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Bug #41714 Problem with intval()
Submitted: 2007-06-16 21:18 UTC Modified: 2007-06-17 21:19 UTC
From: bartlomiejbak at gmail dot com Assigned:
Status: Not a bug Package: *Math Functions
PHP Version: 5.2.3 OS: ubuntu
Private report: No CVE-ID: None
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From: bartlomiejbak at gmail dot com
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 [2007-06-16 21:18 UTC] bartlomiejbak at gmail dot com
Hi. I've got a little bit problem
this code should show me result of equation 20x + 27y = 1
for ($y = 0; $y < 1000; $y++) 
  $x = (1/20) - (27/20) * $y;
  if ($x == intval($x)) $out .= "x: $x,y: $y\n";

I need only int values of X and Y but intval() sometimes return wrong value. Result should be (-4,3),(-31,23),(-58,43) etc. but php is starting showing from (-31,23)...


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 [2007-06-17 09:34 UTC]
Floating point values have a limited precision. Hence a value might 
not have the same string representation after any processing. That also
includes writing a floating point value in your script and directly 
printing it without any mathematical operations.

If you would like to know more about "floats" and what IEEE
754 is read this:
Thank you for your interest in PHP.

 [2007-06-17 21:19 UTC] bartlomiejbak at gmail dot com
so. if you are right look at results of this code

function isInt($value) { return preg_match("/^-?[0-9]+$/",$value); }
for ($y = 0; $y < 100; $y++)
  $x = (1/20) - (27/20) * $y;
  if (isInt($x)) 
    echo "x = $x; intval(x) = ".intval($x)."; var_dump(x == 
intval(x)) = ";
    var_dump($x == intval($x));
    echo "\n";

first result
// x = -4; intval(x) = -4; var_dump(x == intval(x)) = bool(false)

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