Pointer to pointer

Question

I am trying to set the memory of pointer p through pointer to pointer x

 int *p = 0; int **x = &p; *x = new int[2]; *x[0] = 1; //p[0] = 1 *x[1] = 2; //p[1] = 2 

why does it fail with an access violation error?

Answer 1

why does it fail with an access violation error?

[] take precedence over *. You need to dereference x first

 (*x)[0] = 1; (*x)[1] = 1; 

Answer 2

I think your problem is that

 *x[1] 

Facilities

 *(x[1]) 

And not

 (*x)[1] 

This second version is what you want; it pokes the pointer to get the base array, and then reads the second element. The code you are using now treats your double pointer as a pointer to the second element of an array of pointers, looks at the first pointer in it, and then tries to dereference it. This is not what you want as the pointer points to only one pointer, not two. Explicitly enclosing this code in parentheses should solve this problem.

Answer 3

In an expression of type *A[] variable A first indexed and then dereferenced, and you want the exact opposite, so you need to write (*A)[] .

Answer 4

Unfortunately, a type system will not help you here.

x [0] and x [1] are both int * types, so you can do

 *(x[0]) = 1; // allowed *(x[1]) = 2; // access violation, not through x[1] itself but by accessing it 

Of course your intention was

  (*x)[0] = 1; (*x)[1] = 2; 

Answer 5

FYI, you can also try to access the memory locations indicated by x in the same way -

 x[0][0] = 1; x[0][1] = 2; 

Answer 6

Your problem is that you set p to address 0. You have not actually allocated space for p (and therefore x.). Thus, you are trying to install memory at addresses 0 and 1. If you change it to this, it should work without error.

 int *p = new int[5]; int **x = &p; *x = new int[2]; (*x)[0] = 1; //p[0] = 1 (*x)[1] = 2; //p[1] = 2