Answers for "strict types php"

PHP
1

php declare(strict_types=1)

"Strict types" mode only checks types at specific points in the code; it does not track everything that happens to the variable.

Specifically, it checks:

the parameters given to the function, if type hints are included in the signature; here you are giving two ints to a function expecting two ints, so there is no error
the return value of the function, if a return type hint is included in the signature; here you have no type hint, but if you had a hint of : int, there would still be no error, because the result of $a + $b + $c is indeed an int.
Here are some examples that do give errors:

declare(strict_types=1);
$a = '1';
$b = '2';
function FunctionName(int $a, int $b)
{
    return $a + $b;
}
echo FunctionName($a, $b);
Posted by: Guest on March-06-2021
3

php define strict types

declare(strict_types = 1);
Posted by: Guest on September-11-2021
0

strict types php

declare(strict_types=1);
Posted by: Guest on September-25-2021
0

php 7

<?php
// Fetches the value of $_GET['user'] and returns 'nobody'
// if it does not exist.
$username = $_GET['user'] ?? 'nobody';
// This is equivalent to:
$username = isset($_GET['user']) ? $_GET['user'] : 'nobody';

// Coalescing can be chained: this will return the first
// defined value out of $_GET['user'], $_POST['user'], and
// 'nobody'.
$username = $_GET['user'] ?? $_POST['user'] ?? 'nobody';
?>
Posted by: Guest on May-27-2020
-1

php 7

<?php
  echo "oke";
?>
Posted by: Guest on May-01-2020

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