Is php's 'include' a function or a statement?

Question

There are plenty of examples of both on the web. The php manual says "The include() statement [...]", which seems contradictory - if it's a statement shouldn't it not have parenthesis?

Both of these work:

include('somefile.php');
include 'somefile.php;

So should I or anyone else care?

Answer 1

Quoting from the manual (my emphasis)

Because include() is a special language construct, parentheses are not needed around its argument.

These are also called "special forms", and include such things as echo and return statements. Note that while none of these are functions, you can still speak of expressions and statements, the difference being the former have a value while the latter don't. Since include, include_once, require and require_once all return a value (TRUE if the include was successful), they can be used in expressions. By this reasoning, "include statement" would be incorrect, though includes are almost always used as statements.

Answer 2

include is a statement : Explain by following eg

// won't work, evaluated as include(('vars.php') == 'OK'), i.e. include('')
if (include('vars.php') == 'OK') {
    echo 'OK';
}

// works
if ((include 'vars.php') == 'OK') {
    echo 'OK';
}