| Re: Implied cast differs from explicit cast [message #176667 is a reply to message #176662] |
Mon, 16 January 2012 09:04   |
Curtis Dyer
Messages: 34
Registered: January 2011
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Arno Welzel <usenet(at)arnowelzel(dot)de> wrote:
> Jerry Stuckle, 2012-01-16 04:39:
>
>> On 1/15/2012 10:08 PM, lb wrote:
>>> jstucklex(at)attglobal(dot)net wrote:
>>>> ...
>>>> No, it is not. '0x12' is a hex number, not an integer.
>>>> (int)'0x12' converts this correctly to 0.
As Arno points out, the hex value *is* an integer (once converted
from a string, in this case).
>>> OK, but if (int)'0x12' is 0, why is '0x12'+0 == 18?
>>>
>>> Why are they different? Aren't they both doing 'string
>>> conversion to number'?
>>
>> As I said - '0x12' is a hex value, not an integer. You are
>> trying to convert it as if it were a string representation of
>> an integer. The implicit conversion recognizes it is a hex
>> value and uses the appropriate conversion.
>
> There is no difference between "hex value" and "integer" - both
> 0x12 and 18 are just different representations of the same
> value. The problem is that the explizit cast from string to int
> does not take this into account.
The problem, as far as I can tell, has to do with the implicit
behavior of the "(int)$str" cast. Its default behavior seems to be
identicle to:
/* operates on base ten values; 10 is the default base when the
* second parameter is omitted. */
intval($str, 10);
So, '0x12' is obviously then considered an invalid base ten
integer, thus the expression, "(int)'0x12'" yields 0.
If you want to convert a hex value, either use hexdec() or use:
intval($str, 16);
Note, however, the behavior of hexdec($str) and intval($str, 16)
are not identicle! In some cases, hexdec() may be preferable, as
it will attempt to represent values too large for integers as
floating-point values. It also disregards non-hex value
characters in the string.
RTM here:
<http://php.net/intval>
<http://php.net/hexdec>
--
Curtis Dyer
<?$x='<?$x=%c%s%c;printf($x,39,$x,39);?>';printf($x,39,$x,39);?>
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