Jordi Canals wrote:
Hi,
I'm trying to understand how the intval() function works, but I'm
you'd think offhand that that would be easy....
unable to understand how exactly it works.
hmm. very odd. I took a look at your code and did a few more tests
on PHP 5.0.2 (cli) (built: Oct 21 2004 13:52:27):
<OUTPUT>
$a = (0.1 + 0.7) * 10;var_dump($a);
------------
float(8)
$b = intval($a);var_dump($b);
------------
int(7)
$c = $a;settype($c, "integer");var_dump($c);
------------
int(7)
$d = intval("$a");var_dump($d);
------------
int(8)
$e = floatval($a);var_dump($e);
------------
float(8)
$f = floatval("$a");var_dump($f);
------------
float(8)
$g = (int) floatval($a);var_dump($g);
------------
int(7)
$h = $a;settype("$h", "integer");var_dump($h);
------------
int(8)
------------
var_dump($a,$b,$c,$d,$e,$f,$g,$h,"$a");
------------
float(8)
int(7)
int(7)
int(8)
float(8)
float(8)
int(7)
int(8)
string(1) "8"
</OUTPUT>
<CMD-LINE-CODE>
php -r '
$a = (0.1 + 0.7) * 10;
echo "\n\$a = (0.1 + 0.7) * 10;var_dump(\$a);\n------------\n";
var_dump($a);
$b = intval($a);
echo "\n\$b = intval(\$a);var_dump(\$b);\n------------\n";
var_dump($b);
$c = $a;settype($c, "integer");
echo "\n\$c = \$a;settype(\$c, \"integer\");var_dump(\$c);\n------------\n";
var_dump($c);
$d = intval("$a");
echo "\n\$d = intval(\"\$a\");var_dump(\$d);\n------------\n";
var_dump($d);
$e = floatval($a);
echo "\n\$e = floatval(\$a);var_dump(\$e);\n------------\n";
var_dump($e);
$f = floatval("$a");
echo "\n\$f = floatval(\"\$a\");var_dump(\$f);\n------------\n";
var_dump($f);
$g = (int) floatval($a);
echo "\n\$g = (int) floatval(\$a);var_dump(\$g);\n------------\n";
var_dump($g);
$h = "$a";settype($h, "integer");
echo "\n\$h = \$a;settype(\"\$h\", \"integer\");var_dump(\$h);\n------------\n";
var_dump($h);
echo
"------------\n\nvar_dump(\$a,\$b,\$c,\$d,\$e,\$f,\$g,\$h,\"\$a\");\n------------\n";
var_dump($a,$b,$c,$d,$e,$f,$g,$h,"$a");
'
</CMD-LINE-CODE>
Take a look at this piece of code (Tested on PHP 5.0.3):
<?php
$a = (0.1 + 0.7) * 10;
$b = intval($a);
echo 'a -> '. $a .' -> '. gettype($a); // Prints: a -> 8 -> double
echo '<br>';
echo 'b -> '. $b .' -> '. gettype($b); // Prints: b -> 7 -> integer
sidenote: if you use double quotes in your test code you make it a
little easier for people to run the code off the cmdline ('php -r' on *nix).
?>
I also tested settype() and casting:
settype($a, 'integer'); // New value for $a is 7
$b = (int) $a; // Value for $b is 7
I cannot understand it. If originally the value for $a is 8 (double),
why when converting to integer I get 7?
my guess this is a round error issue - never noticed this problem before...
this is a problem right? anybody?
Any help or comment which helps me to understand that will be really
welcome!
Jordi.
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