C编程中void指针的概念

Question

Is it possible to dereference a void pointer without type-casting in the C programming language?

Also, is there any way of generalizing a function which can receive a pointer and store it in a void pointer and by using that void pointer, can we make a generalized function?

for eg:

void abc(void *a, int b)
{
   if(b==1)
      printf("%d",*(int*)a);     // If integer pointer is received
   else if(b==2)
      printf("%c",*(char*)a);     // If character pointer is received
   else if(b==3)
      printf("%f",*(float*)a);     // If float pointer is received
}

I want to make this function generic without using if-else statements - is this possible?

Also if there are good internet articles which explain the concept of a void pointer, then it would be beneficial if you could provide the URLs.

Also, is pointer arithmetic with void pointers possible?

Answer 1

Is it possible to dereference the void pointer without type-casting in C programming language...

No, void indicates the absence of type, it is not something you can dereference or assign to.

is there is any way of generalizing a function which can receive pointer and store it in void pointer and by using that void pointer we can make a generalized function..

You cannot just dereference it in a portable way, as it may not be properly aligned. It may be an issue on some architectures like ARM, where pointer to a data type must be aligned at boundary of the size of data type (eg pointer to 32-bit integer must be aligned at 4-byte boundary to be dereferenced).

For example, reading uint16_t from void* :

/* may receive wrong value if ptr is not 2-byte aligned */
uint16_t value = *(uint16_t*)ptr;
/* portable way of reading a little-endian value */
uint16_t value = *(uint8_t*)ptr
                | ((*((uint8_t*)ptr+1))<<8);

Also, is pointer arithmetic with void pointers possible...

Pointer arithmetic is not possible on pointers of void due to lack of concrete value underneath the pointer and hence the size.

void* p = ...
void *p2 = p + 1; /* what exactly is the size of void?? */

Answer 2

In C, a void * can be converted to a pointer to an object of a different type without an explicit cast:

void abc(void *a, int b)
{
    int *test = a;
    /* ... */

This doesn't help with writing your function in a more generic way, though.

You can't dereference a void * with converting it to a different pointer type as dereferencing a pointer is obtaining the value of the pointed-to object. A naked void is not a valid type so derefencing a void * is not possible.

Pointer arithmetic is about changing pointer values by multiples of the sizeof the pointed-to objects. Again, because void is not a true type, sizeof(void) has no meaning so pointer arithmetic is not valid on void * . (Some implementations allow it, using the equivalent pointer arithmetic for char * .)

Answer 3

You should be aware that in C, unlike Java or C#, there is absolutely no possibility to successfully "guess" the type of object a void* pointer points at. Something similar to getClass() simply doesn't exist, since this information is nowhere to be found. For that reason, the kind of "generic" you are looking for always comes with explicit metainformation, like the int b in your example or the format string in the printf family of functions.

Answer 4

void 指针被称为泛型指针，它可以指向任何数据类型的变量。

Answer 5

So far my understating on void pointer is as follows.

When a pointer variable is declared using keyword void – it becomes a general purpose pointer variable. Address of any variable of any data type (char, int, float etc.)can be assigned to a void pointer variable.

main()
{
    int *p;

    void *vp;

    vp=p;
} 

Since other data type pointer can be assigned to void pointer, so I used it in absolut_value(code shown below) function. To make a general function.

I tried to write a simple C code which takes integer or float as a an argument and tries to make it +ve, if negative. I wrote the following code,

#include<stdio.h>

void absolute_value ( void *j) // works if used float, obviously it must work but thats not my interest here.
{
    if ( *j < 0 )
        *j = *j * (-1);

}

int main()
{
    int i = 40;
    float f = -40;
    printf("print intiger i = %d \n",i);
    printf("print float f = %f \n",f);
    absolute_value(&i);
    absolute_value(&f);
    printf("print intiger i = %d \n",i);
    printf("print float f = %f \n",f);
    return 0;
}   

---

But I was getting error, so I came to know my understanding with void pointer is not correct :(. So now I will move towards to collect points why is that so.

The things that i need to understand more on void pointers is that.

We need to typecast the void pointer variable to dereference it. This is because a void pointer has no data type associated with it. There is no way the compiler can know (or guess?) what type of data is pointed to by the void pointer. So to take the data pointed to by a void pointer we typecast it with the correct type of the data holded inside the void pointers location.

void main()

{

    int a=10;

    float b=35.75;

    void *ptr; // Declaring a void pointer

    ptr=&a; // Assigning address of integer to void pointer.

    printf("The value of integer variable is= %d",*( (int*) ptr) );// (int*)ptr - is used for type casting. Where as *((int*)ptr) dereferences the typecasted void pointer variable.

    ptr=&b; // Assigning address of float to void pointer.

    printf("The value of float variable is= %f",*( (float*) ptr) );

}

A void pointer can be really useful if the programmer is not sure about the data type of data inputted by the end user. In such a case the programmer can use a void pointer to point to the location of the unknown data type. The program can be set in such a way to ask the user to inform the type of data and type casting can be performed according to the information inputted by the user. A code snippet is given below.

void funct(void *a, int z)
{
    if(z==1)
    printf("%d",*(int*)a); // If user inputs 1, then he means the data is an integer and type casting is done accordingly.
    else if(z==2)
    printf("%c",*(char*)a); // Typecasting for character pointer.
    else if(z==3)
    printf("%f",*(float*)a); // Typecasting for float pointer
}

Another important point you should keep in mind about void pointers is that – pointer arithmetic can not be performed in a void pointer.

void *ptr;

int a;

ptr=&a;

ptr++; // This statement is invalid and will result in an error because 'ptr' is a void pointer variable.

So now I understood what was my mistake. I am correcting the same.

References :

http://www.antoarts.com/void-pointers-in-c/

http://www.circuitstoday.com/void-pointers-in-c .

The New code is as shown below.

---

#include<stdio.h>
#define INT 1
#define FLOAT 2

void absolute_value ( void *j, int *n)
{
    if ( *n == INT) {
        if ( *((int*)j) < 0 )
            *((int*)j) = *((int*)j) * (-1);
    }
    if ( *n == FLOAT ) {
        if ( *((float*)j) < 0 )
            *((float*)j) = *((float*)j) * (-1);
    }
}


int main()
{
    int i = 0,n=0;
    float f = 0;
    printf("Press 1 to enter integer or 2 got float then enter the value to get absolute value\n");
    scanf("%d",&n);
    printf("\n");
    if( n == 1) {
        scanf("%d",&i);
        printf("value entered before absolute function exec = %d \n",i);
        absolute_value(&i,&n);
        printf("value entered after absolute function exec = %d \n",i);
    }
    if( n == 2) {
        scanf("%f",&f);
        printf("value entered before absolute function exec = %f \n",f);
        absolute_value(&f,&n);
        printf("value entered after absolute function exec = %f \n",f);
    }
    else
    printf("unknown entry try again\n");
    return 0;
}   

---

Thank you,

Answer 6

No, it is not possible. What type should the dereferenced value have?

Answer 7

void abc(void *a, int b) {
  char *format[] = {"%d", "%c", "%f"};
  printf(format[b-1], a);
}

Answer 8

Here is a brief pointer on void pointers: https://www.learncpp.com/cpp-tutorial/613-void-pointers/

6.13 — Void pointers

Because the void pointer does not know what type of object it is pointing to, it cannot be dereferenced directly! Rather, the void pointer must first be explicitly cast to another pointer type before it is dereferenced.

If a void pointer doesn't know what it's pointing to, how do we know what to cast it to? Ultimately, that is up to you to keep track of.

Void pointer miscellany

It is not possible to do pointer arithmetic on a void pointer. This is because pointer arithmetic requires the pointer to know what size object it is pointing to, so it can increment or decrement the pointer appropriately.

Assuming the machine's memory is byte-addressable and does not require aligned accesses, the most generic and atomic (closest to the machine level representation) way of interpreting a void* is as a pointer-to-a-byte, uint8_t* . Casting a void* to a uint8_t* would allow you to, for example, print out the first 1/2/4/8/however-many-you-desire bytes starting at that address, but you can't do much else.

uint8_t* byte_p = (uint8_t*)p;
for (uint8_t* i = byte_p; i < byte_p + 8; i++) {
  printf("%x ",*i);
}

Answer 9

I want to make this function generic, without using ifs; is it possible?

The only simple way I see is to use overloading .. which is not available in C programming langage AFAIK.

Did you consider the C++ programming langage for your programm ? Or is there any constraint that forbids its use?

Answer 10

You can easily print a void printer

int p=15;
void *q;
q=&p;
printf("%d",*((int*)q));

Answer 11

Because C is statically-typed, strongly-typed language, you must decide type of variable before compile. When you try to emulate generics in C, you'll end up attempt to rewrite C++ again, so it would be better to use C++ instead.

Answer 12

void pointer is a generic pointer.. Address of any datatype of any variable can be assigned to a void pointer.

int a = 10;
float b = 3.14;
void *ptr;
ptr = &a;
printf( "data is %d " , *((int *)ptr)); 
//(int *)ptr used for typecasting dereferencing as int
ptr = &b;
printf( "data is %f " , *((float *)ptr));
//(float *)ptr used for typecasting dereferencing as float

Answer 13

您不能在不指定类型的情况下取消引用指针，因为不同的数据类型在内存中将具有不同的大小，即 int 为 4 个字节，char 为 1 个字节。

Answer 14

Void pointers are pointers that has no data type associated with it.A void pointer can hold address of any type and can be typcasted to any type. But, void pointer cannot be directly be dereferenced.

int x = 1;
void *p1;
p1 = &x;
cout << *p1 << endl; // this will give error
cout << (int *)(*p) << endl; // this is valid

Answer 15

Fundamentally, in C, "types" are a way to interpret bytes in memory. For example, what the following code

struct Point {
  int x;
  int y;
};

int main() {
  struct Point p;
  p.x = 0;
  p.y = 0;
}

Says "When I run main, I want to allocate 4 (size of integer) + 4 (size of integer) = 8 (total bytes) of memory. When I write '.x' as a lvalue on a value with the type label Point at compile time, retrieve data from the pointer's memory location plus four bytes. Give the return value the compile-time label "int.""

Inside the computer at runtime, your "Point" structure looks like this:

00000000 00000000 00000000 00000000 00000000 00000000 00000000

And here's what your void* data type might look like: (assuming a 32-bit computer)

10001010 11111001 00010010 11000101

Answer 16

这行不通，但 void * 可以在定义指向函数的泛型指针并将其作为参数传递给另一个函数（类似于 Java 中的回调）或将其定义为类似于 oop 的结构方面有很大帮助。