CODEWITHSUNDEEP
php
Alberto Padilla
Alberto Padilla

Reputation: 101

why a just defined array of array count = 1?

Sorry i it is duplicated, but I don't seem to find the exact scenario I need to clarify.

So my question is why this:

var = array ();
echo count (var);

prints 0.

and this:

var = array (array());
echo count (var);

prints 1?

Thanks!

Upvotes: 2

Views: 63

Answers (4)

Bhavya Shaktawat
Bhavya Shaktawat

Reputation: 2512

it is because there exist one value in the array and it does not matter whether the inner array is empty or not.

On index 0 an empty array exists which implies array is not blank so count results 1.

Upvotes: 1

pradeepcep
pradeepcep

Reputation: 932

In the first case, you create an empty array.

$var = array();

The contents of this array may look like this:

[ ]

There is nothing here. So, count($var) is zero.

But if you create a nested array, you would have

$var = array(array());

The contents of $var would now be something like this:

[ [] ]

The inner array doesn't have anything inside it. But, the outer array has an empty array inside it. And therefore, its count is 1.

Further explanation:

Consider an array to be a plastic box.

In the first case, you have nothing inside the box, and so the count is 0.

In the second case, however, you have an empty box inside the box. So, count is 1.

Upvotes: 1

Adem
Adem

Reputation: 9429

when you do this var = array (array()); you have two dimensional array. if you familiar java, it seems like 

Object var[][] = {
   {},
};

so, var.length is ", but var[0].length is 0

Upvotes: -1

Marc B
Marc B

Reputation: 360672

Because you've put an array into an array. Even if that inner array is empty, it's still SOMETHING.

It's like putting an empty plastic bag into another plastic bag. That outer bag now contains one item: another plastic bag.

Upvotes: 5

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