Lecture 2 Scribe Notes

UCLA CS 111 - April 2, 2014 - Professor Eggert

Notes by Aman Agarwal, Nicholas van Hoff, Zhao Yang, Kyle Gronich

From Last Lecture

Avoidance of OS

• Incommensurate scaling
• Waterbed effect
• propagation of effects
• Complexity

Complexity

Complexity is the biggest challenge in OS design. Professor Eggert: Complexity bites us harder than bites elsewhere

• Moore's Law: the number of transistors on integrated circuits doubles approximately every two years
• 
• Kryder's Law: the number of bytes on a hard drive doubles every 13 months
• 

Problem

We need a way of counting the number of words in a proposal. We need to be very secure. We cannot trust an operating system, because it might contain codes written by a fellow professor to steal his proposal.

Solution

Write a standalone program that is OS-agnostic.

Assumptions

File Characteristics

• ASCII text
• Ends with Null Byte '\0'
• A word matches the regular expression '[A-Za-z]+'

Machine

• x86
• Intel Core i3
• 3 MB cache
• 3.4 GHz
• 4GB RAM
• 1TB HD
• Intel HD integrated graphic
• Not connected to the network (air gap)
• The user interface consists of plugging it in to run and unplugging to stop

Disk Controller

How to Tell Disk Controller a Command

1. Wait for it to become ready
2. Store number of sectors you want to read into
3. Store which sector you want to read into
4. Store READ command into first slot
5. Wait for ready
6. Slurp data from disk controller cache through CPU into RAM

Bootstrapping

Most software on a computer is loaded using other software already running on that computer. Bootstrapping is the process of putting the initial software onto the computer.

ROM

Read-only memory (ROM) is a solution to the bootstrapping problem by allowing computers to be shipped with a start up program.

• survives power outage
• hard to change/tamper with

BIOS

Basic Input/Output System (BIOS) resided on the ROM and servers to:

1. Self-test
2. Look for devices
3. looks for a special pattern in the first sector indicating device is bootable

MBR

• The first 512 bytes of a bootable device
• The first 446 bytes are for the bootstrap code
• The next 64 consist of 4 partition descriptors, each 16 bytes, each including
  • the status of the partition (is it bootable?)
  • the start sector
  • the size
  • the type
• The last 2 Bytes indicate the boot signature
• The first of these last two bytes should be 0x55 and the last byte should be 0xAA
• The boot program will read in the actual (word count) program to run it

Bootable device

• The MBR is followed by an OS-specific Volume Boot Record (VBR)

Actual Code

mbr.c

Converting mbr.c to the executable

mbr.c -> compiler -> mbr.o -> ld -> mbr exec

Use dd to perform a low level copy onto the disk

dd if=mbr of=/dev/disk1/0 count=1 size=512

mbr.c

int main(void)
{
    for (int i = 0; i <= 20; i++) {
        read_sector(1 + i, 0x100000 + i * 512);
    }
    goto *0x100000;
}

static void read_sector(int s, char *a, int ns)
{
    wait_for_ready();

    outb(0x1f2, ns);
    for (int i = 0x1f6; i >= 0x1f3; i--) {
        outb(i, s & 0xff);
        s >>= 8;
    }

    outb(0x1f7, 0x20); //READ command

    wait_for_ready();

    insl(0x1f0, (int*)a, 128);
}

static void wait_for_ready()
{
    while((inb(0x1f7) & 0xc0) != 0x40) // wait for the disk controller to be ready.
    continue;
}

static unsigned char inb(int addr) {
    asm("inb ..."); //Assembly instruction
}

wc.c

void main(void)
{
    unsigned long nw = 0;
    bool inw = false;
    int s = 21;
    for(;;)
    {
        char buf[512];
        read_sector(s, buf);
        for(int i=0; i<512; i++)
        {
            if(buf[i] == '\0')
            {
                finish(nw);
                return;
            }
            bool inw1 = (buf[i]|('a'-'A'))-'a' < 26u; 
	    // or: bool inw1 = ('a' ≤ buf[i] && buf[i] ≤ 'z' || 'Z');  
            nw += inw1 & inw;
            inw = inw1;
        }
    }
}

void finish(int n)
{
        short *p = (short *) 0xb8014;
        do {
            *p-- = '0' + n % 10;
            n /= 10;
        }while(n);
}