Why can't I put 2 asterisks for ptr2?

Question

int main()
{
float arr[5] = {12.5, 10.0, 13.5, 90.5, 0.5};
float *ptr1 = &arr[0];
float *ptr2 = ptr1 + 3; //Why not **ptr2?

printf("%f ", *ptr2);
printf("%d", ptr2 - ptr1); //Why the output is 3??

return 0;
}

Answer 1

Konybud , ~ swim ~ has all your questions answered.

Answer 2

I’m still learning C myself so will welcome anyone correcting me if I’m wrong:

Firstly, an array variable name acts as a pointer to the array’s first element. You can write

float *ptr1 = arr;

and that should be fine

Next, when doing

ptr2 - ptr1

this is taking the memory address of ptr2 and subtracting from ptr1

you can do printf(“%x”... for both ptr2, ptr1 to see you are not accessing the array’s elements, you are working with the memory addresses themselves. I don’t know how 3 is calculated but even with replacing %d with %x in the subtraction printf line, I got 3 too

If I’m guessing correctly you want to subtract elements, use

*ptr2 - *ptr1

and that should do your 4th element minus the 1st element

If your question is out of curiosity how 3 is calculated, I’m not sure

But yes, you can point to a pointer.

**ptr would work if you declared

float **ptr = &ptr1

as the double asterisk means I am pointing to a pointer and &ptr1 is saying here is the address of that variable/pointer

The subtraction will then be

*ptr2 - **ptr

with the double asterisk being used to access the value (not using the memory address as is)

Without declaring **ptr, this would be using ptr1 as follows

float *ptr2 = &ptr1 + 3;

and

*ptr2 - **(&ptr1)

in the subtraction

Answer 3

Thank you everybody.

Answer 4

How about the ptr2? Why not **ptr2? (fifth line).