How does PHP determine whether a variable is an array or a string?

Question

I'm confused as to how PHP determines whether a variable is a string or an array. It seems to depend on the operators being used.

Here's an example:

<?php
$z1 = "abc";
$out = "";
for ($i = 0; $i < strlen($z1); $i++)
{
    // $out[$i] = $z1[$i]; 
    $out = $out.$z1[$i];
}
print $out;
?>

In the above version $out becomes a string (print $z1 shows "abc"). However, if I use the first line $out[$i] = $z1[$i];, $out becomes an array.

Can someone please clarify why this happens, and if its possible to access a string's characters with square brackets without converting the output to an array?

Interestingly, if you initialise $out as a non-empty string, e.g. $out = 'a', it stays a string... — deceze
– deceze ♦, Commented May 18, 2014 at 19:00
From the PHP Manual: String access and modification by character. As a more general note, the PHP Manual is quite good: you should take a look. — Whymarrh
– Whymarrh, Commented May 18, 2014 at 19:00

Eric Schwarz · Accepted Answer · 2015-01-12 15:45:27Z

The definition of a string in PHP is considered a set of data writen in linear format (i.e: $var = "username=SmokeyBear05,B-day=01/01/1980";)

An array however is a set of data broken down into several parts. A sort of list format if you will. As an example I've written the data string from before, into an array format...

Array(['username']=>"SmokeyBear05", ['B-day']=>"01/01/1980")

Now strings are generally defined as such: $var="Your String"; Arrays however can be written in three different formats:

$var1 = array('data1','data2','data3');
$var2 = array('part A'=>'data1','part B'=>'data2','part C'=>'data3');

The output of var1 starts the index value at 0. The output of var2 however, sets a custom index value. Now the third way to write an array (least common format) is as such:

$var[0]="data1";
$var[1]="data2";
$var[2]="data3";

This takes more work, but allows you to set the index.

Most web developers working with PHP will set data from an external source as a string to deliver it to another PHP script, and then break it down into an array using the explode() function.

Lewis · Accepted Answer · 2014-05-18 19:01:32Z

0

When you define variable $out = "", for loop doesn't understand this variable as string value. If you set $out[$i] value, by default, it was treated as an array.

If you want to get the output result as string value, you can define $out = "a" to make sure it's a string variable.

answered May 18, 2014 at 19:01

Lewis

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3 Comments

deceze Over a year ago

While this appears to be the answer, it'd be interesting why PHP doesn't treat the string "" as a string but does "a".

user51726 Over a year ago

Thanks for replying. So how do I define an empty string?. Based on the replies here, writing $out="" will create an array.

Lewis Over a year ago

@user51726 It doesn't seem to be possible in your case. If you hate the a string, try with a white space $out=' '.

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