将结构内使用的 void * 指针类型转换为 int * 指针时出现问题！

Question

My code is as follows,

#include<stdio.h>
struct data
{
    int a ;
    void *b;
};

int main()
{
    struct data *d;
    int *ptr;

    int key=10000;
    d->b=&key;

    ptr=(int *)d->b;
    printf("%d\n",*ptr);
}

And i get a segmentation fault?? Any idea why?? Thanks in advance for any help

Answer 1

struct data *d merely declares a pointer. You have not allocated this struct anywhere. You need to either malloc it or declare it just as struct data d on the stack or globally.

The former can be done like this:

d = malloc(sizeof(struct data));

If you choose the latter, accessing b has to be written as db .

Answer 2

You are not allocating any memory for d . It likely points to an invalid memory area and so - segmentation fault.

You can solve this like so:

struct data *d = malloc(sizeof(*d));

Answer 3

You are getting segmentation fault at the line d->b=&key; Note that you have not allocated any memory location to the structure variable d . So d contains some garbage value, and d->b it trying to use that garbage address to dereference the pointer and get the component b . Here is where you get the segfault. Either statically allocate the struct variable, or use malloc to dynamically allocate it.

int main()
{
    struct data *d;
    int *ptr;

    /* Here you are allocating memory to the
     * pointer variable, which will be used to
     * point to the structure type data
     */
    d = malloc (sizeof (struct data)); 
    int key=10000;

    /* Now you can dereference the pointer 
     * and get any of the components of the
     * structure, because 'd' contains a valid
     * address.
     */ 
    d->b=&key;

    ptr=(int *)d->b;
    printf("%d\n",*ptr);

    /* Good practice to free the memory location
     * you have allocated. Not freeing will lead to
     * memory leak in larger applications. After you 
     * free the memory location denoted by the address
     * stored in 'd', you will not be anymore access 
     * the contents of it.
     */
    free (d);

    /* d->b; or d->a; is no more possible at this point
     * as we have freed the memory pointed by 'd'
     */
}

Or you can use:

int main()
{
    /* Not a pointer, statically allocated */
    struct data d;
    int *ptr;

    int key=10000;
    d.b=&key;

    ptr=(int *)d.b;
    printf("%d\n",*ptr);
}

So, it is not the typecasting of void * to int * that causes the segfault. Its the illegal memory reference of the pointer variable which you have used but not allocated/initialized.

Answer 4

The problem is that you didn't allocate memory for ad pointer: struct data *d; . This lines only creates a pointer, it doesn't alloc memory for it. Please try the following code:

int main()
{
    struct data *d = (struct data*)malloc(sizeof(struct data));
    int *ptr;
    int key=10000;
    d->b=&key;
    ptr=(int *)d->b;
    printf("%d\n",*ptr);
    free(d);
}