Basic pointer arithmetic for arrays

Question

What would be the value of array and p after executing

int array[] = {1,2,3}, *p = array;

a. *p++;
b. (*p)++;
c. *p++; (*p)++

I know the answer is

a) array = {1,2,3} and *p = 2 
b) array = {2,2,3} and *p = 2
c) array = {1,3,3} and *p = 3

but I just don't understand how. And explanation would be great!

@mclaassen: I don't think that was the intent. The question specifically asks for the value of p. I think that was an incorrect edit... I think he wants p = array+1 and similar there. — Mooing Duck
– Mooing Duck, Commented Sep 11, 2014 at 20:22
Well it matches the results. I think he was clearly interested in the value pointed to by p and not the memory address itself. — mclaassen
– mclaassen, Commented Sep 11, 2014 at 20:22
I'm... not so sure, since the third element of the second array switched to 2 and back. Though I think that was a typo. I begin to think you're correct. — Mooing Duck
– Mooing Duck, Commented Sep 11, 2014 at 20:23
The questions states "I know the answer is.." so the question makes a lot more sense if you update it so that the answer really is what is shown. — mclaassen
– mclaassen, Commented Sep 11, 2014 at 20:24

betteroutthanin · Accepted Answer · 2014-09-11 20:19:35Z

1

p initially points to the first element of the array, so

a. Move the pointer to point to the next element, which is 2.
b. Dereference p, which is 1, then increase it by 1, which becomes 2.
c. The first part is same as a. so the pointer points to the second element, then you dereference it, you get 2, and increase it by 1, you get 3.

answered Sep 11, 2014 at 20:19

betteroutthanin

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