Hi all,
I working in VC++. I have a void pointer. I am assigning a memory block to it using malloc().
void* buf_ptr = NULL;
buf_ptr = malloc(1428480);

I am filling this buffer using fread(). and I want to index this buffer (buf_ptr) to acces its data. How can I do that.
I am facing the following errors.

Error 24 error C2036: 'void *' : unknown size c:\AMT_DLL_tester.cpp
Error 25 error C2440: '=' : cannot convert from 'void' to 'void *' c:\AMT_DLL_tester.cpp

Here is my source code

FILE *f_in = NULL;
f_in = fopen(argv[f],"rb"
( f_in == NULL )
            fprintf(fptr,"ERROR: Could not open '%s'\n"
, argv[f] );

 buf_size = 1428480;
 *buf_ptr = NULL;

buf_ptr = malloc(buf_size+1);
items = fread(buf_ptr,buf_size,1,f_in);

* ptr = buf_ptr[20]; //----------->error

Please guide me how can I access the data of a void type buffer?

Your help will be highly appreciated..

Many Thanks.

Asif Javaid

7 Years
Discussion Span
Last Post by asifjavaid

>Please guide me how can I access the data of a void type buffer?

Normally [B]malloc[/B] is used in a way like this:

int *p;
p = (int *) malloc( 50 * sizeof( int ) );  // allocate space for 50 integers

As you can see in the example above/below, the void pointer which is returned by malloc is casted to an integer pointer

int *p;
p = [B](int *)[/B] malloc( 50 * sizeof( int ) );

If malloc couldn't allocate the memory, then it returns a NULL-pointer, but using a NULL-pointer will almost certainly crash your whole program, that's why it's generally recommended to check the return value of malloc before using the pointer where you've tried to assign memory to...

Checking whether the allocation has succeeded can be achieved by something like this:

int *p;
p = (int *) malloc( 50 * sizeof( int ) );  // allocate space for 50 integers

if( !p ) {
    // out of memory
    // the memory could not be allocated

BTW, Don't forget to free the allocated memory as well, otherwise you'll have memory leaks in your program :)

Hope this helps!

Votes + Comments
well put

In C you can assign void* type pointer value to the pointer of any type without casting. So simply declare a proper type pointer and set it to the buffer:

char* pchar = buf_ptr;
char* ptr = pchar + 20;
...pchar[20] the same as *ptr...

It's impossible to subscript void* pointer because sizeof(void) is not defined (type void has an empty set of values).

void* ptr = buf_ptr[20]; //----------->error

what i want to say is "buf_ptr[20]" is not a pointer,it's
the value of the 20th in buf_ptr

Votes + Comments
The actuall thread solver

Your beautiful post has nothing to do with OP's problem. The construct use for allocating space using malloc is absolutely valid in C.

The only problem is on line:
void* ptr = buf_ptr[20];
as Beair.GQ said, you are assigning a void to void*. The type of buf_ptr is not that of a pointer.
You should perhaps try this: void* ptr=but_ptr+20;//this is good


>You should perhaps try this:
>void* ptr=but_ptr+20;//this is good

Perhaps you should try it first. You have no excuse for giving an answer that will never compile on a conforming C compiler.

ArkM gave the correct answer (and corrected tux's misconception about casting malloc). Alternatively, you can cast the pointer to void to a pointer to an appropriate type, but regardless of how you do it, void* has to be converted to T* to do anything meaningful with it, where T is a non-void type.


int main()
    void* buf_ptr = NULL;
    buf_ptr = malloc(1428480);

        printf("%p \n",buf_ptr);
        void* p;
        printf("%p \n",p);

    return 0;
siddhant3s@Xion:~$ gcc testerc.c -o testerc.exe -ansi -pedantic  
testerc.c: In function ‘main’:
testerc.c:11: warning: ISO C90 forbids mixed declarations and code
testerc.c:12: warning: pointer of type ‘void *’ used in arithmetic
siddhant3s@Xion:~$ ./testerc.exe

You win....... The code compiles though. But you are correct.


>The code compiles though.
What you're seeing is a silent extension where the compiler treats the size of void as 1. GCC is wrong to allow it in pedantic mode as it's a constraint violation in the standard.

>warning: ISO C90 forbids mixed declarations and code
This should be an error as well.


Hi all, Thanks very much for alls participation.
Arkm, yes you are right I can access the contents of void* buf_ptr by doing the following.

void* buf_ptr = NULL;
buf_ptr = malloc(1428480);
strcpy((char*)buf_ptr,"Asif Javaid");
char *p = (char*)buf_ptr;

siddhant3s, your code can compile in gcc compiler but I am using Microsoft VC++9.0 comipler and it is generating an error of unknown size, as I do void*p = buf_ptr+1;

Many Thanks all.

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