You must be doing something else wrongly:

<?php
	function foo($a, $b) {
		var_dump ($a, $b);
	}

	foo(1.0, 1.8);
?>

[derick@kossu derick]$ php bug21537.php 
float(1)
float(1.8)


As you see it works fine...

Derick

$a = 2.0;
$b = '2.0';
xxx($a, $b);
produces
2 2.0
2.0 is still converted to int(2) instead of float(2.0).

I need to know the exact input and thought the easiest way to check the input would be to get the input as an unconverted string.

In the example of 2.0, a float of 2.0 would tell me the type and accuracy. I would then get 3.10 as 3.10 instead of 3.1.

I have not tested floats with e notation or very long numbers. If someone types a long number, I want to detect the long number, recognise the number exceeds the accuracy of the default maths, and switch to arbitrary precision mathematics.

I do not want to tell the person using my function or class, a whole lot of rules about enclosing numbers in quotations when they contain certain values.

Hi Peter,  
  
One minor note here. The first argument is not converted  
to type integer. Rather, it is trimmed of non-numerically  
significant zeros.  
  
  function xxx($a,$b){  
	var_dump($a,$b);  
  }  
  xxx(2.0, "2.0");  
  
Outputs:  
  float(2)  
  string(3) "2.0"   
  
The type of the value is still float.  
  
This behaviour might be difficult to change, considering  
that it happens in all contexts - not just when formal  
parameters are passed to a function.  
  
$a = 2.0;  
var_dump($a);  
var_dump(2.0);  
  
Outputs:  
  float(2);  
  float(2);  
 
Cheers!