Re: Understanding intval() and types conversion

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.