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Anonymous Coward
Asked: 2020-07-02 03:40:50 +0800 CST 2020-07-02 03:40:50 +0800 CST

What does an ampersand mean in a function declaration?

  • 772

The following code seems to do the same thing with and without ampersand( & ):

<?php

class Cartera {

    public $dinero = 93;

    public function &muestraLaPasta() { // Si quito este ampersand
        return $this->dinero;
    }

}

$carteraDeMaria = new Cartera;
$miPastaGansa = &$carteraDeMaria->muestraLaPasta(); // y este
echo $miPastaGansa;                // el resultado es el mismo

?>

Whether I leave them or remove them the result is the same, it prints 93. What are they for?

Thanks to Ivan Botero's answer, I know that this function returns a reference.
But it seems that it is not a reference capable of surviving on its own. If I assign it to a variable, that variable contains the value, not the reference, despite the fact that what I have assigned to it is the result of that function, which is supposed to be a function that returns a reference.

<?php
class Cartera {
    public $dinero = 93;

    public function &muestraLaPasta() { // Esta función devuelve una referencia
        return $this->dinero;
    }
}

$carteraDeMaria = new Cartera;
$miPastaGansa = $carteraDeMaria->muestraLaPasta(); // Parecería que almaceno esa 
                                                   //referencia en una variable
echo $miPastaGansa . "\n";                // el resultado es el mismo

// Cuanto dinero tiene Maria
echo $carteraDeMaria->dinero . "\n";

// Cuanto dinero tiene Maria ahora
$miPastaGansa = 5000; // Pero las modificaciones a través de esa referencia 
                  // almacenada en esta variable
echo $carteraDeMaria->dinero . "\n";  // No se propagan

I think there is still some fine detail of that reference returned by the function that is escaping me.
How come I have to use another & before assigning it? It would seem that then I have a reference to the reference to money. Instead of just a reference to money, which is what I want.

php

2 Answers

  1. Best Answer

    To get started.

    We must be clear about what a reference is in PHP.

    According to a small snippet in the official PHP documentation .

    PHP references allow you to make two variables refer to the same content.

    Let's look at this example:

    <?php

$variableA = "Hola mundo!";
$variableB = $variableA;

$variableA = "Hola amigos!";

print "A: " . $variableA . "\n";
print "B: " . $variableB . "\n";

?>

    The result we will get will be:

    A: Hola amigos!
B: Hola mundo!

    But now, let's add an ampersand (&) to our code and see what happens:

    <?php

$variableA = "Hola mundo!";
$variableB =& $variableA;       /* Agregamos el & */

$variableA = "Hola amigos!";

print "A: " . $variableA . "\n";
print "B: " . $variableB . "\n";

?>

    The result we will get will be:

    A: Hola amigos!
B: Hola amigos!

    What does this tell us?

    We have used an operator to access the same value of another variable and use it, we have modified it, but modifying it modified the value of the original variable, in this case $variableA.

    $variableB =& $variableA;

    This is known as Allocation by Reference .

    In that order of ideas

    We can see a simple example of a function using references :

    <?php

function caminar($X){
    $X++;
}

$pasos = 0;

caminar($pasos);
caminar($pasos);
caminar($pasos);

print "He caminado $pasos pasos\n";

?>

    The result of this program will be:

    He caminado 0 pasos

    And we use the ampersand operator (&) again and we will see how the magic happens.

    <?php

function caminar(&$X){             /* Agregamos el & */
    $X++;
}

$pasos = 0;

caminar($pasos);
caminar($pasos);
caminar($pasos);

print "He caminado $pasos pasos\n";

?>

    And the new result is:

    He caminado 3 pasos

    Like the previous example, what we have done here is to pass a reference to an external variable to the function, and through that reference we have modified it , increasing its value in each call to the function caminar().

    We know this as Pass by Reference .

    Finally

    The ampersand operator ( & ), in front of a function name, returns a reference , rather than a value.

    This is known as return referrals.

    What does this mean?

    Let's look at this example:

    <?php

$variable = "StackOverflow";

function &obtenerReferencia(){
   global $variable;   
   return $variable;
}

$puntero = &obtenerReferencia();
$puntero = "Stack Overflow Rules!";

print $variable;

?>

    We have initialized the value of $variablewith the text "StackOverflow", but when we run the program, we get the following output:

    Stack Overflow Rules!

    What has happened is that in the variable $punterowe have saved a reference to the memory location of the variable $variable, which contains our original text. Obtaining the reference to said variable makes any change we make to it affect the original variable .

    I have made a modification to your example, answering the question why it prints the same thing :

    <?php
class Cartera {
    public $dinero = 93;

    public function &muestraLaPasta() { // Si quito este ampersand
        return $this->dinero;
    }
}

$carteraDeMaria = new Cartera;
$miPastaGansa = &$carteraDeMaria->muestraLaPasta(); // y este
echo $miPastaGansa . "\n";                // el resultado es el mismo

// Cuanto dinero tiene Maria
echo $carteraDeMaria->dinero . "\n";

// Cuanto dinero tiene Maria ahora
$miPastaGansa = 5000;
echo $carteraDeMaria->dinero . "\n";

?>

    We get as a result:

    93
93
5000

    What happened here?

    When printing the return value of a reference function muestraLaPasta(), the value will always be left intact. What changes is the origin of that value, in that case, I have added this couple of lines:

    $miPastaGansa = 5000;
echo $carteraDeMaria->dinero . "\n";

    Modifying their reference $miPastaGansa, which in this case is a "pointer" to the attribute $dinero. Which will give us the value of 5000 later, when we get the value of the attribute $dinero.

    • 30

  2. Little can actually be added to the excellent answer given.

    I want to provide a visual explanation of what happens with references.

    As said, references in PHP allow you to use two variables to refer to the same content. For this, the ampersand sign ( &) is used, when I studied typing they called it: "and" commercial .

    References in PHP are not pointers , as they are in other languages, but would be like symbol table aliases.

    We begin...

    Caso 1:

    enter image description here

    Código:

    //Caso 1:

$x = "original"; //Creamos una variable $x con el valor "original"
$y = &$x;        //Creamos una referencia de $x llamada $y

//Resultado

echo "\nCaso 1:\n\n";
echo "Valor de x: ".$x."\n";
echo "Valor de y: ".$y."\n";

    Resultado:

    Caso 1:

Valor de x: original
Valor de y: original

    Caso 2:

    enter image description here

    Código:

    $x = "original";     //Creamos una variable $x con el valor "original"
$y = &$x;            //Creamos una referencia de $x llamada $y
$y = "NO original";  //Cambiamos el valor de $y, y en consecuencia, el de $x

echo "\n\nCaso 2:\n\n";
echo "Valor de x: ".$x."\n";
echo "Valor de y: ".$y."\n";

    Resultado:

    Caso 2:

Valor de x: NO original
Valor de y: NO original

    Caso 3:

    enter image description here

    Código:

    //Caso 3:

$x = "original";     //Creamos una variable $x con el valor "original"
$y = &$x;            //Creamos una referencia de $x llamada $y (aquí $y="original")
$z = "extraño";      //Creamos una variable $z con el valor "extraño"
$y = &$z;            //Hacemos que $z haga referencia a $y. Y $x recupera su valor "original"


//Prueba

echo "\n\nCaso 3:\n\n";

echo "Valor de x: ".$x."\n";
echo "Valor de x: ".$x."\n";
echo "Valor de z: ".$z."\n";

    Resultado:

    Caso 3:

Valor de x: original
Valor de y: extraño
Valor de z: extraño

    When the reference is deactivated, the link between the variable name and the variable content is simply broken. This does NOT mean that the content of the variable will be destroyed, therefore, it $xrecovers its original value when the reference is broken.

    Caso 4:

    If you want to break the reference, the correct way is to useunset

    enter image description here

    Código:

    //Caso 4:

$x = "original";     //Creamos una variable $x con el valor "original"
$y = &$x;            //Creamos una referencia de $x llamada $y (aquí $y="original")
unset ($y);          //Rompemos la referencia
//echo $y;           //Si hacemos esto tendríamos un aviso: PHP Notice:  Undefined variable: y

//Prueba

echo "\n\nCaso 4:\n\n";

echo "Valor de x: ".$x."\n";
echo (isset ($y) ? "y está definida\n": "y no está definida\n");

    Resultado:

    Caso 4:

Valor de x: original
y no está definida

    Caso 5:

    We will try to break the reference by assigning nullto the variable $xand then assigning it a new value. Let's see what happens:

    Código:

    //Caso 5:

echo "\nCaso 5:\n\n";

$x = "original";      //Creamos una variable $x con el valor "original"
$y = &$x;             //Creamos una referencia de $x llamada $y (aquí $y="original")
$x=null;              //Asignamos null a $x

echo "\nCon null la referencia se rompe 'temporalmente':\n";
echo (isset ($x) ? "x está definida\n": "x no está definida\n"); 
echo (isset ($y) ? "y está definida\n": "y no está definida\n"); 
echo "\n";


$x="OTRO original";   //Asignamos un nuevo valor a $x
echo "\nVeremos que y adquirirá el nuevo valor asignado a x:\n\n";

echo (isset ($x) ? "x está definida\n": "x no está definida\n");
echo "Valor de x: ".$x."\n";

echo (isset ($y) ? "y está definida\n": "y no está definida\n"); 
echo "Valor de y: ".$y."\n";

    Resultado:

    Caso 5:

Con null la referencia se rompe 'temporalmente':
x no está definida
y no está definida


Veremos que y adquirirá el nuevo valor asignado a x:

x está definida
Valor de x: OTRO original
y está definida
Valor de y: OTRO original

    The complete demo of the different cases can be found here .

    • 8