在线编译代码时，php中未捕获TypeError

Question

I am trying to run a very simple code on a php online compiler but everytime it gives me an error: Here is the link

<?php

function value(integer $h) {
    if(!is_int($h)) {
        throw new InvalidArgumentException('integer expected');
    }
    if($h <= 7) {
        throw new RangeException('integer should be greater than 7');
    }
    echo $h;
    return $return;
}

$something = value(11111);
echo $something;

Below is the error:

Fatal error: Uncaught TypeError: Argument 1 passed to value() must be an instance of integer, integer given,

Is there anything wrong I am doing here?

Answer 1

Use type int instead integer in function.

In PHP type named int not integer