Pointer Arithmetics and casting

Question

Somehow these pointer arithmetics are very confusing to me. An example:

uint16_t *a = (uint16_t *)0x200;
a += 4 * sizeof(uint32_t);

When calculating the new value of a, what is your thought process?

This is how I am trying to figure it out:

The pointer (a) points to an address, which has a value of 0x200 of uint16_t type.
The second operation moves the pointer a to a + 4 * 8 bytes to a location 32 bytes (20 in hex) further up, which is apparently 0x220. How come this is not &a + 32? I think this is where I confuse things... Why is the pointer pointing to 0x200 and not &a?

Choose between `C++` and `C`. They are different, so please tag your question with only one of them. — Box Box Box Box, Feb 25 '16 at 15:03
I thought in this case it was valid in both languages...I guess I'll just stick c++ in there. — Space, Feb 25 '16 at 15:06
The compiler already knows how to do pointer arithmetic. You don't need to help it out. `a += 4` will increment the pointer to the fourth `uint16_t`. — Cody Gray - on strike, Feb 25 '16 at 15:06
`&a` is the address *of* `a`, not the address stored *in* `a`. — molbdnilo, Feb 25 '16 at 15:52

score 2 · Accepted Answer · answered Feb 25 '16 at 15:07

2

Pointer arithmetic is in units of the type being pointed to.

So if you just did ++a;, the new value of a would be 0x202, not 0x201.

So, when you add 4 * sizeof (uint32_t), that is the same as adding 2 * 4 * 4, i.e. 32, which is hex 0x20. So the new value is 0x220.

answered Feb 25 '16 at 15:07

unwind

Thank you. This was exactly the answer I was looking for. – Space Feb 26 '16 at 09:11

1 Answers1