Question

is it possible that a variable is referenced without using the $?

For example:

if ($a != 0 && a == true) {
...
}

I don’t think so, but the code (not written by me) doesn’t show an error and I think it’s weird. I’ve overlooked the code and a is not a constant either.

Answer 1

In PHP, a constant can be defined, which would then not have a $, but a variable must have one. However, this is NOT a variable, and is not a substitute for a variable. Constants are intended to be defined exactly once and not changed throughout the lifetime of the script.

define('a', 'some value for a');

Additionally, you cannot interpolate the value of a constant inside a double-quoted or HEREDOC string:

$a = "variable a"
define('a', 'constant a');

echo "A string containing $a";
// "A string containing variable a";

// Can't do it with the constant
echo "A string containing a";
// "A string containing a";

Finally, PHP may issue a notice for an Use of undefined constant a - assumed 'a' and interpret it as a mistakenly unquoted string "a". Look in your error log to see if that is happening. In that case, "a" == TRUE is valid, since the string "a" is non-empty and it is compared loosely to the boolean TRUE.

echo a == TRUE ? 'true' : 'false';
// Prints true
// PHP Notice:  Use of undefined constant a - assumed 'a'