CODEWITHSUNDEEP
cpointersvoid
Nark
Nark

Reputation: 47

C - Memcpy two int's to a single Void pointer

I am trying to copy two ints into a void pointer, and be able to access them, so far no luck.

void *buffer;
buffer = malloc (sizeof(int) + sizeof(int))
memcpy (&buffer, &int1, sizeof(int));
memcpy (&buffer + sizeof(int), &int2, sizeof(int));
printf ("%d", *(int*)buffer);

When printing, all I get is the value in the first 4 bytes. Any help would be appreciated, thanks.

Upvotes: 0

Views: 2027

Answers (2)

selbie
selbie

Reputation: 104539

You are copying to pointer of the pointer.

Change these two lines:

memcpy (&buffer, &int1, sizeof(int));
memcpy (&buffer + sizeof(int), &int2, sizeof(int));

To be this:

memcpy (buffer, &int1, sizeof(int));
memcpy (buffer + sizeof(int), &int2, sizeof(int));

Better form:

void* buffer = malloc(sizeof(int) * 2);
int* intbuffer = (int*)buffer;

memcpy(intbuffer, &int1, sizeof(int));
memcpy(intbuffer+1, &int2, sizeof(int));

You can even avoid the memcpy altogether:

void* buffer = malloc(sizeof(int) * 2);
int* intbuffer = (int*)buffer;

intbuffer[0] = int1;
intbuffer[1] = int2;

The corresponding printf("%d", *(int*)buffer); works for all three examples above.

Upvotes: 4

Some programmer dude
Some programmer dude

Reputation: 409176

The major problem here is this part

memcpy (&buffer, ...);
//      ^

First of all, if you want an "array" or int then why don't you use a pointer to int? You can always cast it to a generic void * later. This will solve one problem, which is the pointer arithmetic you attempt to do in the second memcpy call.

Going back to my highlighted part, using the address-of operator for buffer is wrong and will pass a pointer to where the variable buffer is stored, not the address where buffer is pointing. Drop the address-of operator there.

To sum it up:

int *int_buffer = malloc(sizeof *int_buffer * 2);
memcpy(int_buffer, &int1, sizeof int1);     // or &int_buffer[0]
memcpy(int_buffer + 1, &int2, sizeof int2); // or &int_buffer[1]

void *buffer = int_buffer;

For your printing, you only tell printf to print the very first int element. If you want to print both int elements you need to tell printf to do that.

Upvotes: 1

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