3

I was doing some research but wasn't able to find an answer (probably beacause I did not searched it right)

Consider this piece of code:

public function foo(?array $optionalParam);

And then this one:

public function foo(array $optionalParam = null);

What differs between them? Using PHPstorm I noticed that when I use the ?, it creates a PHPdoc and mark the variable type as type|null. But when I call the function without that argument, PHP screams on my face "you kidding me? where is $optionalParam". In the other side, I managed to use with no problems the =null option.

Sorry if this question is too simple, but i did not find any answers online.

Alexandre Elshobokshy
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João Pedro Martins de Paula
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2 Answers2

8

First of all, the ? goes before the type, not after... other than this:

Using

public function foo(?array $optionalParam);

you are forced to pass something, that can be either null or an array, infact:

<?php
function foo(?array $optionalParam){
  echo "test";
}

foo(); // doesn't work
foo(null); // works
foo([]); // works

where instead using

public function foo(array $optionalParam = null);

will accept null, an array, or 0 parameters

<?php
function foo(array $optionalParam = null){
  echo "test";
}

foo(null); // works
foo(); // work
foo([]); // works
Alberto Sinigaglia
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2

It's a PHP 7.1 feature called Nullable Types

Both of the lines you wrote are identical.

array ?$optionalParam : either an array or null

array $optionalParam = null : either an array or null

Tho using ? you'd still need to add the parameter when calling the function.

Alexandre Elshobokshy
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