在 c 中使用带有 ** 的指针

Question

I am very new to C and currently having some trouble with pointers and I am not sure if my logic is correct on this question clarification would be great.

Is the second expression legal? Why or Why not? What does it mean?

int** x = ....;
... **x ...

This is all the question gives and I came up with the following answer (I think its in the ballpark)

The int** x will initialize a pointer x to whatever address/value that is after the equal sign. **x ... will dereference the pointer to a value/variable

The question link that was proposed in the edit was just showing the difference between int* p and int *p which is nothing what i asked i understand this already.

Answer 1

int *x is a pointer to int . int** y defines y as a pointer to pointer to int , so it points to a pointer, and the latter points to an int . Example:

int n = 42;
int* x = &n; // points to n
int** y = &x; // points to x, which points to n

In this case y is a pointer to pointer. First indirection *y gives you a pointer (in this case x ), then the second indirection dereferences the pointer x , so **y equals 42 .

Using typedef s makes everything looks simpler, and makes you realize that pointers are themselves variables as any other variables, the only exception being that they store addresses instead of numbers(values):

typedef int* pINT;

int n = 42;
pINT x = &n;
pINT* y = &x; // y is a pointer to a type pINT, i.e. pointer to pointer to int

Answer 2

If you use double * you are telling the compiler you want a pointer pointing to another pointer:

int x = 4;
int *pointer = &x;
int **pointer_pointer = &pointer;
printf("%d\n",**pointer_pointer); // This should print the 4