$variable OR $variable = $variable in PHP

Question

This is a standalone expression in a function:

$var OR $var = $var

What does this expression exactly do?

Never saw this before and can't find the answer as I have no idea how is this called.

Code itself is from: Сodeigniter Template on Github

public function build($view, $data = array(), $return = FALSE)
{
    // Set whatever values are given. These will be available to all view files
    is_array($data) OR $data = (array) $data;
    // Merge in what we already have with the specific data
    $this->_data = array_merge($this->_data, $data);

...

I'd say that it does nothing relevant... can we see the function? — Federkun
– Federkun, Commented May 28, 2016 at 21:48
Assignment (=) has higher precedence than OR, so it sets a variable value to itself, and then ors it with itself again, and then does absolutely nothing.... it's basically a waste of processing cycles as it stands — Mark Baker
– Mark Baker, Commented May 28, 2016 at 21:51
is_array($data) OR $data = (array) $data; doesn't do the same thing of $var OR $var = $var. Why didn't you post that from the start? — Federkun
– Federkun, Commented May 28, 2016 at 21:58
it's equivalent to if (!is_array($data)) { $data = (array) $data; } — Federkun
– Federkun, Commented May 28, 2016 at 22:00

johannes · Accepted Answer · 2016-05-28 22:31:52Z

3

Given an expression like yours

is_array($data) OR $data = (array) $data;

we check the precedence table on http://php.net/operators.precedence first. There we see = is above OR. Thus the line is equal to this:

is_array($data) OR ($data = (array) $data);

Next we have to check the table for associativity. or is left associative so it checks is_array($data). If that returns true the orexpression as a whole returns trueand nothing else happens.

If is_array($data) returns false we have to evaluate further, so $data = (array) $data is executed.

In the end this is a "smart" way to write

if (!is_array($data)) {
    $data = (array) $data;
}

Actually in this specific case we could also write only

$data = (array) $data;

As the (array) cast is a no-op if $data already is an array.

answered May 28, 2016 at 22:31

johannes

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