Cannot convert string to int in PHP

Question

I'm having a tough time converting a value I read from socket connection into an integer. It's value tells me how many bytes are going to be received next.

$ar = fread($client, 4);
$binaryInt = unpack('N', $ar);
debug_to_console($binaryInt); //displays 16 (as expected)
$bred = fread($client, intval($binaryInt)); //intval returns 1

For some reason intval always evaluates to 1 and I can't figure out why. Btw the $ar variable is sent an integer from a java server I have running via this code sample

dOut.writeInt(data.length);

Thanks

edit

I finally got it. user2587326 below helped but instead of using intval(binaryInt[1]) it was just binaryInt[1].

unpack returns an array. Calling intval on an array will always return 1 and emit a notice, so what you are seeing is perfectly normal. What are you trying to do actually? — Pevara
– Pevara, Commented Dec 2, 2014 at 0:08
This value tells me how many bits I'll be receiving after the 4 byte integer value I'm trying to evaluate. "$bred = fread($client, ...)" is where it's not evaluating to what I want — bustedware
– bustedware, Commented Dec 2, 2014 at 0:28

Palo · Accepted Answer · 2014-12-19 06:54:40Z

1

I think you are reading a 4-byte integer sent from your Java program over DataOutputStream to your PHP program over socket.

First, the fread($client, 4) call reads those 4 bytes as a string, then you convert them to an array with unpack(), taking 4-bytes at a time (format specifier 'N') thus resulting with an array with one element, so you just use:

~~intval(~~$binaryInt[1]) instead of intval($binaryInt) to obtain that number.

edit: but it works with intval too, it won't hurt it.

edited Dec 19, 2014 at 6:54

answered Dec 2, 2014 at 0:24

Palo

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2 Comments

bustedware Over a year ago

I got it, instead of using intval($binaryInt[1]) it's just $binaryInt[1]. $binaryInt[0] is the string value of the int in case anybody was wondering. php.net/manual/en/function.unpack.php under example #1 it is explained

Palo Over a year ago

You are correct, the intval() is not needed there. but $binaryInt[0] will not be available. You were confused by the example, which uses a different format specifier. In that case the array indexes are chars and int, not 0, and 1. The following example: <?php $x="1234"; $b=unpack('N', $x); foreach ($b as $key => $value) print "\$b[$key] = $value\n"; print gettype($b[1]) . "\n"; print $b[1] . "\n"; print $b[0]; ?> prints the following output: $b[1] = 825373492 integer 825373492 PHP Notice: Undefined offset: 0 in a.php on line 8

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