Weird PHP String Integer Comparison and Conversion

Question

I was working on some data parsing code while I came across the following.

$line = "100 something is amazingly cool";
$key = 100;

var_dump($line == $key);

Well most of us would expect the dump to produce a false, but to my surprise the dump was a true!

I do understand that in PHP there is type conversion like that:

$x = 5 + "10 is a cool number"; // as documented on PHP manual
var_dump($x); // int(15) as documented.

But why does a comparison like how I mentioned in the first example converts my string to integer instead of converting the integer to string.

I do understand that you can do a === strict-comparison to my example, but I just want to know:

Is there any part of the PHP documentation mentioning on this behaviour?
Can anyone give an explanation why is happening in PHP?
How can programmers prevent such problem?

== does numerical comparison I think, and when you take and string and convert it to a number if it begins with a number, it uses that number. If it doesn't probably return something like NaN. — rubixibuc, Feb 13 '12 at 08:53
Most of us would expect this.... when casting a string to an integer (as happens when comparing a string with a number), PHP stops at the first non-numeric digit it encounters, in this case the space between 100 (all digits) and "something". SO 100 is the resultant value. — Mark Baker, Feb 13 '12 at 08:54
@mark yes most of us would expect a conversion when performing arithmetical operations, but in a comparison one? — mauris, Feb 13 '12 at 08:55
This is all fully documented in the PHP manual, see the section entitled "String conversion to numbers" on page http://www.php.net/manual/en/language.types.string.php for details of string->numeric casting, and the table of "comaprison with various types" on page http://php.net/manual/en/language.operators.comparison.php for the explanation that string/numeric comparison converts the string to a numeric — Mark Baker, Feb 13 '12 at 08:58

score 4 · Accepted Answer · answered Feb 13 '12 at 08:57

4

If I recal correcly PHP 'casts' the two variables to lowest possible type. They call it type juggling.

try: var_dump("something" == 0); for example, that'll give you true . . had that bite me once before.

More info: http://php.net/manual/en/language.operators.comparison.php

answered Feb 13 '12 at 08:57

the JinX

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The PHP manual now describes type juggling at http://php.net/manual/en/language.types.type-juggling.php which in turn links to a section about string conversion to numbers at http://php.net/manual/en/language.types.string.php#language.types.string.conversion – Josip Rodin Oct 02 '15 at 14:30

score 1 · Answer 2 · answered Feb 14 '14 at 22:10

I know this is already answered and accepted, but I wanted to add something that may help others who find this via search.

I had this same problem when I was comparing a post array vs. keys in a PHP array where in my post array, I had an extra string value.

$_POST["bar"] = array("other");

$foo = array(array("name"=>"foobar"));

foreach($foo as $key=>$data){
    $foo[$key]["bar"]="0";
    foreach($_POST["bar"] as $bar){
        if($bar==$key){
            $foo[$key]["bar"]="1";
        }
    }
}

From this you would think that at the end $foo[0]["bar"] would be equal to "0" but what was happening is that when $key = int 0 was loosely compared against $bar = string "other" the result was true to fix this, I strictly compared, but then needed to convert the $key = int 0 into a $key = string "0" for when the POST array was defined as array("other","0"); The following worked:

$_POST["bar"] = array("other");

$foo = array(array("name"=>"foobar"));

foreach($foo as $key=>$data){
    $foo[$key]["bar"]="0";
    foreach($_POST["bar"] as $bar){
        if($bar==="$key"){
            $foo[$key]["bar"]="1";
        }
    }
}

The result was $foo[0]["bar"]="1" if "0" was in the POST bar array and $foo[0]["bar"]="0" if "0" was not in the POST bar array.

Remember that when comparing variables that your variables may not being compared as you think due to PHP's loose variable typing.

Maybe it's just me, but your example is too convoluted for its own good. The issue seems to be that you define `$foo` as an associative array nested within a regular array, but then use the associative array expression `($key=>$data)` to parse the regular array. If you have to have this nesting in the data, why not also nest that loop within another one that would handle the regular array? — Josip Rodin, Oct 31 '15 at 14:23

Weird PHP String Integer Comparison and Conversion

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