function programming: optional arguments

[blue928]

Hi, i am currently coding a php function, LIST($str) with an argument. however, I would like for this argument to be optional. is this possible? as it stands now, each time i try to call LIST() w/ no argument i get a php error "Missing argument 1 for list() .. etc.." if the argument is passed, then an IF statement reacts on the basis of that, and if not, then of course, another action is taken. so again, how do i make arguments optional? thanks

blue

[ianholmes01]

I asume you have some way of deciding if the argument is passed?

if so set the value of $string to a value that is never going to appear "natrually" in $string then test for that.

i.e.

if pass arg is true

{

passedarg = $string;

}else{

passedarg = $nevervalue

}

link(passedarg);

then in the function check for $nevervalue.

hope that helps. not sure how good this reasoning is but its a way

ian

[vandana]

u can define a default parameter for this

eg:

function makecoffee ($type = "cappucino") {
return "Making a cup of $type.\n";
}

echo makecoffee ();
echo makecoffee ("espresso");

//The output from the above snippet is:

Making a cup of cappucino.
Making a cup of espresso.

// when u do not sprcify a value a function assumes the default value when a default value is specified

for details refer php manual

[Anon]

hi,

what have i to do, to declare a parameter which is a reference as optional?

i tried this:
function CTreeElement($szName, &$parent = NULL)

but the compiler didn't accept it.

is there a solution?

matze

[arktkbear]

i assume you've figured this out by now, but you can't set a reference parameter to a default value because of the way reference parameters are handled. when you pass a parameter by reference, no storage is set aside for that parameter. the variable you deal with in your function is the variable you pass it. so you can't set the variable to a default value because unless you pass something into your function, there is no variable. understand?

doug