How to force arguments to be integer/string How to force arguments to be integer/string php php

How to force arguments to be integer/string


Scalar TypeHints are available as of PHP 7:

Scalar type declarations come in two flavours: coercive (default) and strict. The following types for parameters can now be enforced (either coercively or strictly): strings (string), integers (int), floating-point numbers (float), and booleans (bool). They augment the other types introduced in PHP 5: class names, interfaces, array and callable.

There is no Type Hints for scalars before PHP7. PHP 5.3.99 did have scalar typehints but it wasn't finalised at that point if they stay and how they will work then.

Nevertheless, there is options for enforcing scalar arguments before PHP7.

There is a couple of is_* functions that let you do that, e.g.

To raise a Warning, you'd use

with an E_USER_WARNING for $errorType.

Example

function setInteger($integer){    if (FALSE === is_int($integer)) {        trigger_error('setInteger expected Argument 1 to be Integer', E_USER_WARNING);    }    // do something with $integer}

Alternative

If you want to use Scalar Type Hints desperately, have a look at

which shows a technique for enforcing scalar typehints via a custom Error Handler.


You can use "Type Juggling" like (int)$height.

For example:

function setImage($target, $source_path, integer $width, $height) {    $height = (int)$height;    ...}


PHP does not (yet) implement strong typing, therefore you cannot force a parameter to be an integer. It does only apply to classes (the error you get imply that $width should be an instance of class integer) and arrays.

Type hinting for classes is available in PHP 5, type hinting for arrays starting with 5.1, and apparently scalar type hinting may (or may not) be available in the future.

You can of course as others have pointed out, check for type within your function/method, but that is fundamentally different from strong typing. The desired effect will of course be present either way.