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I need help. what is difference between a parameter passed by reference, and a parameter passed by value? Could you give me some examples please?

+8  A: 

It's a way how to pass arguments to functions. Passing by reference means the called functions' parameter will be the same as the callers' passed argument (not the value, but the identity - the variable itself). Pass by value means the called functions' parameter will be a copy of the callers' passed argument. The value will be the same, but the identity - the variable - is different. Thus changes to a parameter done by the called function in one case changes the argument passed and in the other case just changes the value of the parameter in the called function (which is only a copy). In a quick hurry:

  • Java only supports pass by value. Always copies arguments, even though when copying a reference to an object, the parameter in the called function will point to the same object and changes to that object will be see in the caller. Since this can be confusing, here is what Jon Skeet has to say about this.
  • C# supports pass by value and pass by reference (keyword ref used at caller and called function). Jon Skeet also has a nice explanation of this here.
  • C++ supports pass by value and pass by reference (reference parameter type used at called function). You will find an explanation of this below.

Codes

Since my language is C++, i will use that here

// passes a pointer (called reference in java) to an integer
void call_by_value(int *p) { // :1
    p = NULL;
}

// passes an integer
void call_by_value(int p) { // :2
    p = 42;
}

// passes an integer by reference
void call_by_reference(int & p) { // :3
    p = 42;
}

// this is the java style of passing references. NULL is called "null" there.
void call_by_value_special(int *p) { // :4
    *p = 10; // changes what p points to ("what p references" in java)
    // only changes the value of the parameter, but *not* of 
    // the argument passed by the caller. thus it's pass-by-value:
    p = NULL;
}

int main() {
    int value = 10;
    int * pointer = &value;

    call_by_value(pointer); // :1
    assert(pointer == &value); // pointer was copied

    call_by_value(value); // :2
    assert(value == 10); // value was copied

    call_by_reference(value); // :3
    assert(value == 42); // value was passed by reference

    call_by_value_special(pointer); // :4
    // pointer was copied but what pointer references was changed.
    assert(value == 10 && pointer == &value);
}

And an example in Java won't hurt:

class Example {
    int value = 0;

    // similar to :4 case in the c++ example
    static void accept_reference(Example e) { // :1
        e.value++; // will change the referenced object
        e = null; // will only change the parameter
    }

    // similar to the :2 case in the c++ example
    static void accept_primitive(int v) { // :2
        v++; // will only change the parameter
    }        

    public static void main(String... args) {
        int value = 0;
        Example ref = new Example(); // reference

        // note what we pass is the reference, not the object. we can't 
        // pass objects. The reference is copied (pass-by-value).
        accept_reference(ref); // :1
        assert ref != null && ref.value == 1;

        // the primitive int variable is copied
        accept_primitive(value); // :2
        assert value == 0;
    }
}

Wikipedia

http://en.wikipedia.org/wiki/Pass_by_reference#Call_by_value

http://en.wikipedia.org/wiki/Pass_by_reference#Call_by_reference

This guy pretty much nails it:

http://javadude.com/articles/passbyvalue.htm

Johannes Schaub - litb
why the downvote? if anything is wrong or leads to misunderstandings please leave a comment.
Johannes Schaub - litb
Wasn't my vote, but on a minor point (you know, the kind taken up in the presidential debates) I'd say it's more of a "tactic" than a "strategy."
harpo
+1 for Completeness. Don't worry about downvotes - people do it for weird reasons. In an opinion question about calculators everyone was downvoted by a guy who didn't think programmers should use calculators! Anyway, I thought your answer was very good.
Mark Brittingham
+3  A: 

When passing by ref you are basically passing a pointer to the variable. Pass by value you are passing a copy of the variable. In basic usage this normally means pass by ref changes to the variable will seen be the calling method and pass by value they wont.

Craig
+8  A: 

Here is an example:

#include <iostream>
using namespace std;

void by_val(int arg) { arg += 2; }
void by_ref(int&arg) { arg += 2; }

int main()
{
    int x = 0;
    by_val(x); cout << x << endl;  // prints 0
    by_ref(x); cout << x << endl;  // prints 2

    int y = 0;
    by_ref(y); cout << y << endl;  // prints 2
    by_val(y); cout << y << endl;  // prints 2
}
Eduardo León
+1 for using C++ to explain :)
Perpetualcoder
A: 

Examples:

class Dog 
{ 
public:

    barkAt( const std::string& pOtherDog ); // const reference

    barkAt( std::string pOtherDog ); // value
};

const & is generally best. You don't incur the construction and destruction penalty. If the reference isn't const your interface is suggesting that it will change the passed in data.

+2  A: 

Pass by value sends a COPY of the data stored in the variable you specify, pass by reference sends a direct link to the variable itself. So if you pass a variable by reference and then change the variable inside the block you passed it into, the original variable will be changed. If you simply pass by value, the original variable will not be able to be changed by the block you passed it into but you will get a copy of whatever it contained at the time of the call.

shogun
A: 

A major difference between them is that value-type variables store values, so specifying a value-type variable in a method call passes a copy of that variable's value to the method. Reference-type variables store references to objects, so specifying a reference-type variable as an argument passes the method a copy of the actual reference that refers to the object. Even though the reference itself is passed by value, the method can still use the reference it receives to interact with—and possibly modify—the original object. Similarly, when returning information from a method via a return statement, the method returns a copy of the value stored in a value-type variable or a copy of the reference stored in a reference-type variable. When a reference is returned, the calling method can use that reference to interact with the referenced object. So, in effect, objects are always passed by reference.

In c#, to pass a variable by reference so the called method can modify the variable's, C# provides keywords ref and out. Applying the ref keyword to a parameter declaration allows you to pass a variable to a method by reference—the called method will be able to modify the original variable in the caller. The ref keyword is used for variables that already have been initialized in the calling method. Normally, when a method call contains an uninitialized variable as an argument, the compiler generates an error. Preceding a parameter with keyword out creates an output parameter. This indicates to the compiler that the argument will be passed into the called method by reference and that the called method will assign a value to the original variable in the caller. If the method does not assign a value to the output parameter in every possible path of execution, the compiler generates an error. This also prevents the compiler from generating an error message for an uninitialized variable that is passed as an argument to a method. A method can return only one value to its caller via a return statement, but can return many values by specifying multiple output (ref and/or out) parameters.

see c# discussion and examples here link text

Tina Endresen
+3  A: 

Best explanation I ever heard of this.

Say I want to share a web page with you.

If I tell you the URL, I'm passing by reference. You can use that URL to see the same web page I can see. If that page is changed, we both see the changes. If you delete the URL, all you're doing is destroying your reference to that page - you're not deleting the actual page itself.

If I print out the page and give you the printout, I'm passing by value. Your page is a disconnected copy of the original. You won't see any subsequent changes, and any changes that you make (e.g. scribbling on your printout) will not show up on the original page. If you destroy the printout, you have actually destroyed your copy of the object - but the original web page remains intact.

Dylan Beattie