D75P 34 – HNC Computer Architecture Week 5 Memory Maps. © C Nyssen/Aberdeen College 2003 All images © C Nyssen/Aberdeen College except where stated Prepared 10/9/03

When a computer is first booted up, the operating system and any device drivers are loaded from the Hard Drive into the RAM.

The OS and any drivers will “claim” bits of RAM for their own use. Any other applications to be run from this point on, have to fit themselves in to the available RAM space left.

Memory is assigned to running applications by various algorithms –  First Fit - Take the first thing that comes along.  Best Fit - Allocate the block closest in size.  Worst Fit - Allocate the largest block (Reduces small holes)  Quick Fit - Make several lists, each has a common size block in it.

The number of addresses, or bytes, that can be pointed to by the Address Bus is 2 width of bus. An 8-bit bus can reference 2 8, or 256, addresses. If the locations are 1 byte in size, this is equivalent to ¼ kilobyte addressable memory. The address numbers referenced will run from 0 – 255 (00h – FFh).

Old 16-bit applications used 16-bit data paths, meaning that only the first 64KB of memory could be addressed. The locations referenced would be numbered 0 to (0-FFFFh). This is still a very common configuration. 32-bit applications can use 4GB of addressable memory.

What happens if a system can only reference the first 64KB, but the memory installed is much bigger? The higher addresses have to be referenced indirectly. This means setting up a “stepping stone” within the range that can be directly addressed.

Sometimes it is essential to know which areas of memory are “pre-claimed” by the boot code, Operating System and peripheral drivers. This information is particularly useful to hardware manufacturers and assembly code developers, especially when writing device driver software. We depict this information graphically as a memory map.

You will find many widely differing examples of “memory maps” on the Internet, many written by vintage computer enthusiasts. A true memory map will show the whole of the addressable memory, the spaces claimed and the remaining free space. They are usually labelled in hexadecimal notation.

When drawing your Memory Map for your Outcome question, you will be given the following information.  The width of the address bus  The width of the locations (it might not always be 1 byte!)  The spaces already claimed. You will be given the dimensions in either decimal or hexadecimal notation or by the size in bytes.

We will look at the specification of a very early business computer from Sharp, and draw a partial map of the processor chip.

First work out the size of addressable memory using 2 address bus width. Draw a box and mentally divide it up into that number of boxes. They will be labelled from 0 to (2 address bus width – 1).

Now fill in the claimed space. We are told that the Stack and Monitor Working Store starts at 0 and extends for 4.5 kilobytes. 4.5 kb = 4608 bytes. As each location is 1 byte in size, this gives us 4608 locations. This section of claimed memory therefore occupies boxes 0 – 4607 (0 –11FFh).

The disk system occupies the top 4KB of memory. This is 4096 bytes or locations. The top address was (FFFFh). Counting down from 65535, this gives a starting address of (F000h). We now fill in the disk system space.

The I/O Ports and Timers start from E000h and are 64 bits wide. 64 bits is equivalent to 8 bytes or locations. Counting up from E000h (or decimal 57344), the top address will be E007h (decimal 57351).

RAM started from (4806h) and was 22KB (or bytes) in size. This gives us a top address of (A005h). Immediately after this (no free space between) came 512 bytes of Video RAM. Video RAM therefore starts from (A006h) and ends on (A205h).

Note that if the space occupied is an even number of addresses, and it starts from an odd number, it must always end on an even one (and vice versa).

Lastly, label all the used blocks and free space.

Summary Portions of the RAM will be claimed by the boot code, drivers and operating system on startup. This can be graphically depicted as a Memory Map. The top address will be (2 address bus width – 1). Fill in the used memory as directed in the question. Finally, label all the used blocks and free space.