Why does `intval(19.9 * 100)` equal `1989`?
Boy, this one is really weird. I expect the following code to print 1990, but it prints 1989!
开发者_StackOverflow中文版$val = '$19.9';
$val = preg_replace('/[^\d.]/','',$val);
$val = intval($val * 100);
echo $val;
Why on earth is this happening?
Edit: and this code:
$val = '$19.9';
$val = preg_replace('/[^\d.]/','',$val);
echo $val . "<br>";
$val = $val * 100;
echo $val . "<br>";
$val = intval($val);
echo $val;
Prints:
19.9
1990
1989
Why does intval(1990)
equal 1989
???
This is a precision issue inherent to floating point numbers in PHP, and lots of other languages. This bug report discusses it a bit, in the context of casting as an int:
http://bugs.php.net/bug.php?id=33731
Try round($val * 100)
instead.
The usual answer to this kind of question is to read What Every Computer Scientist Should Know About Floating-Point Arithmetic.
Why does intval(1990) equal 1989???
Because you're not taking intval(1990)
. You're taking intval($val * 100)
where $val
is a number close to, but slightly smaller than, 19.9.
Read The Floating-Point Guide to understand why this is so.
As for how to fix it: don't ever use floating-point values for money. In PHP, you should use BCMath instead.
i was facing similar issue with my code but got solution php.net
need to convert variable to string for intval operation e.g:
intval( 9.62 * 100 ) //gives 961
intval( strval( 9.62 * 100 ) ) //gives 962
$val
is a floating point number - the result of "19.9" * 100
. Floating point numbers are not 100% accurate in any language (this is by design). If you need 100% decimal accuracy for dollar values, you should use integers and perform all calculations using cents (E.g., "$19.90"
should be 1990
).
精彩评论