C-将真正的int值分配给void指针，而不是地址/指针值？

Question

I'm having to deal with a generalised linked list so I cannot just turn the data format from a void * to an int .

Miniscule version:

int main(int argc, char **argv) {
    void *data;
    int i = 100;
    *((int*)data) = i;
    printf("%d", (int)data);
    return 0;
}

Keeps printing an address. Likewise, data = &i also returns an address. data = i just results in an error. I've tried all that I can and just cannot insert the value i into data .

Answer 1

You need to make data point to a valid memory address first:

int main(int argc, char **argv) {
    void *data = malloc(sizeof (int));
    int i = 100;
    *((int*)data) = i;
    printf("%d", *((int*)data));
    return 0;
}

Without that malloc() , double undefined behaviour(accessing an uninitialised object && writing data to a random address) will be invoked.

Also, change (int)data to *((int*)data) in your printf() statement. If sizeof (int) > sizeof (void *) , data will be truncated when being cast.

Answer 2

If you need to use a void * as a generalised container.

Got two options for integer or other data types

 void * data = malloc(sizeof(DATATYPE)); // Datatype can be int in this case

Then

 *((int *)data) = SOME VALUE;

Then free it at some point

Otherwise one can usually assume an internet "fits" into a void *.

I would not go down this avenue.