functions

[Anon]

I am using this function:

<?
function bgcolor ($type = "1") {
if ($type=="1")
{
$type = "CCEEFF";
}
if ($type=="-1")
{
$type = "WHITE";
}
return "$type";
}
?>

Using the line:

bgcolor ("-1");

returns WHITE as it should, however using

bgcolor ($num);

does not return a value at all.

What is the problem with the way i am calling the function?

[knickerlas]

Hi
Try this

$num=-1;
print(bgcolor($num));

Nick

[rrijnders]

Rest assured your function works perfectly. It is returning $type; it's just that $type is blank.

In your function definition, you default $type to "1". This default happens ONLY IF NO VALUE IS PASSED TO THE FUNCTION.

bgcolor("-1") returns "WHITE" because of the:
if ($type=="-1")
{
$type = "WHITE";
}

but if you invoke the function with
bgcolor($num);
the function returns $type, which is BLANK. This is because you are passing a variable ($num) into the function. Since the function default is $type = "1" the uninitialized variable $num is cast into a string of '' (blank). Since your function does not have any default handling for $type = '', the function just returns $type as is; blank.

What you were probably expecting was "WHITE" to be returned; which would be the case if you invoked the function as:

bgcolor(); // note NO variable passed!

Note that NO VARIABLE is not the same as a BLANK VARIABLE or ZERO.

Ahhhh, the subtleties of syntax and type casting...

HTH

-- Rich Rijnders
-- Irvine, CA US