Question

if not, then how to declare a double type of number?

function testFloat(float $f)
{
    return $f;
}

echo testFloat(1.2);

Catchable fatal error: Argument 1 passed to testFloat() must be an instance of float, double given

Answer 1

Update:

Regarding type hinting:

Type Hints can only be of the object and array (since PHP 5.1) type. Traditional type hinting with int and string isn’t supported.

So I don’t know, but probably you get the error because only array and object types are supported.

---

I am not exactly sure what you want, but there is only float:

Floating point numbers (also known as “floats”, “doubles”, or “real numbers”) can be specified using any of the following syntaxes:

<?php
$a = 1.234; 
$b = 1.2e3; 
$c = 7E-10;
?>

and there you find also:

Converting to float

For information on converting strings
to float , see String conversion to numbers. For values of other types,
the conversion is performed by
converting the value to integer first
and then to float . See Converting to integer for more information. As of
PHP 5, a notice is thrown if an object
is converted to float .