Creating functions with Array as parameter not work with me

Question

In short, I have a function like the following:

function plus($x, $y){
   echo $x+$y;
}

I want to tell the function its parameters as array like the following:

$parms = array(20,10);
plus($parms);

But unfortunately, not work. I'm tired by using another way as the following:

$parms = array(20,10);
$func_params = implode(',', $parms);
plus($func_params);

And also not work, and gives me Error message: Warning: Missing argument 2 for plus(), called in.....

And now, I'm at a puzzled. What can I do to work ?

donutdan4114 · Accepted Answer · 2012-09-19 00:42:27Z

1

There is a couple things you can do. Firstly, to maintain your function definition you can use call_user_func_array(). I think this is ugly.

call_user_func_array('plus', $parms);

You can make your function more robust by taking a variable number of params:

function plus(){
  $args = func_get_args();
  return $args[0] + $args[1];
}

You can simply accept an array and add everything up:

function plus($args){
  return $args[0] + $args[1];
}

Or you could sum up all arguments:

function plus($args){
  return array_sum($args);
}

This is PHP, there are 10 ways to do everything.

answered Sep 19, 2012 at 0:42

donutdan4114

1,2721 gold badge8 silver badges16 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Lion King Over a year ago

Thanks, the first way is what I want.

jeroen · Accepted Answer · 2012-09-19 00:39:42Z

0

You need to adapt your function so that it only accepts one parameter and then in the function itself, you can process that parameter:

Very simple example:

function plus($arr){
   echo $arr[0]+$arr[1];
}

$parms = array(20,10);
plus($parms);

You can easily adapt that to loop through all elements, check the input, etc.

answered Sep 19, 2012 at 0:39

jeroen

91.8k22 gold badges117 silver badges132 bronze badges

Comments

bartvanraaij · Accepted Answer · 2012-09-19 00:40:40Z

0

Heh? The error message is very clear: you ask for two parameters in your function, but you only provide one.

If you want to pass an array, it would be a single variable.

function plus($array){
    echo ($array[0]+$array[1]);
}
$test = array(1,5);
plus($test); //echoes 6

answered Sep 19, 2012 at 0:40

bartvanraaij

3531 silver badge8 bronze badges

Comments

Hernan Velasquez · Accepted Answer · 2012-09-19 00:40:44Z

0

Use this:

function plus($arr){
   $c = $arr[0]+$arr[1];
   echo $c;
}

And the you can invoke:

$parms = array(20,10);
plus($parms);

answered Sep 19, 2012 at 0:40

Hernan Velasquez

2,82017 silver badges22 bronze badges

Collectives™ on Stack Overflow

Creating functions with Array as parameter not work with me

4 Answers 4

1 Comment

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

1 Comment

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related