Crashing code(pointers C99)

Question

I am trying to understand why that code crash:(CodeBlocks C99)

int a;
int **b=0;
*b=&a;

I know that * b is of type int* and &a is also int* so what's the problem here?

Answer 1

Let's take this apart:

int a; /* make a an integer, assuming this is in a function on the stack */
int **b=0; /* make b a pointer to a pointer to an integer, but make it point to 0 */
*b=&a; /* store the address of a at the address pointed to by b, which is 0 */

IE you are explicitly writing the address of a to a zero location. The problem is not type compatibility, it's that you are trying to store something at location zero, which will cause a seg fault.

To fix it do something like:

int a; /* make a an integer, assuming this is in a function on the stack */
int *c = 0; /* make an integer pointer, initialise it to NULL */
int **b=&c; /* make b a pointer to a pointer to an integer, make it point to c */
*b=&a; /* store the address of a at the address pointed to by b, which is c */

Answer 2

b指向一个指针，您将指针指向0 / NULL，这意味着当您执行* b =您正在将值分配给地址0时，它将在大多数OS上消失（在嵌入式系统上，这取决于处理器）

Answer 3

You are dereferencing the NULL-Pointer. b points to NULL and in the next line you are dereferencing it and assigning it a new value.

You are not allowed to write to memory you don't own, and you are especially not allowed to write to NULL.

Answer 4

You cannot set pointers to values like that (*b=&a) because they are at the time not pointing to anything; in order to set their value they must be pointing to something.

int a;
int *tmp;
int **b = &tmp;
*b = &a; //this should work because you are setting a real variable (in this case tmp) to &a

Answer 5

Your int **b=0 is a pointer to int * , that is initialized to NULL , and isn't pointing at an allocated storage area. When you write to *b , you are attempting to dereference a null pointer, which is illegal.

If you were to allocate storage, for example using b=malloc(sizeof(*b)) , you would then be able to write into the area pointer at by b using *b=&a , because then *b would be a valid address to write into.

Answer 6

#include <stdlib.h>

int
main()
{
  int a;
  int **b = (int**)malloc(1 * sizeof(int));
  *b = &a;
  free(b);
  return 0;
}

Your *b = &a has the same effect of b[0] = &a .