Is there an easier way to understand negation of a function result? - php

Question

I recently had to write a piece of code where I used the following; but it was very confusing for me and took me longer than I should
I keep getting lost by the negation and false turning into true and then back maybe. I end up solving these problems by trial and error.
Here is the code for getting rid of empty elements. NB, perhaps there are better code but I'm interested in knowing how to think this through
array_filter($array, "eliminate");
function eliminate($v)
{
return !(empty($v));
}

Answer 1

array_filter takes two arguments. The first is an array. The second is a callback. Here you are using the name of a function contained within a string "eliminate" to tell PHP which function you want to invoke.
The purpose of array_filter is to remove items from the array if you don't want them. The function from the second argument is called on every item in the array. The item is passed as the first argument to that function -- so $v in your above code is set to the value of the element we're looking at. If you return true from your callback, the item is kept in the array. If you return false, it is removed.
Your function returns true or false based on the following:
return !(empty($v));
empty returns true if the element is empty (e.g. an empty string, an empty array, the integer 0, among others). The ! operator turns this around. So the function will return false (and remove the item from the array) if the item is empty. Otherwise, it will return true and keep it.
Your function is approximately equivalent to the following loop:
$newArray = array();
foreach ($oldArray as $key=>$value) {
if (!empty($value)) {
$newArray[$key] = $value;
}
}

Answer 2

Iterates over each value in the input array passing them to the callback function. If the
callback function returns true, the current value from input is returned into the result array.
Array keys are preserved.
Source: http://php.net/array_filter
Each element is passed to eliminate function. Now inside eliminate if the value passed is empty it will return false and it will not belong in the next array.
Please note that your code does nothing because you should do:
$newArray=array_filter($array, "eliminate"); //< note the newarray
function eliminate($v)
{
return !(empty($v));
}
Or a better code with PHP 5.3
$newArray=array_filter($array, function ($v){
return !(empty($v));
} );

Answer 3

The best way to think about this code is by "talking" it out:
Let's start with eliminate, eliminate returns true when $v is not empty, false otherwise
Now, array_filter applies the designated function, here it's eliminate, to every element in $array, one at a time.
Finally, we need to understand that if, for some single element $v in $array, eliminate($v) returns true, then that element is removed.

Answer 4

array_filter runs the callback function on each array element. If the callback returns true, the element is kept, if false it's removed.
So, for every element in $array, it checks if it's empty(). If it's not, it gets to stay in the array, if it is it leaves. empty() returns true if it's empty (which means we want it removed from the array), so we need to return false.
If you call array_filter without a callback, it remove things that convert to false, so in your case $array = array_filter($array); should work.
array_filter returns you the modified array.

Answer 5

Just use:
array_filter($array);
This will eliminate every empty element of the array - the result should be identical to the code you have shown (with 'eliminate' function and !(empty($v)) code).