expression evaluation : help!

[manan.kathuria]

hi all ,
the expression in question is

++i&&++j||++k

most sources say that since the result of the || operation is decided
by the LHS itself , the right side is not computed

my point of thinking is that since the unary operator has higher
precedence than || , it will be evaluated before || in any
case...........where am im going wrong

thanks in anticipation

Manan

[Joona I Palaste]

ma************@gmail.com scribbled the following:

hi all ,
the expression in question is ++i&&++j||++k most sources say that since the result of the || operation is decided
by the LHS itself , the right side is not computed my point of thinking is that since the unary operator has higher
precedence than || , it will be evaluated before || in any
case...........where am im going wrong

Where you're going wrong is thinking precedence and order of evaluation
are the same thing. They're nothing of the sort. Precedence dictates how
a string of C tokens is compiled into an expression. Order of evaluation
dictates how that expression is evaluated at run-time.

--
/-- Joona Palaste (pa*****@cc.helsinki.fi) ------------- Finland --------\
\-------------------------------------------------------- rules! --------/
"C++. C++ run. Run, ++, run."
- JIPsoft

[Michael Mair]

ma************@gmail.com wrote:

hi all ,
the expression in question is

++i&&++j||++k

most sources say that since the result of the || operation is decided
by the LHS itself , the right side is not computed

my point of thinking is that since the unary operator has higher
precedence than || , it will be evaluated before || in any
case...........where am im going wrong

See FAQ 3.5 and 3.8, best read the whole section 3:
http://www.eskimo.com/~scs/C-faq/s3.html

Note: The ASCII text version posted here by Steve Summit is
more complete/up to date than the above HTML version but can
be obtained following the links from
http://www.eskimo.com/~scs/C-faq/top.html
Cheers
Michael
--
E-Mail: Mine is an /at/ gmx /dot/ de address.

[Joona I Palaste]

Michael Mair <Mi**********@invalid.invalid> scribbled the following:

ma************@gmail.com wrote:

hi all ,
the expression in question is

++i&&++j||++k

most sources say that since the result of the || operation is decided
by the LHS itself , the right side is not computed

my point of thinking is that since the unary operator has higher
precedence than || , it will be evaluated before || in any
case...........where am im going wrong

See FAQ 3.5 and 3.8, best read the whole section 3:
http://www.eskimo.com/~scs/C-faq/s3.html Note: The ASCII text version posted here by Steve Summit is
more complete/up to date than the above HTML version but can
be obtained following the links from
http://www.eskimo.com/~scs/C-faq/top.html

ITYM also see FAQ 3.4, which answers the OP's question pretty much
directly.

--
/-- Joona Palaste (pa*****@cc.helsinki.fi) ------------- Finland --------\
\-------------------------------------------------------- rules! --------/
"I wish someone we knew would die so we could leave them flowers."
- A 6-year-old girl, upon seeing flowers in a cemetery

[Clark S. Cox III]

On 2005-02-22 13:55:49 -0500, ma************@gmail.com said:

hi all ,
the expression in question is

++i&&++j||++k

most sources say that since the result of the || operation is decided
by the LHS itself , the right side is not computed
This is true. If the left-hand side of || is true, then the right-hand
side will never be evaluated. So, in your above expression, if (++i)
evaluates to a non-zero value, and (++j) does as well, then k will
never be incremented.
my point of thinking is that since the unary operator has higher
precedence than || , it will be evaluated before || in any
case...........where am im going wrong

The point that you're missing is that precedence has nothing to do with
order of evaluation.

--
Clark S. Cox, III
cl*******@gmail.com

[Michael Mair]

Joona I Palaste wrote:

Michael Mair <Mi**********@invalid.invalid> scribbled the following:

ma************@gmail.com wrote:

hi all ,
the expression in question is

++i&&++j||++k

most sources say that since the result of the || operation is decided
by the LHS itself , the right side is not computed

my point of thinking is that since the unary operator has higher
precedence than || , it will be evaluated before || in any
case...........where am im going wrong

See FAQ 3.5 and 3.8, best read the whole section 3:
http://www.eskimo.com/~scs/C-faq/s3.html

Note: The ASCII text version posted here by Steve Summit is
more complete/up to date than the above HTML version but can
be obtained following the links from
http://www.eskimo.com/~scs/C-faq/top.html

ITYM also see FAQ 3.4, which answers the OP's question pretty much
directly.

No, I plainly oversaw it when I was quickly scanning for special
questions to point out -- some people are scared by
"read the whole section" ;-)
Cheers
Michael
--
E-Mail: Mine is an /at/ gmx /dot/ de address.

[Joona I Palaste]

Michael Mair <Mi**********@invalid.invalid> scribbled the following:

Joona I Palaste wrote:

Michael Mair <Mi**********@invalid.invalid> scribbled the following:

ma************@gmail.com wrote:
hi all ,
the expression in question is

++i&&++j||++k

most sources say that since the result of the || operation is decided
by the LHS itself , the right side is not computed

my point of thinking is that since the unary operator has higher
precedence than || , it will be evaluated before || in any
case...........where am im going wrong

See FAQ 3.5 and 3.8, best read the whole section 3:
http://www.eskimo.com/~scs/C-faq/s3.html

Note: The ASCII text version posted here by Steve Summit is
more complete/up to date than the above HTML version but can
be obtained following the links from
http://www.eskimo.com/~scs/C-faq/top.html

ITYM also see FAQ 3.4, which answers the OP's question pretty much
directly.

No, I plainly oversaw it when I was quickly scanning for special
questions to point out -- some people are scared by
"read the whole section" ;-)

Does your reply "no" to my "ITYM" mean "no, that's just plain wrong" or
"no, I didn't think of that"?

--
/-- Joona Palaste (pa*****@cc.helsinki.fi) ------------- Finland --------\
\-------------------------------------------------------- rules! --------/
"I am looking for myself. Have you seen me somewhere?"
- Anon

[Michael Mair]

Joona I Palaste wrote:

Michael Mair <Mi**********@invalid.invalid> scribbled the following:

Joona I Palaste wrote:

Michael Mair <Mi**********@invalid.invalid> scribbled the following:

ma************@gmail.com wrote:

>hi all ,
>the expression in question is
>
>++i&&++j||++k
>
>most sources say that since the result of the || operation is decided
>by the LHS itself , the right side is not computed
>
>my point of thinking is that since the unary operator has higher
>precedence than || , it will be evaluated before || in any
>case...........where am im going wrong

See FAQ 3.5 and 3.8, best read the whole section 3:
http://www.eskimo.com/~scs/C-faq/s3.html

Note: The ASCII text version posted here by Steve Summit is
more complete/up to date than the above HTML version but can
be obtained following the links from
http://www.eskimo.com/~scs/C-faq/top.html

ITYM also see FAQ 3.4, which answers the OP's question pretty much
directly.

No, I plainly oversaw it when I was quickly scanning for special
questions to point out -- some people are scared by
"read the whole section" ;-)

Does your reply "no" to my "ITYM" mean "no, that's just plain wrong" or
"no, I didn't think of that"?

The latter :-)
Cheers
Michael
--
E-Mail: Mine is an /at/ gmx /dot/ de address.

[manan.kathuria]

thanks a lot everyone ....cheers
manan