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Why,

  $b = [];
  echo $b[0]; //gives E_NOTICE Undefined offset: 0

gives notice?

But,

$b = false;
echo $b[0]; // no notice or warnings

var_dump($b[0]) // returns NULL

Could this be due to php.ini settings?

Andrey _
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  • I don't have a direct answer, but I can give you a tips on using `is_array($myVar);` to check if your variable is an array, and then checking if your column is set using `if (isset($b[0])) {}`. – Anwar Sep 14 '16 at 15:32
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    Because of type inference. Using a variable as certain type makes the interpreter to try to cast it to this type and use it as one. – Royal Bg Sep 14 '16 at 15:34

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