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Sorry, been doing heavy perl and haven't written any PHP in 3 years so a tad
rusty.

Can someone explain why the second expression in this code snippet evaluates
to 7 and not 8?

$a = (int) (0.1 +0.7);

echo "$a\n";

$x = (int) ((0.1 + 0.7) * 10);

echo "$x\n";

$y = (int) (8);

echo "$y\n";

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On Thu, 2010-02-18 at 09:47 -0600, Chuck wrote:

> Sorry, been doing heavy perl and haven't written any PHP in 3 years so a tad
> rusty.
>
> Can someone explain why the second expression in this code snippet evaluates
> to 7 and not 8?
>
> $a = (int) (0.1 +0.7);
>
> echo "$a\n";
>
> $x = (int) ((0.1 + 0.7) * 10);
>
> echo "$x\n";
>
> $y = (int) (8);
>
> echo "$y\n";


It works as expected if you take out the int() parts in each line. I'm
not sure why, but the use of int() seems to be screwing around with the
results. That's why the first line outputs a 0.

Thanks,
Ash
http://www.ashleysheridan.co.uk



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On Thu, Feb 18, 2010 at 10:50 AM, Ashley Sheridan
<ash [at] ashleysheridan.co.uk> wrote:
> On Thu, 2010-02-18 at 09:47 -0600, Chuck wrote:
>
>> Sorry, been doing heavy perl and haven't written any PHP in 3 years so a tad
>> rusty.
>>
>> Can someone explain why the second expression in this code snippet evaluates
>> to 7 and not 8?
>>
>> $a = (int) (0.1 +0.7);
>>
>> echo "$a\n";
>>
>> $x = (int) ((0.1 + 0.7) * 10);
>>
>> echo "$x\n";
>>
>> $y = (int) (8);
>>
>> echo "$y\n";
>
>
> It works as expected if you take out the int() parts in each line. I'm
> not sure why, but the use of int() seems to be screwing around with the
> results. That's why the first line outputs a 0.
>
> Thanks,
> Ash
> http://www.ashleysheridan.co.uk
>

Another fine example of floating point math.

<?php

$x = ((0.1 + 0.7) * 10);

echo "((0.1 + 0.7) * 10) === $x\n";
// ((0.1 + 0.7) * 10) === 8


var_dump(8 == $x);
// bool(false)

var_dump($x - (int) $x);
// float(1)

var_dump(8 - $x);
// float(8.8817841970013E-16)

?>

Andrew

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On Thu, Feb 18, 2010 at 16:47, Chuck <chuck.carson [at] gmail.com> wrote:
> Sorry, been doing heavy perl and haven't written any PHP in 3 years so a tad
> rusty.
>
> Can someone explain why the second expression in this code snippet evaluates
> to 7 and not 8?
>
> $a = (int) (0.1 +0.7);
>
> echo "$a\n";
>
> $x = (int) ((0.1 + 0.7) * 10);
>
> echo "$x\n";
>
> $y = (int) (8);
>
> echo "$y\n";
>

The reason why you get 7 instead of 8 is because you are using
floating point arithmetic. 0.1 (i.e. the fraction 1/10) does not have
a finite representation in base 2 (like you cannot finitely represent
1/3 in base 10). So the number 0.1 is represented in the computer as a
number that is strictly less than 0.1 so when you do 0.1+0.7=x then
you have x<0.8 in the computer (think 7.9999999...). When you cast to
int you just truncate the number, i.e. you chop off the fractional
part leaving you with 7.

--
Daniel Egeberg

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According to the PHP manual using the same expression, "Never cast an
unknown fraction to integer, as this can sometimes lead to unexpected
results". My guess is that since it is an expression of floating
points, that the result is not quite 8 (for whatever reason).
Therefore, it is rounded towards 0. Of course, that is only a guess,
and I have no true documentation on it.

Ashley Sheridan wrote:
> On Thu, 2010-02-18 at 09:47 -0600, Chuck wrote:
>
>
>> Sorry, been doing heavy perl and haven't written any PHP in 3 years so a tad
>> rusty.
>>
>> Can someone explain why the second expression in this code snippet evaluates
>> to 7 and not 8?
>>
>> $a = (int) (0.1 +0.7);
>>
>> echo "$a\n";
>>
>> $x = (int) ((0.1 + 0.7) * 10);
>>
>> echo "$x\n";
>>
>> $y = (int) (8);
>>
>> echo "$y\n";
>>
>
>
> It works as expected if you take out the int() parts in each line. I'm
> not sure why, but the use of int() seems to be screwing around with the
> results. That's why the first line outputs a 0.
>
> Thanks,
> Ash
> http://www.ashleysheridan.co.uk
>
>
>
>

--------------000707080504020206050703--

Daniel Egeberg wrote:
> On Thu, Feb 18, 2010 at 16:47, Chuck <chuck.carson [at] gmail.com> wrote:
>> Sorry, been doing heavy perl and haven't written any PHP in 3 years so a tad
>> rusty.
>>
>> Can someone explain why the second expression in this code snippet evaluates
>> to 7 and not 8?
>>
>> $a = (int) (0.1 +0.7);
>>
>> echo "$a\n";
>>
>> $x = (int) ((0.1 + 0.7) * 10);
>>
>> echo "$x\n";
>>
>> $y = (int) (8);
>>
>> echo "$y\n";
>>
>
> The reason why you get 7 instead of 8 is because you are using
> floating point arithmetic. 0.1 (i.e. the fraction 1/10) does not have
> a finite representation in base 2 (like you cannot finitely represent
> 1/3 in base 10). So the number 0.1 is represented in the computer as a
> number that is strictly less than 0.1 so when you do 0.1+0.7=x then
> you have x<0.8 in the computer (think 7.9999999...). When you cast to
> int you just truncate the number, i.e. you chop off the fractional
> part leaving you with 7.
>

yup as Daniel pointed out; this is correct - love floats & casting!

see:

$y = ((0.1 + 0.7) * 10);
$x = (int)$y;
$z = intval($y);
var_dump($y);
var_dump(serialize($y));
var_dump($x);
var_dump($z);

outputs:
float(8)
string(55) "d:7.99999999999999911182158029987476766109466552734375;"
int(7)
int(7)

lovely! - note: serializing gives us the exact value of the float

regards,

Nathan



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