What Does "Dereferencing" a Pointer Mean

What does dereferencing a pointer mean?

Reviewing the basic terminology

It's usually good enough - unless you're programming assembly - to envisage a pointer containing a numeric memory address, with 1 referring to the second byte in the process's memory, 2 the third, 3 the fourth and so on....

What happened to 0 and the first byte? Well, we'll get to that later - see null pointers below.
For a more accurate definition of what pointers store, and how memory and addresses relate, see "More about memory addresses, and why you probably don't need to know" at the end of this answer.

When you want to access the data/value in the memory that the pointer points to - the contents of the address with that numerical index - then you dereference the pointer.

Different computer languages have different notations to tell the compiler or interpreter that you're now interested in the pointed-to object's (current) value - I focus below on C and C++.

A pointer scenario

Consider in C, given a pointer such as p below...

const char* p = "abc";

...four bytes with the numerical values used to encode the letters 'a', 'b', 'c', and a 0 byte to denote the end of the textual data, are stored somewhere in memory and the numerical address of that data is stored in p. This way C encodes text in memory is known as ASCIIZ.

For example, if the string literal happened to be at address 0x1000 and p a 32-bit pointer at 0x2000, the memory content would be:

Memory Address (hex)    Variable name    Contents
1000                                     'a' == 97 (ASCII)
1001                                     'b' == 98
1002                                     'c' == 99
1003                                     0
...
2000-2003               p                1000 hex

Note that there is no variable name/identifier for address 0x1000, but we can indirectly refer to the string literal using a pointer storing its address: p.

Dereferencing the pointer

To refer to the characters p points to, we dereference p using one of these notations (again, for C):

assert(*p == 'a');  // The first character at address p will be 'a'
assert(p[1] == 'b'); // p[1] actually dereferences a pointer created by adding
                     // p and 1 times the size of the things to which p points:
                     // In this case they're char which are 1 byte in C...
assert(*(p + 1) == 'b');  // Another notation for p[1]

You can also move pointers through the pointed-to data, dereferencing them as you go:

++p;  // Increment p so it's now 0x1001
assert(*p == 'b');  // p == 0x1001 which is where the 'b' is...

If you have some data that can be written to, then you can do things like this:

int x = 2;
int* p_x = &x;  // Put the address of the x variable into the pointer p_x
*p_x = 4;       // Change the memory at the address in p_x to be 4
assert(x == 4); // Check x is now 4

Above, you must have known at compile time that you would need a variable called x, and the code asks the compiler to arrange where it should be stored, ensuring the address will be available via &x.

Dereferencing and accessing a structure data member

In C, if you have a variable that is a pointer to a structure with data members, you can access those members using the -> dereferencing operator:

typedef struct X { int i_; double d_; } X;
X x;
X* p = &x;
p->d_ = 3.14159;  // Dereference and access data member x.d_
(*p).d_ *= -1;    // Another equivalent notation for accessing x.d_

Multi-byte data types

To use a pointer, a computer program also needs some insight into the type of data that is being pointed at - if that data type needs more than one byte to represent, then the pointer normally points to the lowest-numbered byte in the data.

So, looking at a slightly more complex example:

double sizes[] = { 10.3, 13.4, 11.2, 19.4 };
double* p = sizes;
assert(p[0] == 10.3);  // Knows to look at all the bytes in the first double value
assert(p[1] == 13.4);  // Actually looks at bytes from address p + 1 * sizeof(double)
                       // (sizeof(double) is almost always eight bytes)
++p;                   // Advance p by sizeof(double)
assert(*p == 13.4);    // The double at memory beginning at address p has value 13.4
*(p + 2) = 29.8;       // Change sizes[3] from 19.4 to 29.8
                       // Note earlier ++p and + 2 here => sizes[3]

Pointers to dynamically allocated memory

Sometimes you don't know how much memory you'll need until your program is running and sees what data is thrown at it... then you can dynamically allocate memory using malloc. It is common practice to store the address in a pointer...

int* p = (int*)malloc(sizeof(int)); // Get some memory somewhere...
*p = 10;            // Dereference the pointer to the memory, then write a value in
fn(*p);             // Call a function, passing it the value at address p
(*p) += 3;          // Change the value, adding 3 to it
free(p);            // Release the memory back to the heap allocation library

In C++, memory allocation is normally done with the new operator, and deallocation with delete:

int* p = new int(10); // Memory for one int with initial value 10
delete p;

p = new int[10];      // Memory for ten ints with unspecified initial value
delete[] p;

p = new int[10]();    // Memory for ten ints that are value initialised (to 0)
delete[] p;

Losing and leaking addresses

Often a pointer may be the only indication of where some data or buffer exists in memory. If ongoing use of that data/buffer is needed, or the ability to call free() or delete to avoid leaking the memory, then the programmer must operate on a copy of the pointer...

const char* p = asprintf("name: %s", name);  // Common but non-Standard printf-on-heap

// Replace non-printable characters with underscores....
for (const char* q = p; *q; ++q)
    if (!isprint(*q))
        *q = '_';

printf("%s\n", p); // Only q was modified
free(p);

...or carefully orchestrate reversal of any changes...

const size_t n = ...;
p += n;
...
p -= n;  // Restore earlier value...
free(p);

C++ smart pointers

In C++, it's best practice to use smart pointer objects to store and manage the pointers, automatically deallocating them when the smart pointers' destructors run. Since C++11 the Standard Library provides two, unique_ptr for when there's a single owner for an allocated object...

{
    std::unique_ptr<T> p{new T(42, "meaning")};
    call_a_function(p);
    // The function above might throw, so delete here is unreliable, but...
} // p's destructor's guaranteed to run "here", calling delete

...and shared_ptr for share ownership (using reference counting)...

{
    auto p = std::make_shared<T>(3.14, "pi");
    number_storage1.may_add(p); // Might copy p into its container
    number_storage2.may_add(p); // Might copy p into its container    } // p's destructor will only delete the T if neither may_add copied it

Null pointers

In C, NULL and 0 - and additionally in C++ nullptr - can be used to indicate that a pointer doesn't currently hold the memory address of a variable, and shouldn't be dereferenced or used in pointer arithmetic. For example:

const char* p_filename = NULL; // Or "= 0", or "= nullptr" in C++
int c;
while ((c = getopt(argc, argv, "f:")) != -1)
    switch (c) {
      case f: p_filename = optarg; break;
    }
if (p_filename)  // Only NULL converts to false
    ...   // Only get here if -f flag specified

In C and C++, just as inbuilt numeric types don't necessarily default to 0, nor bools to false, pointers are not always set to NULL. All these are set to 0/false/NULL when they're static variables or (C++ only) direct or indirect member variables of static objects or their bases, or undergo zero initialisation (e.g. new T(); and new T(x, y, z); perform zero-initialisation on T's members including pointers, whereas new T; does not).

Further, when you assign 0, NULL and nullptr to a pointer the bits in the pointer are not necessarily all reset: the pointer may not contain "0" at the hardware level, or refer to address 0 in your virtual address space. The compiler is allowed to store something else there if it has reason to, but whatever it does - if you come along and compare the pointer to 0, NULL, nullptr or another pointer that was assigned any of those, the comparison must work as expected. So, below the source code at the compiler level, "NULL" is potentially a bit "magical" in the C and C++ languages...

More about memory addresses, and why you probably don't need to know

More strictly, initialised pointers store a bit-pattern identifying either NULL or a (often virtual) memory address.

The simple case is where this is a numeric offset into the process's entire virtual address space; in more complex cases the pointer may be relative to some specific memory area, which the CPU may select based on CPU "segment" registers or some manner of segment id encoded in the bit-pattern, and/or looking in different places depending on the machine code instructions using the address.

For example, an int* properly initialised to point to an int variable might - after casting to a float* - access memory in "GPU" memory quite distinct from the memory where the int variable is, then once cast to and used as a function pointer it might point into further distinct memory holding machine opcodes for the program (with the numeric value of the int* effectively a random, invalid pointer within these other memory regions).

3GL programming languages like C and C++ tend to hide this complexity, such that:

If the compiler gives you a pointer to a variable or function, you can dereference it freely (as long as the variable's not destructed/deallocated meanwhile) and it's the compiler's problem whether e.g. a particular CPU segment register needs to be restored beforehand, or a distinct machine code instruction used
If you get a pointer to an element in an array, you can use pointer arithmetic to move anywhere else in the array, or even to form an address one-past-the-end of the array that's legal to compare with other pointers to elements in the array (or that have similarly been moved by pointer arithmetic to the same one-past-the-end value); again in C and C++, it's up to the compiler to ensure this "just works"
Specific OS functions, e.g. shared memory mapping, may give you pointers, and they'll "just work" within the range of addresses that makes sense for them
Attempts to move legal pointers beyond these boundaries, or to cast arbitrary numbers to pointers, or use pointers cast to unrelated types, typically have undefined behaviour, so should be avoided in higher level libraries and applications, but code for OSes, device drivers, etc. may need to rely on behaviour left undefined by the C or C++ Standard, that is nevertheless well defined by their specific implementation or hardware.

Why is dereferencing a pointer called dereferencing?

A pointer refers to an object. Ergo, we dereference the pointer (or, get the referent of the pointer) to get the object pointed-to.

The de- prefix most likely comes from the Latin preposition meaning from; I suppose you could think of dereference as meaning "to obtain the referent (or object) from the reference."

Meaning of referencing and dereferencing in C

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.

What is the difference between derefencing and assigning the address of a variable to pointer variable in C?

In your first piece of code:

int main() {
    int a = 12;
    int *p;
    *p = a;
}

you have a serious case of undefined behaviour because, what you are trying to do is assign the value of a to the int variable that p currently points to. However, p has not been assigned an 'address', so it will have an arbitrary - and invalid - value! Some compilers may initialise p to zero (or NULL) but that is still an invalid address (on most systems).

Your second code snippet is 'sound' but, as it stands, doesn't actually achieve anything:

int main() {
    int a = 12;
    int *p;
    p = &a;
}

Here, you are assigning a value (i.e. an address) to your pointer variable, p; in this case, p now points to the a variable (that is, it's value is the address of a).

So, if you appended code like this (to the end of your second snippet):

*p = 42;

and then printed out the value of a, you would see that its value has been changed from the initially-given 12 to 42.

Feel free to ask for further clarification and/or explanation.

What does the value from dereferencing a function pointer means

In my concept of understanding the func, it would be a pointer to an int passed by value.

func is a pointer to function, which takes an int and returns int.

But in this case I'm passing a lambda which doesn't seem to be called anywhere.

You're passing a lambda without capturing, which could convert to pointer to function implicitly. In PrintTheValue, *func, i.e dereference on the pointer results in a reference to function, and for being passed to operator<< of std::cout, it converts to function pointer again, then converts to bool with value true (as a non-null pointer), then you should get the result 1 (or true with the usage of std::boolalpha). If you want to call on func you could func(42) (or (*func)(42)).

How does dereferencing the pointer is changing the value in this code snippet?

You're not only dereferencing the pointer, you're updating the value as well.

 ptr[2] = 0x05;

updates the value at 3rd octet. So, for a starting value of

 (0x) 04 03 02 01

it gets modified like

 (0x) 04 05 02 01    
         ^--------- ptr[2]

To add, from C11, chapter 6.3.2.3,

[...] When a pointer to an object is converted to a pointer to a character type, the result points to the lowest addressed byte of the object. Successive increments of the result, up to the size of the object, yield pointers to the remaining bytes of the object.

So, in a little-endian system, it looks like

ptr[0] = 01
ptr[1] = 02
ptr[2] = 03
ptr[3] = 04

and you're modifying ptr[2], so the value change is reflected in result.

How does dereferencing of a function pointer happen?

It's not quite the right question. For C, at least, the right question is

What happens to a function value in an rvalue context?

(An rvalue context is anywhere a name or other reference appears where it should be used as a value, rather than a location — basically anywhere except on the left-hand side of an assignment. The name itself comes from the right-hand side of an assignment.)

OK, so what happens to a function value in an rvalue context? It is immediately and implicitly converted to a pointer to the original function value. If you dereference that pointer with *, you get the same function value back again, which is immediately and implicitly converted into a pointer. And you can do this as many times as you like.

Two similar experiments you can try:

What happens if you dereference a function pointer in an lvalue context—the left-hand side of an assignment. (The answer will be about what you expect, if you keep in mind that functions are immutable.)
An array value is also converted to a pointer in an lvalue context, but it is converted to a pointer to the element type, not to a pointer to the array. Dereferencing it will therefore give you an element, not an array, and the madness you show doesn't occur.

Hope this helps.

P.S. As to why a function value is implicitly converted to a pointer, the answer is that for those of us who use function pointers, it's a great convenience not to have to use &'s everywhere. There's a dual convenience as well: a function pointer in call position is automatically converted to a function value, so you don't have to write * to call through a function pointer.

P.P.S. Unlike C functions, C++ functions can be overloaded, and I'm not qualified to comment on how the semantics works in C++.

When to use dereference operator for pointers C++ [duplicate]

A pointer is a variable whose value is the address of another variable or object. We say that the pointer "points to" that other object.

When writing an expression, cp means the pointer variable. *cp means the variable or object whose address the pointer variable holds.

Try to keep clear in your mind the distinction between "the pointer" and "the thing being pointed to". These are two distinct things, each with their own lifetimes and storage requirements.

So to address the code in your question, if (cp) is testing the pointer. It's short for if ( cp != nullptr ), i.e. has the pointer been set to point somewhere? (In other words, does the pointer currently hold the address of another object?)

if (*cp) means if (*cp != 0) , it is asking about the value of the object whose address is stored in the pointer (in other words, the value of the object being pointed to).

When does dereferencing a pointer cause a copy to be made in c++?

Dereferencing alone never makes a copy. Though dereferencing a pointer is just applying the * operator and there is more happening in your code.

Here:

int num = 10;
int *ptr = #
int& num2 = *ptr;

the last line dereferences the pointer and uses the result to initialize the reference num2. num2 now references num. No copies.

Here:

int num = 10;
int *ptr = #
int num2 = *ptr;

in the last line you first derference ptr then use the result to initialize num2. As num2 is a value a copy is made.

You do not need pointers to see the same effect:

int x = 42;
int& y = x;   // no copy, y is an alias for x
int z = x;    // z is a copy of x

What Does "Dereferencing" a Pointer Mean