[RESOLVED] Quick function question

I want to do something like

$value = update_text($value)

function text_process($value){
$value = trim(strip_tags($value));
$value = addslashes($value);
$value = htmlspecialchars($value);
}

Do I need to return $value in order to have it available to the code that calls the function or is it already loaded into the calling value "$value ="

You will need to return it. Or, if desired, you could pass $value by reference by making this change:

Code:

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function text_process(&$value){

---

If you did it that way, no return is necessary, and you'd only need this when calling it:

Code:

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text_process($value);

---

(Your sample function call is different from your function definition here - I'm sticking with "text_process" for both...)

Quote:

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Originally Posted by SambaNeko

You will need to return it. Or, if desired, you could pass $value by reference by making this change:

Code:

---

function text_process(&$value){

---

If you did it that way, no return is necessary, and you'd only need this when calling it:

Code:

---

text_process($value);

---

(Your sample function call is different from your function definition here - I'm sticking with "text_process" for both...)

---

Oops sorry, still playing around with names to get some sort of logic into what functions are called.

So if I modify the variable I pass in the function list the changes are reflected in the calling script?

Quote:

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Originally Posted by KiwiDexter

So if I modify the variable I pass in the function list the changes are reflected in the calling script?

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You're not really modifying the variable, just changing what the function does with the variable...

The default behavior of a function is to take input variables by value. Like this:

Code:

---

function addTwo($number){
  return $number + 2;
}

$num = 1;
$newNum = addTwo($num);

---

At the end of that block of code, $newNum is 3, but $num is still 1, because the function took the value of $num, added two, then put the result into $newNum. It didn't change the value of $num.

Adding the & prior to a variable in the function definition makes the function take the variable by reference instead. Which changes things...

Code:

---

function addTwo(&$number){
  $number += 2;
}

$num = 1;
addTwo($num);

---

In this case, the value of $num at the end of the code is 3, because the function took a reference to the variable $num, and added two to the variable that was referenced.

So, as said, you could do it either way - you can add a return to get the altered value from your function, or add the & to have it alter your variable directly.

Thanks for the answer dude, think I'll settle on the "return" option as it makes it more readable. :)