70
votes

I read different things on the Internet and got confused, because every website says different things.

I read about * referencing operator and & dereferencing operator; or that referencing means making a pointer point to a variable and dereferencing is accessing the value of the variable that the pointer points to. So I got confused.

Can I get a simple but thorough explanation about "referencing and dereferencing"?

7
Note that the official names are address (&) and indirection (*) operator.Daniel Fischer
You have the operators mixed up. * is the derefencing operator.cheznead

7 Answers

107
votes

Referencing means taking the address of an existing variable (using &) to set a pointer variable. In order to be valid, a pointer has to be set to the address of a variable of the same type as the pointer, without the asterisk:

int  c1;
int* p1;
c1 = 5;
p1 = &c1;
//p1 references c1

Dereferencing a pointer means using the * operator (asterisk character) to retrieve the value from the memory address that is pointed by the pointer: NOTE: The value stored at the address of the pointer must be a value OF THE SAME TYPE as the type of variable the pointer "points" to, but there is no guarantee this is the case unless the pointer was set correctly. The type of variable the pointer points to is the type less the outermost asterisk.

int n1;
n1 = *p1;

Invalid dereferencing may or may not cause crashes:

  • Dereferencing an uninitialized pointer can cause a crash
  • Dereferencing with an invalid type cast will have the potential to cause a crash.
  • Dereferencing a pointer to a variable that was dynamically allocated and was subsequently de-allocated can cause a crash
  • Dereferencing a pointer to a variable that has since gone out of scope can also cause a crash.

Invalid referencing is more likely to cause compiler errors than crashes, but it's not a good idea to rely on the compiler for this.

References:

http://www.codingunit.com/cplusplus-tutorial-pointers-reference-and-dereference-operators

& is the reference operator and can be read as “address of”.
* is the dereference operator and can be read as “value pointed by”.

http://www.cplusplus.com/doc/tutorial/pointers/

& is the reference operator    
* is the dereference operator

http://en.wikipedia.org/wiki/Dereference_operator

The dereference operator * is also called the indirection operator.
21
votes

I've always heard them used in the opposite sense:

  • & is the reference operator -- it gives you a reference (pointer) to some object

  • * is the dereference operator -- it takes a reference (pointer) and gives you back the referred to object;

13
votes

For a start, you have them backwards: & is reference and * is dereference.

Referencing a variable means accessing the memory address of the variable:

int i = 5;
int * p;
p = &i; //&i returns the memory address of the variable i.

Dereferencing a variable means accessing the variable stored at a memory address:

int i = 5;
int * p;
p = &i;
*p = 7; //*p returns the variable stored at the memory address stored in p, which is i.
//i is now 7
9
votes

find the below explanation:

int main()
{
    int a = 10;// say address of 'a' is 2000;
    int *p = &a; //it means 'p' is pointing[referencing] to 'a'. i.e p->2000
    int c = *p; //*p means dereferncing. it will give the content of the address pointed by 'p'. in this case 'p' is pointing to 2000[address of 'a' variable], content of 2000 is 10. so *p will give 10. 
}

conclusion :

  1. & [address operator] is used for referencing.
  2. * [star operator] is used for de-referencing .
5
votes

The context that * is in, confuses the meaning sometimes.

  // when declaring a function
int function(int*); // This function is being declared as a function that takes in an 'address' that holds a number (so int*), it's asking for a 'reference', interchangeably called 'address'. When I 'call'(use) this function later, I better give it a variable-address! So instead of var, or q, or w, or p, I give it the address of var so &var, or &q, or &w, or &p.   

//even though the symbol ' * ' is typically used to mean 'dereferenced variable'(meaning: to use the value at the address of a variable)--despite it's common use, in this case, the symbol means a 'reference', again, in THIS context. (context here being the declaration of a 'prototype'.) 


    //when calling a function
int main(){ 
    function(&var);  // we are giving the function a 'reference', we are giving it an 'address'
  }

So, in the context of declaring a type such as int or char, we would use the dereferencer ' * ' to actually mean the reference (the address), which makes it confusing if you see an error message from the compiler saying: 'expecting char*' which is asking for an address.

In that case, when the * is after a type (int, char, etc.) the compiler is expecting a variable's address. We give it this by using a reference operator, alos called the address-of operator ' & ' before a variable. Even further, in the case I just made up above, the compiler is expecting the address to hold a character value, not a number. (type char * == address of a value that has a character)

int* p;
int *a;   // both are 'pointer' declarations. We are telling the compiler that we will soon give these variables an address (with &).

int c = 10;  //declare and initialize a random variable
//assign the variable to a pointer, we do this so that we can modify the value of c from a different function regardless of the scope of that function (elaboration in a second)

p = c; //ERROR, we assigned a 'value' to this 'pointer'. We need to assign an 'address', a 'reference'.
p = &c; // instead of a value such as: 'q',5,'t', or 2.1 we gave the pointer an 'address', which we could actually print with printf(), and would be something like
//so
p = 0xab33d111; //the address of c, (not specifically this value for the address, it'll look like this though, with the 0x in the beggining, the computer treats these different from regular numbers)
*p = 10; // the value of c

a = &c; // I can still give c another pointer, even though it already has the pointer variable "p"

*a = 10;
 a = 0xab33d111;

Think of each variable as having a position (or an index value if you are familiar with arrays) and a value. It might take some getting used-to to think of each variable having two values to it, one value being it's position, physically stored with electricity in your computer, and a value representing whatever quantity or letter(s) the programmer wants to store.

//Why it's used
int function(b){
    b = b + 1; // we just want to add one to any variable that this function operates on.
} 

int main(){

    int c = 1;  // I want this variable to be 3.

    function(c); 
    function(c);// I call the function I made above twice, because I want c to be 3.

     // this will return c as 1. Even though I called it twice.
     // when you call a function it makes a copy of the variable.
     // so the function that I call "function", made a copy of c, and that function is only changing the "copy" of c, so it doesn't affect the original
}
  //let's redo this whole thing, and use pointers

int function(int* b){ // this time, the function is 'asking' (won't run without) for a variable that 'points' to a number-value (int). So it wants an integer pointer--an address that holds a number.
*b = *b + 1; //grab the value of the address, and add one to the value stored at that address
}

int main(){
    int c = 1; //again, I want this to be three at the end of the program
    int *p = &c; // on the left, I'm declaring a pointer, I'm telling the compiler that I'm about to have this letter point to an certain spot in my computer. Immediately after I used the assignment operator (the ' = ') to assign the address of c to this variable (pointer in this case) p. I do this using the address-of operator (referencer)' & '.
    function(p); // not *p, because that will dereference. which would give an integer, not an integer pointer ( function wants a reference to an int called int*, we aren't going to use *p because that will give the function an int instead of an address that stores an int.

    function(&c); // this is giving the same thing as above, p = the address of c, so we can pass the 'pointer' or we can pass the 'address' that the pointer(variable) is 'pointing','referencing' to. Which is &c. 0xaabbcc1122...


      //now, the function is making a copy of c's address, but it doesn't matter if it's a copy or not, because it's going to point the computer to the exact same spot (hence, The Address), and it will be changed for main's version of c as well.

}

Inside each and every block, it copies the variables (if any) that are passed into (via parameters within "()"s). Within those blocks, the changes to a variable are made to a copy of that variable, the variable uses the same letters but is at a different address (from the original). By using the address "reference" of the original, we can change a variable using a block outside of main, or inside a child of main.

3
votes

Referencing

& is the reference operator. It will refer the memory address to the pointer variable.

Example:

int *p;
int a=5;
p=&a; // Here Pointer variable p refers to the address of integer variable a.

Dereferencing

Dereference operator * is used by the pointer variable to directly access the value of the variable instead of its memory address.

Example:

int *p;
int a=5;
p=&a;
int value=*p; // Value variable will get the value of variable a that pointer variable p pointing to.
0
votes

Reference of the de-referenced pointer is also same as the address of the pointed variable.

Explanation :-

int var = 3; int *p;

p = &var;

so, let's think address of var is : ABCDE

then,

p = ABCDE and &*p = ABCDE;

that means put &* together ,neutral the referencing and de-referencing.


also when declaring a function ,

the function's arguments should be the pointers,

and in the arguments of the this function when calling it in main method are should been with & operator.

it's bit confusing. But remember that int *p = &var; is also correct as the above pointer declaration.