I don't understand this;
PS> $x=0;while ($x++ -lt 10) {$x}
1
2
3
4
5
6
7
8
9
10
PS>
Why doesn't the loop stop at 9.

On the other hand
$x=0;while ($x+=1 -lt 10) {$x}
never stops looping.

Fred Jacobowitz

For your first. I believe it is because your adding 1 before the write $x,
thus its starting at 1 not 0.
PS>$x = 0
PS>$x++
PS>$x
1

For the second: This is because its $x += 1 doesnt return anything. It just
modifies the value of $x.
PS> $x = 0
PS> $x+=1
PS> $x
1

man about_while would suggest
PS> $x=0;while ($x -lt 10) {$x;$x++}
0
1
2
3
4
5
6
7
8
9

p.s. you are getting back 10.

--
Brandon Shell
---------------
Stop by my blog some time
Blog: http://www.bsonposh.com/
PSH Scripts Project: www.codeplex.com/psobject
--------------------------------------

"Fred Jacobowitz" <fjacobow@OptOnline.Net> wrote in message
news:Ot%23X0JwZHHA.4856@TK2MSFTNGP03.phx.gbl...
>I don't understand this;
> PS> $x=0;while ($x++ -lt 10) {$x}
> 1
> 2
> 3
> 4
> 5
> 6
> 7
> 8
> 9
> 10
> PS>
> Why doesn't the loop stop at 9.
>
> On the other hand
> $x=0;while ($x+=1 -lt 10) {$x}
> never stops looping.
>
> Fred Jacobowitz
>
>

"Brandon Shell" <tshell.mask@gmail.com> wrote in message
news:u76ckqwZHHA.4772@TK2MSFTNGP05.phx.gbl...
> For your first. I believe it is because your adding 1 before the write $x,
> thus its starting at 1 not 0.
> PS>$x = 0
> PS>$x++
> PS>$x
> 1

To be even more precise, "$x++ -lt 10" compares first and then adds one to
$x.

PS> $x = 0
PS> $x++ -lt 1
True
PS> $x
1

>
> For the second: This is because its $x += 1 doesnt return anything. It
> just modifies the value of $x.
> PS> $x = 0
> PS> $x+=1
> PS> $x
> 1

As shown by this:

PS> $x = 0
PS> $x+=1 -lt 1
PS> $x
1

You will note that as opposed to the above example with $x++ the comparison
on $x+=1 doesn't return any result.

Jacques

> To be even more precise, "$x++ -lt 10" compares first and then adds one to
> $x.

Yep. To add first and THEN compare you could use: "++$x -lt 10"

PS> $x = 0
PS> ++$x -lt 1
False
PS> $x
1

I think its more common to use the "for" statement for this pattern ->
"$x=0;while ($x++ -lt 10)"

for($x = 0; $x -lt 10; $x++)
{
$x
}


"Jacques Barathon [MS]" <jbaratho@online.microsoft.com> wrote in message
news:%23G1gozwZHHA.5080@TK2MSFTNGP04.phx.gbl...
> "Brandon Shell" <tshell.mask@gmail.com> wrote in message
> news:u76ckqwZHHA.4772@TK2MSFTNGP05.phx.gbl...
>> For your first. I believe it is because your adding 1 before the write
>> $x, thus its starting at 1 not 0.
>> PS>$x = 0
>> PS>$x++
>> PS>$x
>> 1
>
> To be even more precise, "$x++ -lt 10" compares first and then adds one to
> $x.
>
> PS> $x = 0
> PS> $x++ -lt 1
> True
> PS> $x
> 1
>
>>
>> For the second: This is because its $x += 1 doesnt return anything. It
>> just modifies the value of $x.
>> PS> $x = 0
>> PS> $x+=1
>> PS> $x
>> 1
>
> As shown by this:
>
> PS> $x = 0
> PS> $x+=1 -lt 1
> PS> $x
> 1
>
> You will note that as opposed to the above example with $x++ the
> comparison on $x+=1 doesn't return any result.
>
> Jacques

Thank you for your responses. I completely understand it now.
Actually, Bruce has the following on page 23 of his book;
ps > (1) $i=0
ps >(2) while ($i++ -lt 10) { if ($i % 2) {"$i is odd"}}
1 is odd
3 is odd
5 is odd
7 is odd
9 is odd
ps (3) >

Normally, I would increment the counter in the statement block.
However, this example had me scratching my head.

Fred Jacobowitz

Actually, the += operator *does* return the value. The confusion is related
to operator precedence. Take a look at the following:

PS> $x = 0
PS> $x+=1 -lt 10
PS> $x
1
PS> $x+=10 -lt 10
PS> $x
1

What's happening here is that += has a higher operator precedence than -lt,
so you're getting "$x += (1 -lt 10)", not "($x += 1) -lt 10" as you might
expect. (1 -lt 10) interpreted as an integer comes out as 1 (just as (10 -lt
10) interpreted as an integer is zero, so $x doesn't change) , so the end
result of the statement is that $x will keep getting incremented until it
loops all the way around and results in a zero, which is the only integer
that will be considered "false" when interpreted as a boolean. As long as
the number to the right of the += is less than 10, $x will always be
incremented by one.

Hope this helps clear up some of the confusion.

-- Ryan Milligan

"Jacques Barathon [MS]" <jbaratho@online.microsoft.com> wrote in message
news:%23G1gozwZHHA.5080@TK2MSFTNGP04.phx.gbl...
> "Brandon Shell" <tshell.mask@gmail.com> wrote in message
> news:u76ckqwZHHA.4772@TK2MSFTNGP05.phx.gbl...
>> For your first. I believe it is because your adding 1 before the write
>> $x, thus its starting at 1 not 0.
>> PS>$x = 0
>> PS>$x++
>> PS>$x
>> 1
>
> To be even more precise, "$x++ -lt 10" compares first and then adds one to
> $x.
>
> PS> $x = 0
> PS> $x++ -lt 1
> True
> PS> $x
> 1
>
>>
>> For the second: This is because its $x += 1 doesnt return anything. It
>> just modifies the value of $x.
>> PS> $x = 0
>> PS> $x+=1
>> PS> $x
>> 1
>
> As shown by this:
>
> PS> $x = 0
> PS> $x+=1 -lt 1
> PS> $x
> 1
>
> You will note that as opposed to the above example with $x++ the
> comparison on $x+=1 doesn't return any result.
>
> Jacques

"Ryan Milligan" <Ceiled@hotmail.com> wrote in message
news:7didnUUER6zL3WfYnZ2dnUVZ_q-vnZ2d@comcast.com...
> Actually, the += operator *does* return the value. The confusion is
> related to operator precedence. Take a look at the following:
>
> PS> $x = 0
> PS> $x+=1 -lt 10
> PS> $x
> 1
> PS> $x+=10 -lt 10
> PS> $x
> 1
>
> What's happening here is that += has a higher operator precedence
> than -lt, so you're getting "$x += (1 -lt 10)", not "($x += 1) -lt 10" as
> you might expect. (1 -lt 10) interpreted as an integer comes out as 1
> (just as (10 -lt 10) interpreted as an integer is zero, so $x doesn't
> change) , so the end result of the statement is that $x will keep getting
> incremented until it loops all the way around and results in a zero, which
> is the only integer that will be considered "false" when interpreted as a
> boolean. As long as the number to the right of the += is less than 10, $x
> will always be incremented by one.

Great input, you're absolutely right. So the initial example provided should
be rewritten as:

PS> $x=0; while (($x+=1) -lt 10) {$x}
1
2
3
4
5
6
7
8
9

Thanks,
Jacques