We have learned in chapter Pointer Basics in C that if a pointer is of type pointer to int or (int *) then it can hold the address of the variable of type int only. It would be incorrect, if we assign an address of a float variable to a pointer of type pointer to int. But void pointer is an exception to this rule. A void pointer can point to a variable of any data type. Here is the syntax of void pointer.
Syntax: void *vp;
Let’s take an example:
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void *vp; int a = 100, *ip; float f = 12.2, *fp; char ch = 'a'; |
Here vp is a void pointer, so you can assign the address of any type of variable to it.
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vp = &a; // ok vp = ip; // ok vp = fp; // ok ip = &f; // wrong since type of ip is pointer to int fp = ip; // wrong since type of fp is pointer to float |
A void pointer can point to a variable of any data type and void pointer can be assigned to a pointer of any type.
Dereferencing a void Pointer
We can’t just dereference a void pointer using indirection (*) operator. For example:
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void *vp; int a = 100; vp = &a; printf("%d", *vp); // wrong |
It simply doesn’t work that way!. Before you dereference a void pointer it must be typecasted to appropriate pointer type. Let me show you what I mean.
For example: In the above snippet void pointer vp is pointing to the address of integer variable a. So in this case vp is acting as a pointer to int or (int *). Hence the proper typecast in this case is (int*).
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(int *)vptr |
Now the type of vptr temporarily changes from void pointer to pointer to int or (int*) , and we already know how to dereference a pointer to int, just precede it with indirection operator (*)
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*(int *)vptr |
Note: typecasting changes type of vp temporarily until the evaluation of the expression, everywhere else in the program vp is still a void pointer.
The following program demonstrates how to dereference a void pointer.
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#include<stdio.h> #define SIZE 10 int main() { int i = 10; float f = 2.34; char ch = 'k'; void *vptr; vptr = &i; printf("Value of i = %d\n", *(int *)vptr); vptr = &f; printf("Value of f = %.2f\n", *(float *)vptr); vptr = &ch; printf("Value of ch = %c\n", *(char *)vptr); // signal to operating system program ran fine return 0; } |
Expected Output:
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Value of i = 10 Value of f = 2.34 Value of ch = k |
Pointer Arithmetic in Void Pointers
Another important point I want to mention is about pointer arithmetic with void pointer. Before you apply pointer arithmetic in void pointers make sure to provide a proper typecast first otherwise you may get unexcepted results.
Consider the following example:
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int one_d[5] = {12, 19, 25, 34, 46}, i; void *vp = one_d; printf("%d", one_d + 1); // wrong |
Here we have assigned the name of the array one_d to the void pointer vp. Since the base type of one_d is a pointer to int or (int*), the void pointer vp is acting like a pointer to int or (int*). So the proper typecast is (int*).
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int one_d[5] = {12, 19, 25, 34, 46}, i; void *vp = one_d; printf("%d", (int *)one_d + 1); // correct |
The following program demonstrates pointer arithmetic in void pointers.
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#include<stdio.h> #define SIZE 10 int main() { int one_d[5] = {12, 19, 25, 34, 46}, i; void *vp = one_d; for(i = 0; i < 5; i++) { printf("one_d[%d] = %d\n", i, *( (int *)vp + i ) ); } // signal to operating system program ran fine return 0; } |
Expected Output:
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one_d[0] = 12 one_d[1] = 19 one_d[2] = 25 one_d[3] = 34 one_d[4] = 46 |
The void pointers are used extensively in dynamic memory allocation which we will discuss next.
a few typo in snippets under “Pointer Arithmetic in Void pointers”:
int one_d[5] = {12, 19, 25, 34, 46}, i;
void *vp = one_d;
printf(“%d”, arr + 1); // wrong => printf(“%d”, one_d+ 1);
Same issue, for the next snippet.
Thanks for posting! 🙂
Nice article on the void pointer.
can i use an array of void pointer? Can i assign the value of a void pointer at an another void pointer? Thanks
very nice
I have searched a lot about dereferencing concept but at last, I got ur article in which my all doubts are clear. Thank you so much to share this useful information.