&& and || Operator precedence enforcement

[Vivek N]

Hi Folks,
This question may well have been asked and answered before. But, sorry
that I couldn't find one from the archives.

I typed up this program and compiled it with gcc 3.3.2
main() {
int i = -3,j= 2,k=0,m;

m = ++i || ++j && ++k;
printf("\n%d %d %d %d\n",i,j,k,m);
}
and got the output to be
-2 2 0 1

I presumed that the answer should have well been
-3 3 1 1

as ++j and ++k would first be evaluated and && operator applied to
obtain the result to be TRUE (3 && 1). Hence, ++i wouldn't be evaluated
as one part of || goes TRUE.

The ANSI C specification clearly states that && has precedence over ||.

Could someone explain why this strange "-2 2 0 1" output for the program
is obtained?

Cheers
Vivek

[pete]

Vivek N wrote:

Hi Folks,
This question may well have been asked and answered before. But, sorry
that I couldn't find one from the archives.

I typed up this program and compiled it with gcc 3.3.2
main() {
int i = -3,j= 2,k=0,m;

m = ++i || ++j && ++k;
printf("\n%d %d %d %d\n",i,j,k,m);
}
and got the output to be
-2 2 0 1

I presumed that the answer should have well been
-3 3 1 1

as ++j and ++k would first be evaluated and && operator applied to
obtain the result to be TRUE (3 && 1).
Hence, ++i wouldn't be evaluated
as one part of || goes TRUE.

The ANSI C specification clearly states that &&
has precedence over ||.
(a || b && c), is the same as (a || (b && c)).
Could someone explain why this strange "-2 2 0 1"
output for the program is obtained?

j and k are not evaluated at all.

The left operand of || is evaluted before the right.
The right operand of || is evaluated conditionally only if
the left operand is equal to zero.
Your left operand (++i), is not equal to zero.
Your right operand (++j && ++k) is not evaluated.

--
pete

[Christopher Benson-Manica]

Vivek N <ke***********@ yahoo.com> spoke thus:

main() {
Allowing the return type of a function to default to int is illegal in
C99 and dubious otherwise.
int i = -3,j= 2,k=0,m; m = ++i || ++j && ++k;
printf("\n%d %d %d %d\n",i,j,k,m);
Where is <stdio.h>? Turn up gcc's warnings (-Wall -pedantic -ansi).
Could someone explain why this strange "-2 2 0 1" output for the program
is obtained?

Not strange at all, thanks to short-circuit evaluation. The fact that
&& has precedence over || only means that the assignment to m is
equivalent to

m=++i || (++j && ++k);

++i is computed first, and if it is nonzero (true), then the entire
statement is guaranteed to be true regardless of what (++j && ++k) is.
In C, this means that (++j && ++k) will not be evaluated. That is
what short-circuit evaluation is, and that is why only i is
incremented in the code you posted.

--
Christopher Benson-Manica | I *should* know what I'm talking about - if I
ataru(at)cybers pace.org | don't, I need to know. Flames welcome.

[Richard Bos]

ke***********@y ahoo.com (Vivek N) wrote:

as ++j and ++k would first be evaluated and && operator applied to
obtain the result to be TRUE (3 && 1). Hence, ++i wouldn't be evaluated
as one part of || goes TRUE.

The ANSI C specification clearly states that && has precedence over ||.

The ISO C specification states no such thing, since "precedence " is
ambiguous.
What it _does_ state is this:
* To begin with, all logical operators are evaluated left-to-right, with
short-circuit evaluation. This means that it's possible to do this:

if (ptr!=NULL && *ptr!='A') ...

and be certain that your pointer is tested before being dereferenced.
* The logical and (&&) operator binds more closely than the logical or
(||). That is, x || y && z means x || (y && z), not (x || y) && z. The
order of evaluation remains the same, however.

Richard

[Grumble]

Richard Bos wrote:

<snip>

Richard,

I think your clock is incorrectly set.

The original message is dated 14:34:18 GMT (1) and your reply is dated
13:56:05 GMT (2).

(1) Date: 19 Jan 2004 06:34:18 -0800
(2) Date: Mon, 19 Jan 2004 13:56:05 GMT

[Richard Bos]

Grumble <in*****@kma.eu .org> wrote:

Richard Bos wrote:

I think your clock is incorrectly set.
This is true (I've been looking for a gizmo that keeps it set to an
atomic clock, but I'm not paying $39.95 to some vague shareware firm for
it, so in the mean time I'm correcting it by hand every now and then)...
The original message is dated 14:34:18 GMT (1) and your reply is dated
13:56:05 GMT (2).

....but this doesn't necessarily indicate that. Usenet is weird.

Richard

[Sean Kenwrick]

"Vivek N" <ke***********@ yahoo.com> wrote in message
news:e7******** *************** *@posting.googl e.com...

Hi Folks,
This question may well have been asked and answered before. But, sorry
that I couldn't find one from the archives.

I typed up this program and compiled it with gcc 3.3.2
main() {
int i = -3,j= 2,k=0,m;

m = ++i || ++j && ++k;
printf("\n%d %d %d %d\n",i,j,k,m);
}
and got the output to be
-2 2 0 1

I presumed that the answer should have well been
-3 3 1 1

as ++j and ++k would first be evaluated and && operator applied to
obtain the result to be TRUE (3 && 1). Hence, ++i wouldn't be evaluated
as one part of || goes TRUE.

The ANSI C specification clearly states that && has precedence over ||.

The precedence of an operator only defines where brackets are implicitely
placed to avoid abiguity. Thus a || b && c evaluates to a || (b && c)
rather than (a || b) && c. But after that the conditional statement is
executed left to right and will abort as soon as it is know that the
condition is false, which is why the ++j and ++k are not evaluated. I've
seen some compilers with an option to force all expressions in a conditional
statement to be evaluated so check your compiler if you really need this ..

Sean

[Dan Pop]

In <bu**********@n ews-rocq.inria.fr> Grumble <in*****@kma.eu .org> writes:

Richard,

I think your clock is incorrectly set.

The original message is dated 14:34:18 GMT (1) and your reply is dated
13:56:05 GMT (2).

(1) Date: 19 Jan 2004 06:34:18 -0800
(2) Date: Mon, 19 Jan 2004 13:56:05 GMT

How did you decide that the original message was correctly time stamped?

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de

[Nudge]

Dan Pop wrote:

Grumble wrote:

I think your clock is incorrectly set.

The original message is dated 14:34:18 GMT (1) and your reply is dated
13:56:05 GMT (2).

(1) Date: 19 Jan 2004 06:34:18 -0800
(2) Date: Mon, 19 Jan 2004 13:56:05 GMT

How did you decide that the original message was correctly time stamped?

The original message showed up on my system when I downloaded new
messages at approximately 14:40:00 GMT.

Then Pete's and Christopher's messages showed up on my system when I
downloaded new messages at approximately 14:55:00 GMT.

When Richard's message showed up 5 minutes later, I figured it was
more probable that the first 3 were correctly time-stamped, and it
was Richard's clock which was incorrectly set.

Are you satisfied with my answer? :-)

[Jarno A Wuolijoki]

On 19 Jan 2004, Vivek N wrote:

m = ++i || ++j && ++k;

as ++j and ++k would first be evaluated and && operator applied to
obtain the result to be TRUE (3 && 1). Hence, ++i wouldn't be evaluated
as one part of || goes TRUE.
The ANSI C specification clearly states that && has precedence over ||.

Your reasoning doesn't make sense as ++ has even higher precedence than &&.