PHP function Error : Notice: Undefined offset

Question

I have simple function but got the error message : Notice: Undefined offset: 3 in

i checked the problem is in line : $newValue += $value[$i];

the answer of below function is "6" but error message still displayed.

<?php
function countMe($value) {

    $newValue = 0;

    for ($i=0; $i<=count($value); $i++) 
    {
        $newValue += $value[$i];
    }   
    return $newValue;
}   

$value[0] = 1;
$value[1] = 2;
$value[2] = 3;

echo countMe($value);
?>

I know the solution is just Suppress the error with the @ operator such as @countMe($value) but i want to know What's wrong with this function ?? Any help would be appreciated. Thanks

change this for ($i=0; $i<=count($value); $i++) to for ($i=0; $i — Maninderpreet Singh, Apr 10 '16 at 13:23
Possible duplicate of [PHP: "Notice: Undefined variable" and "Notice: Undefined index"](http://stackoverflow.com/questions/4261133/php-notice-undefined-variable-and-notice-undefined-index) — Gerald Schneider, Apr 10 '16 at 13:23

score 2 · Accepted Answer · answered Apr 10 '16 at 13:26

2

Indexes are from 0 to 2... your function should count to count() - 1:

 for ($i=0; $i<=count($value)-1; $i++)

or

 for ($i=0; $i<count($value); $i++)

By the way... just use array_sum($value) to get the same result :)

answered Apr 10 '16 at 13:26

Flash Thunder

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score 1 · Answer 2 · answered Apr 10 '16 at 13:23

Please see that the value of count($value) is 3.

When you're mentioning <= in the loop, it's trying to access $value[3] but there's no such index available.

You could simply change it to <.

Another approach would be to use foreach instead. That provides more flexibility.

 for ($i = 0; $i< count($value); $i++)

OR

foreach ($value as $v) {
   $newValue += $v;
}

Hope this helps.

PHP function Error : Notice: Undefined offset

2 Answers2