Hi All,
I have one question regarding the bellow mentioned code
#include<stdio.h>
int main(void)
{
int x = 0;
int y = 0;
if ( x++ && ++y)
{
++x;
}
printf("%d %d\n",x, y);
return 0;
}
Output of the program
1 0
But my understanding is the output of the program should be 2 1. I
would like to explain my understanding.
Initially x is 0 . in "if ( x++ && ++y)" x will be 1 as && introduce
sequence point that's why x will have value 1 . Then as per rule C
compiler will not stop evaluating the "if ( x++ && ++y)" as it does
not know the result of & so value of y will be 1
..Now "if ( x++ && ++y)" becomes "if ( 1 &&1 )" is if(1) . So ++x will
be executed .So the final value of x and y is 2,1 .But it is not so .
Where am I going wrong ?
Regards,
Somenath
17  1908 
On Dec 13, 9:56 pm, somenath <somenath...@gmail.comwrote:
Hi All,
I have one question regarding the bellow mentioned code
#include<stdio.h>
int main(void)
{
int x = 0;
int y = 0;
if ( x++ && ++y)
{
++x;
}
printf("%d %d\n",x, y);
return 0;
}
Output of the program
1 0
But my understanding is the output of the program should be 2 1. I
would like to explain my understanding.
Initially x is 0 . in "if ( x++ && ++y)" x will be 1 as && introduce
sequence point that's why x will have value 1 . Then as per rule C
compiler will not stop evaluating the "if ( x++ && ++y)" as it does
not know the result of & so value of y will be 1
.Now "if ( x++ && ++y)" becomes "if ( 1 &&1 )" is if(1) . So ++x will
be executed .So the final value of x and y is 2,1 .But it is not so .
Where am I going wrong ?
x is evaluated as zero and then incremented. That is the meaning of x+
+ as opposed to ++x.
Is it clear?
Regards,
Somenath
On Dec 14, 11:05 am, user923005 <dcor...@connx.comwrote:
On Dec 13, 9:56 pm, somenath <somenath...@gmail.comwrote:
Hi All,
I have one question regarding the bellow mentioned code
#include<stdio.h>
int main(void)
{
int x = 0;
int y = 0;
if ( x++ && ++y)
{
++x;
}
printf("%d %d\n",x, y);
return 0;
}
Output of the program
1 0
But my understanding is the output of the program should be 2 1. I
would like to explain my understanding.
Initially x is 0 . in "if ( x++ && ++y)" x will be 1 as && introduce
sequence point that's why x will have value 1 . Then as per rule C
compiler will not stop evaluating the "if ( x++ && ++y)" as it does
not know the result of & so value of y will be 1
.Now "if ( x++ && ++y)" becomes "if ( 1 &&1 )" is if(1) . So ++x will
be executed .So the final value of x and y is 2,1 .But it is not so .
Where am I going wrong ?
x is evaluated as zero and then incremented. That is the meaning of x+
+ as opposed to ++x.
Is it clear?
My question is why x is evaluated as zero ? Because compiler will stop
evaluating one expression only in sequence point . Is this not true? I
think I missing the concept how "if" statement work
somenath wrote:
On Dec 14, 11:05 am, user923005 <dcor...@connx.comwrote:
>On Dec 13, 9:56 pm, somenath <somenath...@gmail.comwrote:
>>Hi All,
I have one question regarding the bellow mentioned code
#include<stdio.h>
int main(void)
{
int x = 0;
int y = 0;
if ( x++ && ++y)
{
++x;
}
printf("%d %d\n",x, y);
return 0;
}
Output of the program
1 0
>>Where am I going wrong ?
x is evaluated as zero and then incremented. That is the meaning of x+
+ as opposed to ++x.
Is it clear?
My question is why x is evaluated as zero ? Because compiler will stop
evaluating one expression only in sequence point . Is this not true? I
think I missing the concept how "if" statement work
It's nothing to do with the if statement. x++ is a post-increment
operator, the value of x is returned and x is then incremented. If you
were to write ++x, the pre-increment form, x would be incremented and
then evaluated as 1.
--
Ian Collins.
somenath <so*********@gmail.comwrites:
I have one question regarding the bellow mentioned code
#include<stdio.h>
int main(void)
{
int x = 0;
int y = 0;
if ( x++ && ++y)
{
++x;
}
printf("%d %d\n",x, y);
return 0;
}
Output of the program
1 0
But my understanding is the output of the program should be 2 1.
[...]
No, "1 0" is correct.
The left operand of the "&&" is "x++", where the original value of x
is 0. The "x++" causes x to become one, but postfix "++" yields the
*previous* value of its argument. So the result of "x++" is 0. This
short-circuits the "&&" operator, so "++y" is not evaluated and "{
++x; }" is not executed.
If you change the condition from
x++ && ++y
to
++x && ++y
you'll get the result you expected.
--
Keith Thompson (The_Other_Keith) <ks***@mib.org>
Looking for software development work in the San Diego area.
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"
somenath wrote:
Hi All,
I have one question regarding the bellow mentioned code
BELOW. One L. Below, underneath. Bellow, shout.
[Also, said the pedant, the code isn't /mentioned/ below -- it /is/
below.]
#include<stdio.h>
int main(void)
{
int x = 0;
int y = 0;
if ( x++ && ++y)
{
++x;
}
printf("%d %d\n",x, y);
return 0;
}
Output of the program
1 0
But my understanding is the output of the program should be 2 1. I
would like to explain my understanding.
Initially x is 0 . in "if ( x++ && ++y)" x will be 1 as && introduce
sequence point that's why x will have value 1 .
`x` /starts off/ as zero. The tested value is zero. Hence the condition
fails. `x` is incremented as a side-effect.
`x++` and `++x` are different.
--
Two-Sides Hedgehog
"Life is full of mysteries. Consider this one of them." Sinclair, /Babylon 5/
Ian Collins <ia******@hotmail.comwrote:
somenath wrote:
On Dec 14, 11:05 am, user923005 <dcor...@connx.comwrote:
On Dec 13, 9:56 pm, somenath <somenath...@gmail.comwrote:
> if ( x++ && ++y)
x is evaluated as zero and then incremented. That is the meaning of x++
as opposed to ++x.
Nope, that's too simplistic.
My question is why x is evaluated as zero ? Because compiler will stop
evaluating one expression only in sequence point . Is this not true? I
think I missing the concept how "if" statement work
It's nothing to do with the if statement. x++ is a post-increment
operator, the value of x is returned and x is then incremented. If you
were to write ++x, the pre-increment form, x would be incremented and
then evaluated as 1.
Almost, but not quite. In as simple a situation as the above, that's how
you can safely pretend it works, but in fact all the Standard requires
is that:
for the pre-increment ++i,
- some time before the next sequence point, i is incremented;
- this sub-expression evaluates to the _new_ value of i,
and for the post-increment i++,
- some time before the next sequence point, i is incremented;
- this sub-expression evaluates to the _old_ value of i.
Note that in both cases, the actual increment of the object can happen
before, after, or even in parallel with the evaluation of the
expression. This is important to understand why, for example, undefined
behaviour involving ++ operators need not be restricted to two or three
values, but can actually crash in reasonable situations.
Richard
Richard Bos wrote:
Ian Collins <ia******@hotmail.comwrote:
>somenath wrote:
>>On Dec 14, 11:05 am, user923005 <dcor...@connx.comwrote:
On Dec 13, 9:56 pm, somenath <somenath...@gmail.comwrote:
if ( x++ && ++y)
>>>x is evaluated as zero and then incremented. That is the meaning of x++
as opposed to ++x.
Nope, that's too simplistic.
>>My question is why x is evaluated as zero ? Because compiler will stop
evaluating one expression only in sequence point . Is this not true? I
think I missing the concept how "if" statement work
It's nothing to do with the if statement. x++ is a post-increment
operator, the value of x is returned and x is then incremented. If you
were to write ++x, the pre-increment form, x would be incremented and
then evaluated as 1.
Almost, but not quite. In as simple a situation as the above, that's how
you can safely pretend it works,
You can see I've written a lot of operators in C++, where the simplistic
version of events is exactly what happens!
--
Ian Collins.
On Dec 14, 10:56 am, somenath <somenath...@gmail.comwrote:
Hi All,
I have one question regarding the bellow mentioned code
#include<stdio.h>
int main(void)
{
int x = 0;
int y = 0;
if ( x++ && ++y)
{
++x;
}
printf("%d %d\n",x, y);
return 0;
}
Output of the program
1 0
But my understanding is the output of the program should be 2 1. I
would like to explain my understanding.
Initially x is 0 . in "if ( x++ && ++y)" x will be 1 as && introduce
sequence point that's why x will have value 1 . Then as per rule C
compiler will not stop evaluating the "if ( x++ && ++y)" as it does
not know the result of & so value of y will be 1
.Now "if ( x++ && ++y)" becomes "if ( 1 &&1 )" is if(1) . So ++x will
be executed .So the final value of x and y is 2,1 .But it is not so .
Where am I going wrong ?
Regards,
Somenath
hi,
dekh jab tum x++ kar rahe ho tab value increase karegi par jab tum ++y
karo ge tab value zero hi rahegi kyuki ++y ka matlab hota hai ki phale
plus karega aur phir y ki value lega isliye y ki value zero hi rahegi
aur baad mai phir tum ++x kar rahe ho usme bhi yahi hoga x++ karne mai
initial value ab 1 ho gahi hai par baad mai ++x phele 1rakhega phir
plus karega matlab(1+0);
samje isliye final value 2 1 nahi dega.
In article <ec**********************************@e6g2000prf.g ooglegroups.com>,
somenath <so*********@gmail.comwrote:
>Initially x is 0 . in "if ( x++ && ++y)" x will be 1
x will be 1 after the evaluation after the evaluation of x++. But the
value of the expression x++ (which is what you're testing) is the
*old* value of x, so it's 0.
Try printf("%d\n", x++) to convince yourself.
-- Richard
--
:wq
somenath wrote:
Hi All,
I have one question regarding the bellow mentioned code
#include<stdio.h>
int main(void)
{
int x = 0;
int y = 0;
if ( x++ && ++y)
{
++x;
}
printf("%d %d\n",x, y);
return 0;
My question is why x is evaluated as zero ?
(x++) is an expression of object type.
A useful way for to consider expressions of object type
is in terms of:
1 Values
2 Side Effects
The value of (x++), is the original value of (x).
The side effect, is that x gets incremented.
The value of (++x), is the final value of (x).
The side effect, is that x gets incremented.
Whether the increment takes place
prior to or subsequent to
the evaluation, is unspecified by the standard.
These two expressions are interchangable: (x++) and (x++, x - 1)
--
pete
>>>>"CD" == Chris Dollin <eh@electrichedgehog.netwrites:
CDsomenath wrote:
>Hi All, I have one question regarding the bellow mentioned code
CDBELOW. One L. Below, underneath. Bellow, shout.
Maybe he's WISHING HE COULD TYPE THE CODE IN ALL CAPS but he realizes
this is C, not FORTRAN?
Charlton
--
Charlton Wilbur cw*****@chromatico.net
somenath wrote:
>
#include<stdio.h>
int main(void) {
int x = 0;
int y = 0;
if (x++ && ++y) ++x;
printf("%d %d\n",x, y);
return 0;
}
Output of the program
1 0
But my understanding is the output of the program should be 2 1. I
would like to explain my understanding.
Initially x is 0 . in "if ( x++ && ++y)" x will be 1 as && introduce
sequence point that's why x will have value 1 . Then as per rule C
compiler will not stop evaluating the "if ( x++ && ++y)" as it does
not know the result of & so value of y will be 1
.Now "if ( x++ && ++y)" becomes "if ( 1 &&1 )" is if(1) . So ++x will
be executed .So the final value of x and y is 2,1 .But it is not so .
In the if statement, the value of x++ is 0. Thus the condition is
false and there is no need to test ++y, whose execution is
skipped. Result, 1 0.
Change the && to || and you get your expectation.
--
Chuck F (cbfalconer at maineline dot net)
<http://cbfalconer.home.att.net>
Try the download section.
--
Posted via a free Usenet account from http://www.teranews.com
pete wrote:
The value of (x++), is the original value of (x).
The side effect, is that x gets incremented.
The value of (++x), is the final value of (x).
The side effect, is that x gets incremented.
Whether the increment takes place
prior to or subsequent to
the evaluation, is unspecified by the standard.
Good explanation.
These two expressions are interchangable: (x++) and (x++, x - 1)
The resulting values are the same, but they are not interchangeable,
since the second expression introduces a sequence point following the
increment, which affects whether the following expressions are defined:
(x++) * x /* undefined */
(x++, x - 1) * x /* defined for suitable x */
--
Thad
On Sat, 15 Dec 2007 09:09:10 -0700, Thad Smith wrote:
pete wrote:
>These two expressions are interchangable: (x++) and (x++, x - 1)
The resulting values are the same, but they are not interchangeable,
since the second expression introduces a sequence point following the
increment, which affects whether the following expressions are defined:
(x++) * x /* undefined */
(x++, x - 1) * x /* defined for suitable x */
The second expression is undefined has well. The comma operator does
introduce a sequence point, but the final reference to x is not the right
operand to the comma operator. There is no requirement that the
multiplication by x happens after the increment. It is equivalent to
x * (x++, x - 1)
where it is perhaps easier to see the problem.
Harald van Dijk wrote:
On Sat, 15 Dec 2007 09:09:10 -0700, Thad Smith wrote:
>pete wrote:
>>These two expressions are interchangable: (x++) and (x++, x - 1)
The resulting values are the same, but they are not interchangeable,
since the second expression introduces a sequence point following the
increment, which affects whether the following expressions are defined:
(x++) * x /* undefined */
(x++, x - 1) * x /* defined for suitable x */
The second expression is undefined has well. The comma operator does
introduce a sequence point, but the final reference to x is not the right
operand to the comma operator. There is no requirement that the
multiplication by x happens after the increment. It is equivalent to
x * (x++, x - 1)
where it is perhaps easier to see the problem.
Good catch!
I now think Pete is correct. I see no way to take advantage of the
sequence point introduced by the comma operator.
--
Thad
On Dec 15, 12:53 am, pete wrote:
These two expressions are interchangable:
(x++) and (x++, x - 1)
Almost.
If x is an unsigned short with the value USHRT_MAX,
and if USHRT_MAX <= INT_MAX, then the former yields
USHRT_MAX, the latter yields -1. [Also, cf sizeof]
Similarly for unsigned char.
There are also potential differences if x is volatile.
--
Peter
Peter Nilsson wrote:
>
On Dec 15, 12:53 am, pete wrote:
These two expressions are interchangable:
(x++) and (x++, x - 1)
Almost.
If x is an unsigned short
My statement was intended to be in the context
of the quoted code, which you snipped:
int x = 0;
The expressions are also not interchangable
if x is a macro with side effects.
--
pete This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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