What is the cost of memory access?

Question

We like to think that a memory access is fast and constant, but on modern architectures/OSes, that's not necessarily true.

Consider the following C code:

int i = 34;
int *p = &i;

// do something that may or may not involve i and p

{...}

// 3 days later:

*p = 643;

What is the estimated cost of this last assignment in CPU instructions, if

• i is in L1 cache,
• i is in L2 cache,
• i is in L3 cache,
• i is in RAM proper,
• i is paged out to an SSD disk,
• i is paged out to a traditional disk?

Where else can i be?

Of course the numbers are not absolute, but I'm only interested in orders of magnitude. I tried searching the webs, but Google did not bless me this time.

Answer 1

Here's some hard numbers, demonstrating that exact timings vary from CPU family and version to version: http://www.agner.org/optimize/

These numbers are a good guide:

L1           1 ns
L2           5 ns
RAM         83 ns
Disk  13700000 ns

And as an infograph to give you the orders of magnitude:

(src http://news.ycombinator.com/item?id=702713)

Answer 2

Norvig has some values from 2001. Things have changed some since then but I think the relative speeds are still roughly correct.

Answer 3

It could also be in a CPU-register. The C/C++-keyword "register" tells the CPU to keep the variable in a register, but you can't guarantee it will stay or even ever get in there.