if loop

[Roman Töngi]

for (int i = 1; i <= 10; i++)
cout << i << endl;

I expected the following:
1
2
3
4
5
6
7
8
9
10
11

When it comes to the last iteration i is 10.
Then i++ increments, but returns the old value 10.
So, condition is again true.
Now, i in the body of if is 11.

Apparently, I am wrong.
Does the postfix increment operator function differently within a for
construct?

[Ioannis Vranos]

Roman Töngi wrote:

for (int i = 1; i <= 10; i++)

The above can also be written:

for(int i=1; i<11; i++)

cout << i << endl;

I expected the following:
1
2
3
4
5
6
7
8
9
10
11

When it comes to the last iteration i is 10.
Then i++ increments, but returns the old value 10.
So, condition is again true.
Now, i in the body of if is 11.

Apparently, I am wrong.
Does the postfix increment operator function differently within a for
construct?

In both cases the loop body is executed for i values in [1, 10].
--
Ioannis Vranos

http://www23.brinkster.com/noicys

[Stefan Näwe]

Roman Töngi schrieb:

for (int i = 1; i <= 10; i++)
cout << i << endl;

I expected the following:
1
2
3
4
5
6
7
8
9
10
11

When it comes to the last iteration i is 10.
Then i++ increments, but returns the old value 10.
So, condition is again true.
Now, i in the body of if is 11.

Apparently, I am wrong.
Does the postfix increment operator function differently within a for
construct?

No.

Maybe it get's clearer if you re-write the for-loop as a while-loop
(which are semantically equivalent, except for the scope of the loop
variable i):

int i=1;
while(i<=10)
{
cout << i << endl;
i++;
}
HTH
Stefan

[Stefan Näwe]

Roman Töngi schrieb:

for (int i = 1; i <= 10; i++)
cout << i << endl;

I expected the following:
1
2
3
4
5
6
7
8
9
10
11

When it comes to the last iteration i is 10.
Then i++ increments, but returns the old value 10.
So, condition is again true.
Now, i in the body of if is 11.

Apparently, I am wrong.
Does the postfix increment operator function differently within a for
construct?

No.

Maybe it get's clearer if you re-write the for-loop as a while-loop
(which are semantically equivalent, except for the scope of the loop
variable i):

int i=1;
while(i<=10)
{
cout << i << endl;
i++;
}
HTH
Stefan

[Jakob Bieling]

"Roman Töngi" <ro**********@bluewin.ch> wrote in message
news:42**********@news.bluewin.ch

When it comes to the last iteration i is 10.
Then i++ increments, but returns the old value 10.
So, condition is again true.
Now, i in the body of if is 11.

Apparently, I am wrong.
Does the postfix increment operator function differently within a for
construct?

No, but the return value from incrementing is not of interest. It
used nowhere in the code. First, you increment, then you use i to check
the condition. Note: you do not use the return value, but the new value
i has.

hth
--
jb

(reply address in rot13, unscramble first)

[Niels Dybdahl]

> for (int i = 1; i <= 10; i++)

cout << i << endl;

I expected the following:
1
2
3
4
5
6
7
8
9
10
11

When it comes to the last iteration i is 10.
Then i++ increments, but returns the old value 10.
So, condition is again true.
Now, i in the body of if is 11.

Apparently, I am wrong.
Does the postfix increment operator function differently within a for
construct?

The two expressions i<=19 and i++ are evaluated independently, so when i is
incremented from 10 to 11, then 11 is stored and used for the comparison.

Niels Dybdahl

[Roman Töngi]

I understand. Condition and Expression within the if head
are separate.

Thanks

[ulrich]

On Wed, 20 Apr 2005 08:33:01 +0200, Roman Töngi <ro**********@bluewin.ch>
wrote:

for (int i = 1; i <= 10; i++)
cout << i << endl;

I expected the following:
1
2
3
4
5
6
7
8
9
10
11

When it comes to the last iteration i is 10.
Then i++ increments, but returns the old value 10.
So, condition is again true.
Now, i in the body of if is 11.

Apparently, I am wrong.
Does the postfix increment operator function differently within a for
construct?

i would say that the postfix increments i after the whole loop is done. so
when the loop has printed 10, i is incremented to 11, _then_ the loop
condition is evaluated and the loop isn't entered any more. so, no
printing of 11.

:)

[Howard]

"ulrich" <ua********@aon.at> wrote in message
news:opspisc6sin2mgp5@innsbruck-neu...

On Wed, 20 Apr 2005 08:33:01 +0200, Roman Tvngi <ro**********@bluewin.ch>
wrote:

for (int i = 1; i <= 10; i++)
cout << i << endl;

I expected the following:
1
2
3
4
5
6
7
8
9
10
11

When it comes to the last iteration i is 10.
Then i++ increments, but returns the old value 10.
So, condition is again true.
Now, i in the body of if is 11.

Apparently, I am wrong.
Does the postfix increment operator function differently within a for
construct?

i would say that the postfix increments i after the whole loop is done. so
when the loop has printed 10, i is incremented to 11, _then_ the loop
condition is evaluated and the loop isn't entered any more. so, no
printing of 11.

:)

Just to be clear... the same would happen with a prefix operator instead of
a postfix operator. The expression portion of the for statement is *always*
executed after the loop is executed.

-Howard

[Howard]

"Roman Tvngi" <ro**********@bluewin.ch> wrote in message
news:42**********@news.bluewin.ch...

I understand. Condition and Expression within the if head
are separate.

That's a "for" loop, not an "if" statement. Forget your coffee this
morning? :-)

-Howard

[ruksan]

Howard wrote:

"ulrich" <ua********@aon.at> wrote in message
news:opspisc6sin2mgp5@innsbruck-neu...

On Wed, 20 Apr 2005 08:33:01 +0200, Roman Tvngi <ro**********@bluewin.ch> wrote:

for (int i = 1; i <= 10; i++)
cout << i << endl;

I expected the following:
1
2
3
4
5
6
7
8
9
10
11

When it comes to the last iteration i is 10.
Then i++ increments, but returns the old value 10.
So, condition is again true.
Now, i in the body of if is 11.

Apparently, I am wrong.
Does the postfix increment operator function differently within a for construct?
i would say that the postfix increments i after the whole loop is done. so when the loop has printed 10, i is incremented to 11, _then_ the loop condition is evaluated and the loop isn't entered any more. so, no
printing of 11.

:)

Just to be clear... the same would happen with a prefix operator

instead of a postfix operator. The expression portion of the for statement is *always* executed after the loop is executed.

-Howard

Hi,

I'll show you the execution order of a 'for loop' then you'll
understand

for ( int i = 0; i <= 10; i++)
cout << i;

Order Of Execution;

1) i = 0 ----> i initialized

2) i <= 10 condition checked ( if evaluates to true you get in to the
loop else
you exit the loop. your i will remain as 0 if you exit here

3) if in step 2 you did not exist i++ would be evaluated

4) now onwards step 2 and 3 are followed till you exit the loop.
just after you have 10 output i would be incremented to 11 and then the
condtion checked and evaluates to false and you exit the loop without
printing
10. but your i would contain the value 11

rgds

ruksan

[Howard]

"ruksan" <ru****@gmail.com> wrote in message
news:11**********************@l41g2000cwc.googlegr oups.com...

Hi,

I'll show you the execution order of a 'for loop' then you'll
understand

for ( int i = 0; i <= 10; i++)
cout << i;

Order Of Execution;

1) i = 0 ----> i initialized

2) i <= 10 condition checked ( if evaluates to true you get in to the
loop else
you exit the loop. your i will remain as 0 if you exit here

3) if in step 2 you did not exist i++ would be evaluated
That should read:

3) if in step 2 you did not exit the loop, then the loop body would be
executed. After that, the expression (i++, in this case) would be
evaluated.

4) now onwards step 2 and 3 are followed till you exit the loop.
just after you have 10 output i would be incremented to 11 and then the
condtion checked and evaluates to false and you exit the loop without
printing
10. but your i would contain the value 11

The variable i would indeed contain 11 when the check is made, but once the
loop is exited, i is no longer in scope and thus cannot be said to have any
value at all.

-Howard