Posted by The Natural Philosopher on 02/01/08 16:53

Jerry Stuckle wrote:
> The Natural Philosopher wrote:
>> Rik Wasmus wrote:
>>> On Fri, 01 Feb 2008 16:59:18 +0100, The Natural Philosopher <a@b.c>
>>> wrote:
>>>> The question says it all.
>>>>
>>>> I have an input box, which I fill in with a price.
>>>> IOt gets passed to the main form as a variable, then shoved into an
>>>> SQL field vue a print '%d' statement where the argument is
>>>> $price*100.
>>>
>>> Could you give us the exact code & input values? And what database
>>> (&version) are you using?
>>>
>>>> For some reaosn, this particular value goes to 6998.
>>>>
>>>> If I update hee database manually to 6999, it displays as 69.99.
>>>>
>>>> If I enter 69.999 it updates as 6999..not 69999
>>>
>>> Seems to have something to do with float/double precision. However,
>>> the precision isn't that big, and I cannot reproduce it in PHP 5.2
>>> here..
>>>
>>> PHP5.2.4:
>>> <?php echo 69.99*100 ?> => output 6999
>>> MySQL5.0.45-community-nt:
>>> SELECT CAST(69.99*100 AS UNSIGNED); => 6999
>>
>> Rik. Its worse than that NOTHING to do with databases.
>>
>> Here is a code fragment.
>>
>> echo $sale_price."<br>\r\n"; // gives 69.99
>> printf ("%d<br>\r\n",($sale_price*100)); //gives 6998
>> $xxx=$sale_price*100;
>> echo $xxx."<br>\r\n"; // gives 6999
>> printf ("%d<br>\r\n",$xxx); //gives 6998
>>
>>
>> Now $sale_price comes from a text type input via a _POST_ variable, so
>> I guess its 'text' as its raw form PHP invisible casting drives me
>> nuts, so you can tell ME what $xxx is...
>>
>
> According to the PHP manual:
>
> %d - the argument is treated as an integer, and presented as a (signed)
> decimal number.
>
> Which means it is truncating instead of rounding.
>
> 69.99 is not exact. It's approximately 69.98999999999999488409.
>
> So 69.99*100 comes out to be 6998.999999999999488409 and truncated you
> get 6998.
>
Right. So every other php funtion treats 6999 as 6999,. except printf's %d.

If you use %s, it works..

Invisible casting has to be the worst thing ever. That's WHY I did it
all in cents (or pence)to avoid these pissing float issues.

Is there a function that takes 69.99*100 and makes it a properly rounded
integer?

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