PHP constructor arguments?

Question

I'm building a class that needs two arguments, and they can be passed through __constructor, or they can be set using setter methods.

What is the best way to check if arguments are passed through constructor?

I did it like this:

class Services {

    public function __construct(array $configs, array $services)
    {
        if(isset($configs) AND isset($services)) {
            $this->configs = $configs;
            $this->services = $services;
        }
    }

    public function setConfigs(array $configs)
    {
        $this->config = $configs;
    }

    public function setServices(array $services)
    {
        $this->services = $services;
    }
}

Now this works fine, but I'm not 100% if this is the right way. The thing thats bothering me is that if arguments are passed through constructor, I want both of them there, not only one.

How would I prevent user to put only one argument in constructor?

deceze · Accepted Answer · 2012-10-18 12:47:26Z

5

Currently you have to pass two arguments. To make them optional, you need to assign default values. You can then enforce both with a simple check:

public function __construct(array $configs = null, array $services = null) {
    if ($configs === null xor $services === null) {
        throw new InvalidArgumentException('Supply both or none!');
    }
    if ($configs && $services) {
        $this->setConfigs($configs);
        $this->setServices($services);
    }
}

You shouldn't use isset, since the variable always exists, as it's part of the function signature. You only need to check the values.

answered Oct 18, 2012 at 12:47

deceze♦

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Comments

davidkonrad · Accepted Answer · 2012-10-18 12:54:58Z

1

Have you tested it? When you have defined the types in the function header __construct(array $configs, array $services), you ensure that a user MUST pass two arrays!

$test = new Services(array());

fails - Catchable fatal error: Argument 2 passed to Services::__construct() must be an array

$test = new Services('','');

Fails - Catchable fatal error: Argument 1 passed to Services::__construct() must be an array,

only

$test = new Services(array(), array());

is allowed!

answered Oct 18, 2012 at 12:54

davidkonrad

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Comments

LuigiEdlCarno · Accepted Answer · 2012-10-18 12:44:57Z

0

Seperately handle the arguments you passed.

public function __construct(array $configs, array $services)
{
    if(isset($configs)) {
        $this->configs = $configs;
    }
    if(isset($services)) {
        $this->services = $services;
    }
}

answered Oct 18, 2012 at 12:44

LuigiEdlCarno

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Comments

Florent · Accepted Answer · 2012-10-18 12:45:11Z

0

you can add the php is_null() function, so if !is_null(arg1) && !is_null(arg2), then you're fine.

edited Oct 18, 2012 at 12:45

Florent

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answered Oct 18, 2012 at 12:44

voluminat0

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2 Comments

Havelock Over a year ago

How about !(is_null(arg1) || is_null(arg2))?

voluminat0 Over a year ago

It gets the same result, and the code is equally long. So both ways are correct :)

Gregor Menih · Accepted Answer · 2012-10-18 12:47:40Z

0

I would do it like this:

public function __construct(array $configs = false, array $services = false)
    {
        if($configs && $services) {
            $this->configs = $configs;
            $this->services = $services;
        }
    }

answered Oct 18, 2012 at 12:47

Gregor Menih

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1 Comment

Boris Guéry Over a year ago

I would rather not doing this, it is better to don't mix type, you should use array $configs = array() to allow no value to be set.

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