Daniel Egeberg wrote:
> On Thu, Feb 18, 2010 at 16:47, 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);
>>
>> ech
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 t
On Thu, Feb 18, 2010 at 16:47, 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 +
On Thu, Feb 18, 2010 at 10:50 AM, 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
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)
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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