Print values of void pointer

Question

I have a funciton that returns a void pointer. Lets say I know that the block of data pointed at is an array of ints. How can I print them?

From another thread I saw that I cast the void as my desired data type this way:

printf("%i",*((int*)data));

But like I said that data is an array of ints. I tried to do this, but it's not a valid expression:

for(i = 0; i<3; i++){
    printf("%i \n", *((int*)(data+sizeof(int)*i)));
}

What is the proper way of printing this?

Answer 1

Just use a temporary int*:

int *p = data;
for (int i=0; i<3; i++)
    printf("%i\n", p[i]);

Note that there's no explicit cast in the first line, because that's not needed with void* and avoiding explicit casts is considered good C style. Inside the loop, I've used the fact that you can index into a pointer to an array as if it were the array itself.

Answer 2

Your original attempt is incorrect in "formal" C, since pointer arithmetic is not applicable to void * pointers. However, some compilers support pointer arithmetic on void * pointers as a non-standard extension (GCC being a notable example). In such compilers void * pointers are treated as char * pointers for arithmetic purposes. And your

for(i = 0; i < 3; i++)
    printf("%i\n", *((int *) (data + sizeof(int) * i)));

will actually work as intended in such compiler.

With a more pedantic compiler, you can save the day by doing

for(i = 0; i < 3; i++)
    printf("%i\n", *((int *) ((char *) data + sizeof(int) * i)));

But in any case, the above two pieces of code are hopelessly overcomplicated. They have some value as examples when you are just learning pointer arithmetic, but that's about all they are good for.

What you really need is just

for(i = 0; i < 3; i++)
    printf("%i\n", ((int *) data)[i]);

You can introduce an extra int * pointer, as other answers suggested, to make your code even more readable. But that's up to you.