c编程：强制转换“ void指针”以指向结构

Question

I'm writing this program to practice C programming, this is my code:

#include <stdio.h>
#include <stdlib.h>
struct s1 {
        int i;
        void * p;
    };
static struct s1 *dmk;
int main(void) {

    int tong(int a, int b);
    int (*tinh)(int,int);

    struct s2 {
        int num;
        int (*cal)(int a, int b);
    };
    if(dmk->p == NULL)
    {
        printf("NULL ALERT\n");
    }
    struct s2 *cl = dmk->p;
    cl->cal = tong;

    tinh = ((struct s2 *)(dmk->p))->cal;
    printf("tinh 2, 4 ra %d\n",tinh(2,4));
    puts("!!!Hello World!!!"); /* prints !!!Hello World!!! */
    return EXIT_SUCCESS;
}

int tong(int a, int b)
{
    return a + b;
}

When I compiled it, it didn't show any error or warning. But when I ran the program, the terminal told me that "core dumped" and did not show any result accept for the "NULL ALERT". Can anyone explain why I failed ? Thanks alot.

Answer 1

if(dmk->p == NULL)

fails as dmk is uninitialised initialised to NULL , points "no-where", so de-referencing it invokes undefined behaviour. Anything can happen afterwards.

Answer 2

Bottom line is that what you're doing wrong is not assigning things to your various pointers before using them.

Your dmk is a global, so its initial value is NULL . You're never changing that value.

Later, you read dmk->p . You're now in Undefined Behavior (UB) territory. The program could easily have faulted right there, because you were reading from a NULL pointer.

Apparently it didn't, though, because you're seeing your NULL ALERT message, and so we continue.

Then you do this:

struct s2 *cl = dmk->p;
cl->cal = tong;

On the second line there, cl is completely indeterminate. It could be garbage, it could be NULL , whatever, because you're entered UB territory. Then you dereference it and write to the result . You could be trying to write anywhere. Writing to random pointers (or NULL pointers) tends to make core dumps and other Bad Things™ happen.

You need to actually assign values to your pointers before you use them.

Answer 3

dmk is a global (aka. static (this is not related to the static keyword!), variable, thus it is initialized to a null pointer . You do not change this value, so dkms->p dereferences a null pointer , which invokes undefined behaviour (UB).

So from here, it is speculation, as UB means anything can happen .

Apparently the printf passes, probably because the address can be read. The following write OTOH fails, generating the system messsage.

Answer 4

You don't initialize dmk in your code. That dmk->p is NULL is pure luck. Anything could happen.

You need to allocate memory for your struct using, for instance, malloc(), and initialize the members properly.