php – Is isset($var1) the same as (!$var1)?

Question:

Is isset($var1) the same as (!$var1) ?. That is, in the examples that I put below it seems to work the same way: detect if a variable is defined. If they are not the same, when do you use one and the other?

<?php 
  $var1;  
  if(!$var1){
    echo "si";
  }
?>

Same example with isset

<?php
  $var1;
  if(isset($var1)){
    echo "si";
  }
?>

This example contains both isset and ! . As you will see when using the ! not checks if a controller exists but… what happens when it doesn't exist. Will it throw a PHP "undefined variable" error?

class Request{
  private $_controlador;
  private $_metodo;
  private $_argumentos;

  public function __construct(){
    if(isset($_GET['url'])){
      $url = filter_input(INPUT_GET, 'url', FILTER_SANITIZE_URL);
      $url = explode('/', $url);
      $url = array_filter($url);

      //toma array url y coge el primer elemento y lo asigna a _controlador.
      //Lo mismo con _metodo y argumentos.
      $this->_controlador = strtolower(array_shift($url));
      $this->_metodo = strtolower(array_shift($url));
      $this->_argumentos = $url;
    }

    if(!$this->_controlador){
      $this->_controlador = DEFAULT_CONTROLLER;
    }

    if(!$this->_metodo){
      $this->_metodo = 'index';
    }

    if(!isset($this->_argumentos)){
      $this->_argumentos = array();
    }
  }

  public function getControlador()
  {
    return $this->_controlador;
  }

  public function getMetodo()
  {
    return $this->_metodo;
  }

  public function getArgs()
  {
    return $this->_argumentos;
  }

}

Answer:

Well, no, it's not the same.

isset() checks that the variable exists, while ! it just checks that the value is "false".

That is, if you do an isset() to a variable that doesn't exist, it will return false, but if you do !variableInexistente you will get a PHP error for using an undefined variable.