要从C中的结构内存访问类型的结构指针数据

Question

My program dynamically allocates nodes of type struct pointer. I want to access the data of struct pointer via it's memory address. The main purpose of my question is to test b-tree insert with split. Here is the struct:

struct LeafNode
{
    int EmpID[M];
    int Location[M];
    int keys;
};
typedef struct LeafNode* LNodePtr;

Here is my test code:

#include <stdio.h>
#include <stdlib.h>
int main()
{
    int * a = (int *)malloc(sizeof(int));
    *a = 5; 
    printf("int a created at %p\n", (void *) &a);

    void * addr;

    printf("Give the memory address: ");
    scanf("%p", addr);
    void * addr2 = (void *)addr;
    printf("The value at memory is: %d\n",*(int*)addr2);

return 0;
}

Here' s the result:

int a created at 0xbff573b4
Give the memory address: 0xbff583b4
Segmentation fault

The second step is to cast to LNodePtr instead of int. I don't know if we can use double cast, ie

*(int *(LNodePtr))addr 

so that i can access

struct's EmpID[0]

int member through memory. Thanks for help.

Answer 1

void * addr;

Allocates a pointer, but does not initialize it.

scanf("%p", addr);

Reads a pointer into the variable stored at the address held in addr . You did not initialize addr so this is undefined behaviour.

Perhaps you meant to write:

scanf("%p", &addr);

But it's very hard to tell. Not least because you talk at length about a struct and then have code that does not mention the struct and instead works with void* .

---

printf("int a created at %p\n", (void *) &a);

This prints the address of the variable holding the pointer a . If you want to print the value of the pointer, use a instead of &a .

Answer 2

You shouldn't dereference a in your printf line, a itself is already a pointer, *a is the int!

printf("int a created at %p\n", (void *) a);

Also, you should dereference the void* in scanf

scanf("%p", &addr);

As a general tip, try compiling with the C flag -Wall, so you can easily catch these mistakes.

To answer the question in the comments, in case you have the node address 0x1234, you could do this:

MyNode *node = 0x1234;
node->int[0] = 5;

Hope that answers the question