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    Void pointers in C/C++

What is a void pointer

In C is legal to declare a pointer to a void variable. A void* contains an address like a regular pointer, but the pointed memory area has no specific type. void pointers are declared like any other pointer

void *ptr;

Note that is NOT legal to have a void variable. The declaration

void a;

will give an error.

Assigning a void pointer

Since void* has no specific type, it can be used to point to the memory address of any variable type. So

void *ptr;
char *ch;
int *i;
double *dbl;
struct {
    int value;
} my_structure;
ptr = &ch;
ptr = &i;
ptr = &dbl;
ptr = &my_structure;

are all valid assignement.

Dereference a void pointer

But to be dereferenced and hence used, it must be cast to some existing type. Casting is necessary so the compiler knows how to use the pointer. For example

void *ptr;
*ptr = 1.234;          /* WRONG! */
*ptr = 'a';            /* WRONG! */
*(float*)ptr = 1.234;  /* cast to float* CORRECT! */
*(char*)ptr = 'a';     /* cast to char*  CORRECT! */

Uses of a void pointer

Generic structure

void* can be used to create a generic structure used to store any kind of data.

struct {
    void *data;
} generic_str;

Function input parameter

A function can take in a parameter of void* type

void function( int p1, void *p2 ){
    /* ... */

To use p2 we must cast it to the correct type inside the function. This way we can have functions having different behavior with different parameter type.
The function of course must know to which type cast. One way is to pass the type in another parameter. For example

This can be called with


Sometimes a function doesn't need to know the type of a variable to be able to work correctly. It's the case of a generic swap function. Such a function, just swapping the values of a and b, just need to know the size of the data type, but not the data type itself. Everything is cast to a char (which has size 1) and bytes of the input variables are taken and swapped one by one.

void swap( void *a, void *b, size_t size )
    unsigned char *aa = a;
    unsigned char *bb = b;
    unsigned char *tmp;
    int i;
    for ( i=0; i<size; ++i ){
        tmp = aa[i];
        aa[i] = bb[i];
        bb[i] = tmp;

This can be called with

double num1=1.43, num2=0.34;

Function output parameter

A function that return a void* can be defined as

void* function( int p )
    /* ... */

The return type of such a function can be cast to any type.
One example is malloc. malloc allocates memory and returns a void* pointer to it. The memory is still raw since it's allocated but nothing has been constructed in it.

char *string;
string = (char*)malloc(100);

Note that this code is very old style and nowadays is better to use the implicit casting of void pointers

char *string;
string = malloc(100);

Pointer arithmetic

To perform pointer arithmetic on a void* pointer it must be cast. This is because the compiler must know the size of the data which is pointed to to be able of jump of the correct amount of bytes. For example

void *ptr;
ptr = &var;
void *ptr_2 = ptr++;

This won't work since the size of void* is not known without a casting.
Note that GNU C (and hence the gcc compiler) allows pointer arithmetic with void* pointers assuming a size of 1, like a char*. However this a non portable feature and its use should be avoided.

Pointer to pointer to void

It is valid to declare pointers to pointers to void: void**. However a double pointer doesn't have the same property of a normal void*. A void* can point to any variable type, but a void** can point to a void* only. In other words it doesn't have the special meaning that a normal pointer to void has and it is treated like any other pointer.

Explicit casting of C++

In C++ it is forbidden to have an implicit casting when using a void* pointer. For example in C the following is allowed

int value;
void *ptr = &value;
int other = *ptr;

In C++ the following is valid instead

int value;
void *ptr = &value;
int other = *static_cast<int*>ptr;
int *other2 = static_cast<int*>ptr;
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