[Resolved] [Resolved] Function Questing

DuNuNuBatman

I am having trouble returning a value from a function I wrote. Here is an example of it.

function this () {
    $number = 1
    return ($number);
}

this ();
print ($number);

All I am trying to do is return the value of a number and print it outside of the function.
I looked in my books and they don't talk about it too much, any help is appreciated.

lilleman

You have the call the function to be able to get the returned value. What you've done now is just defining the function, not using it.

function this () {
    $number = 1
    return ($number);
}

print this();

Another example:

function calc($num) {
   return ( $num * 5 );
}

print calc(2);

-- lilleman

DuNuNuBatman

I did, I just forgot to include it sorry 😃

Would this format work?

dnast

You need a semicolon after "$number = 1" and when you call the function use something like

$var = this();
print $var;

print this();

to use the returned value. $number won't exist outside of the function.

DuNuNuBatman

I know, that is what I am trying to do. Get $number to exist outside of the function.

lilleman

Declare $number as a global:

<?php

function this () {
   global $number;
   $number = 1;
}

this();
print $number;

?>

-- lilleman

DuNuNuBatman

That was it, thanks lilleman and everyone else.😃

dnast

Heh, I wrote my reply and left the computer for a few minutes, then came back and saw that I never submitted it and pressed the button.

I tried to come back and say "nevermind" on my reply, but the server was having problems. :-/

DuNuNuBatman

that's kew, thanks for trying :smiles: :shrugs:

Weedpacket

Just return it....

function this() {
    $number = 1;
    return ($number);
}

$number = this();
print ($number);

Making it global is generally poor practice because it can lead to all sorts of niggling problems, and fixing them could require hunting through all your code to find all the places where $number might be getting set and/or used before you can find out why it hasn't got the value you expected it to have.

And of course it makes changes easier if you ever want to change this() to use different variable names:

function this() {
    $wacky = 1;
    return ($wacky);
}

$number = this();
print ($number);

still works as advertised.