000cwa.goo glegroups.com.. .

Hi

My question is : Would the expression *p++=c be disallowed by the
complier.
The answer given in the book is: No.Because here even though the vlue
of p is acced twice it is used to modify two different objects p and
*p.
so can anyone explain this answer to me detail

Your question is similar to a FAQ, did you read it before posting?:

----------------------------------------------------
4.3: Does *p++ increment p, or what it points to?

A: Postfix ++ essentially has higher precedence than the prefix
unary operators. Therefore, *p++ is equivalent to *(p++); it
increments p, and returns the value which p pointed to before p
was incremented. To increment the value pointed to by p, use
(*p)++ (or perhaps ++*p, if the order of the side effect doesn't
matter).

References: K&R1 Sec. 5.1 p. 91; K&R2 Sec. 5.1 p. 95; ISO
Sec. 6.3.2, Sec. 6.3.3; H&S Sec. 7.4.4 pp. 192-3, Sec. 7.5 p.
193, Secs. 7.5.7,7.5.8 pp. 199-200.
----------------------------------------------------
A side point:
I would never rely on the compiler to reject a nonsense code fragment.
There are some types of problems that the compiler MUST issue a diagnostic
message for, but there are other things that are totally wrong but which the
compiler does not have to diagnose.