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Typo produces unexpected results in PHP

Posted by: admin July 12, 2020 Leave a comment


Just found a piece of my code that had one original typo in it.

$msg = "Some text";
$msg .= " some more text";
$msg .+ " yet more text!";
$msg .= " last text";

Notice the .+ that should be .=. What surprises me is that the code ran without producing any error, warning or notice and the output was:

Some text some more text last text

I was wondering why it did that. I know full well what .= and += are but how is .+ interpreted especially since there is no equal sign.

How to&Answers:

There’s no .+ operator, so that is . followed by +.

You’re building an expression that consists of $msg concatenated with the result of applying unary + to " yet more text!" (which is 0 due to the cast to integer) … and then discarding the whole thing because you’re not doing anything with the result.

$msg .+ " yet more text!"; 
$msg . +" yet more text!"; // 1. PHP doesn't care about the spacing 
$msg . 0;                  // 2. Conversion to int from unary `+`
$msg . "0";                // 3. Coersion to string for concatenation
                           // 4. Nothing done with value

It’s perfectly valid; it just doesn’t do anything useful.


The + gets interpreted as a unary plus. PHP casts the string into an integer with the value 0 and concatenates it with $msg. However you do not assign $msg anything on that line, so $msg won’t be changed.


This works because the . is the concatenation operator, and + is the addition operator.

The line got interpreted like this:

$msg . (+" yet more text!");

The expression +" yet more text!" converts the string to an int (0 in this case as when PHP converts a string to an int, it stops at the 1st non-number character). It then concated the 0 to $msg, and ignored the result.