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Home/ Questions/Q 6999847
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T20:35:41+00:00 2026-05-27T20:35:41+00:00

<?php /* PHP devs, test & tell me I’m crazy. */ $x[] = ‘1’;

  • 0
<?php 
/* PHP devs, test & tell me I'm crazy. */ 
$x[] = '1'; 
if (empty($x[0]['x'])) {
  echo 'No PHP bug.';
} 
else {
  echo 'PHP bug exists.';
} 
?>

I always get “PHP bug exists.”

<?php 
/* PHP devs, test & tell me I'm crazy. */ 
$x[] = 1; 
if (empty($x[0]['x'])) {
  echo 'No PHP bug.';
} 
else {
  echo 'PHP bug exists.';
} 
?>

Outputs “No PHP bug.”

<?php 
/* PHP devs, test & tell me I'm crazy. */ 
$x[] = '1'; 
if (!isset($x[0]['x'])) {
  echo 'No PHP bug.';
} 
else {
  echo 'PHP bug exists.';
} 
?>

Outputs “PHP bug exists.”

<?php 
/* PHP devs, test & tell me I'm crazy. */ 
$x[] = '1'; 
if (!isset($x[0]['hello world'])) {
  echo 'No PHP bug.';
} 
else {
  echo 'PHP bug exists.';
} 
?>

Outputs “PHP bug exists.”

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-27T20:35:41+00:00Added an answer on May 27, 2026 at 8:35 pm

    This is because you are assigning a string to the array. Because of that, the x in $x[0]['x'] gets auto-cast into 0 – remember, the second indes is not pointing to an array, but a string, which can’t have non-numeric indexes.

    $x[0][0] indeed exists – it’s the first character of the string.

    $x[0][1] does not exist, and your test returns the correct result if you change the index accordingly:

    if (isset($x[0][0])) echo "0 index is set."; 
    if (isset($x[0][1])) echo "1 index is not set."; 
    

    Lesson: Even though strings can be accessed like arrays, they aren’t arrays.

    There’s a warning in the PHP manual page on strings explaining the behaviour:

    Writing to an out of range offset pads the string with spaces. Non-integer types are converted to integer. Illegal offset type emits E_NOTICE. Negative offset emits E_NOTICE in write but reads empty string. Only the first character of an assigned string is used. Assigning empty string assigns NULL byte.

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