4 ms·

> defaulting to 0 doesn't count Why this arbitrary restriction? A forced cast of a string to integer is always guaranteed to produce an integer.

• dvdkon 4y ago
  And a forced cast is therefore a bad way to generally parse a user-provided string to int.
  • Y-bar 4y ago
    So why do you insist on using it that way?
    • dvdkon 4y ago
      I don't. I asked for an int parsing function, got (int)$val as a suggestion, so I had to explain why it doesn't fit the bill.
• samus 4y ago
  And how are we supposed to distinguish that from a legit 0?
  • Y-bar 4y ago
    is_numeric() has been available since forever, and is intended for exactly the thing you appear to want to do. https://www.php.net/manual/en/function.is-numeric.php https://www.php.net/manual/en/function.is-numeric.php
    • deathanatos 4y ago
      At this point you're writing the parser that ships in most language's standard libraries. And, if you naïvely use the suggested function for the purpose you've suggested it for, e.g., like, if(!is_numeric($input)) { error(); } else { return (int)$input; } …you'll parse a lot of strings that still aren't valid integers. (E.g., "2e-2" is "numeric" and would thus parse.)