Hello guys can someone explain why while declaring pointers to pointer we need to use ** why cant we use only single * to point a pointer to another pointer or is it just a syntax related issue E.g
int main() { int a=5,*b,*c; b=&a; c=&b //Why cant this simply doesn't make c point to memory location of b with the above pointer declaration why is there a need to declare c as **c } With the following codes:
int a=5,*b,**c; b=&a; c=&b; We have:
+---+ a | 5 | <-- value +---+ |100| <-- address +---+ +---+ *b |100| <-- value +---+ |200| <-- address +---+ +---+ **c |200| <-- value +---+ |300| <-- address +---+ When you store a's address in b, b's value is a's address. But b has it's own address (200). c can store b's address as it's value. But c has it's own address too (300).
printf("%x", &c); will give you: 300
Deferencing *c will get you down "1 level" and give you 100 (get value of address 200)
Deferencing **c will get you down 1 more level and give you 5 (get value of address 100)
If you try to use *c instead of **c to hold *b, how are you able to deference all the way down to reach value 5?
Testing the codes on a compiler:
printf("Address of a: %x\n", &a); printf("Address of b: %x\n", &b); printf("Address of c: %x\n", &c); printf("Value of a: %d\n", a); printf("Value of b: %x\n", b); printf("Value of c: %x\n", c); Output:
Address of a: 28ff44 Address of b: 28ff40 Address of c: 28ff3c Value of a: 5 Value of b: 28ff44 Value of c: 28ff40 
In this case
int main() { int a=5,*b,*c; b=&a; c=&b; } Here b points to a and c points to b. It is what you have commented in the commented.
c still points to the memory location of b.
The catch is : When you de-reference b i.e *b = a = 5.
But When you de-reference c i.e *c = b = &a. So When you dereference c the output would be address of a instead of the value of the variable a
PS : you will face this warning when compiling the code warning: assignment from incompatible pointer type
Every level of indirection needs a level of dereferencing. So for:
T*** x = ...; you would need:
***x to get to T&.
If you had a pointer to pointer and you saved it in:
T* x = ...; T* y = &x; it would mean that *ptr leads to T&, while it really leads to another T*.
You have your answer in your question only.
pointer to variable , use of *pointers to pointer of a variable , use **Details:
** is not a new operator. it's a combination of * and *. In case 2. as per your terminology, you can think of
only single * to point a pointer to another pointer as in
int * to an inother int * ==> int ** EDIT:
as per your code
int main() { int a=5,*b,*c; b=&a; c=&b; } b is a pointer to int. You can store the address of int there, and a is an int. Perfect.c is a pointer to int. You can store the address of int there, and b is a pointer to int. Not accepted.To make point 2 work, you need to declare c as a pointer to int *, right? The notation for the same is int **.
Here's another way to think of pointers-to-pointers: imagine how it works in memory. Here's a little snippet that shows what I mean:
int TheInteger = 123; int *p = &TheInteger; int **pp = &p; printf("The value at memory location &pp (0x%x) is 0x%x (pp). This value (which we assigned as &p (0x%x) is 0x%x (p). This value, in turn, we assign as &TheInegeter (0x%x) points to the 'instance' of TheInteger, which is %d", &pp, pp, &p, p, &TheInteger, TheInteger); The output of this would be:
The value at memory location &pp (0x657e588) is 0x657e594 (pp). This value (which we assigned as &p (0x657e594) is 0x657e5a0 (p). This value, in turn, we assign as &TheInegeter (0x657e5a0) points to the 'instance' of TheInteger, which is 123 Now, to go back to your original question, you cannot declare a variable as being a pointer when the value you're setting it to is a pointer-to-a-pointer. In other words, in your example, you set 'b' as a pointer to a -- so, you can't tell the compiler that 'c' is just a pointer and then try to set it to a value that the compiler knows is a pointer-to-a-pointer.
