Php – In PHP (>= 5.0), is passing by reference faster

pass-by-referenceperformancePHP

In PHP, function parameters can be passed by reference by prepending an ampersand to the parameter in the function declaration, like so:

function foo(&$bar)
{
    // ...
}

Now, I am aware that this is not designed to improve performance, but to allow functions to change variables that are normally out of their scope.

Instead, PHP seems to use Copy On Write to avoid copying objects (and maybe also arrays) until they are changed. So, for functions that do not change their parameters, the effect should be the same as if you had passed them by reference.

However, I was wondering if the Copy On Write logic maybe is shortcircuited on pass-by-reference and whether that has any performance impact.

ETA: To be sure, I assume that it's not faster, and I am well aware that this is not what references are for. So I think my own guesses are quite good, I'm just looking for an answer from someone who really knows what's definitely happening under the hood. In five years of PHP development, I've always found it hard to get quality information on PHP internals short from reading the source.

Best Answer

In a test with 100 000 iterations of calling a function with a string of 20 kB, the results are:

Function that just reads / uses the parameter

pass by value:      0.12065005 seconds
pass by reference:  1.52171397 seconds

Function to write / change the parameter

pass by value:      1.52223396 seconds
pass by reference:  1.52388787 seconds

Conclusions

Pass the parameter by value is always faster
If the function change the value of the variable passed, for practical purposes is the same as pass by reference than by value

Related Solutions

Java – Is Java “pass-by-reference” or “pass-by-value”

Java is always pass-by-value. Unfortunately, when we deal with objects we are really dealing with object-handles called references which are passed-by-value as well. This terminology and semantics easily confuse many beginners.

It goes like this:

public static void main(String[] args) {
    Dog aDog = new Dog("Max");
    Dog oldDog = aDog;

    // we pass the object to foo
    foo(aDog);
    // aDog variable is still pointing to the "Max" dog when foo(...) returns
    aDog.getName().equals("Max"); // true
    aDog.getName().equals("Fifi"); // false
    aDog == oldDog; // true
}

public static void foo(Dog d) {
    d.getName().equals("Max"); // true
    // change d inside of foo() to point to a new Dog instance "Fifi"
    d = new Dog("Fifi");
    d.getName().equals("Fifi"); // true
}

In the example above aDog.getName() will still return "Max". The value aDog within main is not changed in the function foo with the Dog "Fifi" as the object reference is passed by value. If it were passed by reference, then the aDog.getName() in main would return "Fifi" after the call to foo.

Likewise:

public static void main(String[] args) {
    Dog aDog = new Dog("Max");
    Dog oldDog = aDog;

    foo(aDog);
    // when foo(...) returns, the name of the dog has been changed to "Fifi"
    aDog.getName().equals("Fifi"); // true
    // but it is still the same dog:
    aDog == oldDog; // true
}

public static void foo(Dog d) {
    d.getName().equals("Max"); // true
    // this changes the name of d to be "Fifi"
    d.setName("Fifi");
}

In the above example, Fifi is the dog's name after call to foo(aDog) because the object's name was set inside of foo(...). Any operations that foo performs on d are such that, for all practical purposes, they are performed on aDog, but it is not possible to change the value of the variable aDog itself.

For more information on pass by reference and pass by value, consult the following SO answer: https://stackoverflow.com/a/430958/6005228. This explains more thoroughly the semantics and history behind the two and also explains why Java and many other modern languages appear to do both in certain cases.

Javascript – Is JavaScript a pass-by-reference or pass-by-value language

It's interesting in JavaScript. Consider this example:

function changeStuff(a, b, c)
{
  a = a * 10;
  b.item = "changed";
  c = {item: "changed"};
}

var num = 10;
var obj1 = {item: "unchanged"};
var obj2 = {item: "unchanged"};

changeStuff(num, obj1, obj2);

console.log(num);
console.log(obj1.item);
console.log(obj2.item);

This produces the output:

10
changed
unchanged

If obj1 was not a reference at all, then changing obj1.item would have no effect on the obj1 outside of the function.
If the argument was a proper reference, then everything would have changed. num would be 100, and obj2.item would read "changed". Instead, num stays 10 and obj2.item remains "unchanged".

Instead, the situation is that the item passed in is passed by value. But the item that is passed by value is itself a reference. Technically, this is called call-by-sharing.

In practical terms, this means that if you change the parameter itself (as with num and obj2), that won't affect the item that was fed into the parameter. But if you change the internals of the parameter, that will propagate back up (as with obj1).

Edited for clarification:

JavaScript is pass-by-value. Passing-by-reference and passing a reference are two different things. Pass-by refers to the mechanism for passing a parameter to a function. It's either the value of the variable (pass-by-value) or a pointer (reference) to the location of the variable (pass-by-reference). There's no mechanism within JavaScript to do the latter. It's not as complicated or nuanced as some of these answers try to make it.

Think of a variable as a box that contains information. When you pass it by value, you're just passing a new box with a copy of the information contained in the box. If you change that information within the new box in the function, the original value in the original box is unchanged.

Passing by reference is passing the actual box. If you change what's inside, it will be changed even after the function exits.

In the case of objects, think of the information contained within the box as a map to the locations of other boxes. You can change the contents of those boxes which will be permanent changes after the function exits, but you can't change the contents of the original map.

Assigning a new object to c just put a new map into the new box that is only local to the function. This is also what happened with a.