PHP Undefined offset error repeated

Question

I'm trying to run a PHP script that finds all the numbers divisible by 3 or 5, dumps them into an array, and adds all the values together. However, When I try to run it, I get a number output (I don't know if it's correct or not) and several hundred errors. They start out with:

Notice: Undefined offset: 1 in G:\Computer Stuff\WampServer\wamp\www\findthreesandfives.php on line 18

Then the offset number increases by increments of 1-3 (randomly, I haven't seen a pattern yet). I can't figure out what's wrong. Here's my code:

<?php
function loop($x)
{
$a = array(); //array of values divisible by 3 or 5
$l = 0; //length of the array
$e = 0; //sum of all the values in the array
for ($i=0; $i<=$x; $i++){ //this for loop creates the array
    $n3=$i%3; 
    $n5=$i%5;
    if($n3 === 0 || $n5 === 0){
        $a[$i]=$i;
        $l++;
    }


}
for ($v=0; $v<=$l; $v++){ //this loop adds each value of the array to the total value
    $e=$e + $a[$v];
}
return $e;   
}
echo loop(1000);
?>

Someone please help...

zachjs · Accepted Answer · 2012-10-09 01:11:45Z

4

The problem in your code is the following line:

$a[$i]=$i;

Should be:

$a[count($a)] = $i;

This is because the value of $i is always increasing, so using $i as your pointer will create gaps in the array's indices. count($a) returns the total number of items in the given array, which also happens to be the next index.

EDIT: @pebbl suggested using $a[] = $i; as a simpler alternative providing the same functionality.

EDIT 2: Solving the subsequent problem the OP described in the comments:

The problem seems to be that $l is greater than the number of items in $a. Thus, using count($a) in the for loop should fix your subsequent error.

Try replacing:

for ($v=0; $v<=$l; $v++){

With:

for ($v=0; $v<=count($a); $v++){

edited Oct 9, 2012 at 1:11

answered Oct 9, 2012 at 0:25

zachjs

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6 Comments

Pebbl Over a year ago

+1 for the right answer :) however it's easier to write $a[] = $i; in php, because php will handle incrementing the index for you.

NinJoel Over a year ago

Alright, that worked.. except now it just gives this error (and a bigger final number) Notice: Undefined offset: 468 in G:\Computer Stuff\WampServer\wamp\www\findthreesandfives.php on line 18

codingbiz Over a year ago

$a[] = $i; is safer if OP still want to maintain the sequence of index - no gaps

zachjs Over a year ago

@codingbiz Safer how? $a[count($a)] = $i; and $a[] = $i; should produce the same exact results in the OP's situation and in any case where the array starts out being empty.

zachjs Over a year ago

@user1653552 I edited my post (again) to solve the second issue you ran into.

|

codingbiz · Accepted Answer · 2012-10-09 01:20:01Z

2

I found the same problem as @zsnow said. There are gaps within $a. The if condition allowed the gaps making the assignment skip some indexes. You can also use this

foreach ($a as $v){ //this loop adds each value of the array to the total value
    $e=$e + $a[$v];
}

should actually be

foreach ($a as $v){ //this loop adds each value of the array to the total value
    $e=$e + $v;
}

edited Oct 9, 2012 at 1:20

answered Oct 9, 2012 at 0:37

codingbiz

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2 Comments

anditpainsme Over a year ago

+ 1 for another working answer. the foreach construct deals with situations like this wonderfully... but leaving an array with disfigured indexes like that can only cause trouble later on down the track. bad practice, imo.

NinJoel Over a year ago

alright.. I tried that, but now it gives the same pattern of errors (after I tried the first solution), but it starts at offset 468, which is where the last stream of errors ended.

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