On 04/17/2007 10:16 PM, Sebastian Nohn wrote:
Dmitry Stogov wrote:
It is bad practice to use echo $float or var_dump($float)
because they depends on php.ini settings.
You should use printf() of number_format() for deterministic result.
<?php
$a = 6900000000;
$b = $a.""; <-------------- that's the very same mistake
printf("%d", $a); echo "\n";
printf("%d", $b); echo "\n";
?>
You're converting float to string and THEN trying to output as integer.
My patch returns the old behaviour, though you're code is still wrong.
PHP 5.2.2:
-1689934592
6
#php -r 'printf("%d", 6900000000);'
-1689934592
#php -r 'printf("%d", "6900000000");'
2147483647
#php -v
PHP 5.2.2RC1 (cli) (built: Apr 16 2007 10:01:11)
--
Wbr,
Antony Dovgal
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