1 CS/COE0447 Computer Organization & Assembly Language Pre-Chapter 2

2 “C Program” Down to “Numbers” swap: muli$2, $5, 4 add$2, $4, $2 lw$15, 0($2) lw$16, 4($2) sw$16, 0($2) sw$15, 4($2) jr$31 void swap(int v[], int k) { int temp; temp = v[k]; v[k] = v[k+1]; v[k+1] = temp; } … … … … … … … compiler assembler

3 “Numbers” in Memory … … … … … … …

4 Stored Program Concept processor main memory hard disk program A program B program C program A program B data A data B program fetch data load/store disk I/O program counter

5 Stored Program Concept Programs (instructions) are stored in memory as a stream of bits (numbers) –Indistinguishable from data –More than one program can reside in memory at the same time –Programs can be modified by the processor or I/O just as data can be modified Instructions are fetched by the processor and decoded; they determine processor actions Program Counter determines which instruction is fetched next

6 Stored Program Concept In fact, one of the great ideas in computer science is the idea that programs could be stored just as data is stored. Before that, people envisioned the hardware running a fixed program, and data being stored in memory.

7

8 …

9

10

11

12 Addresses and Contents shown in Hex

13 Number Systems Actual machine code is in binary –O, 1 are high and low signals to hardware Hex (base 16) is often used by humans (code, simulator, manuals, …) because: 16 is a power of 2 (while 10 is not); mapping between hex and binary is easy It’s more compact than binary We can write, e.g., 0x in programs rather than

14 Base 10 (Decimal) Digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 of them) Example: –3217 = (3  10 3 ) + (2  10 2 ) + (1  10 1 ) + (7  10 0 ) –A shorthand form we’ll also use:

15 Numbers and Bases in General Number Base B  B unique values per digit –Base 10: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} –Base 2: {0, 1} –Base 16: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F} (Unsigned) number representation –d 31 d 30 …d 1 d 0 is a 32-digit non-negative number –Value = d 31  B 31 + d 30  B 30 + … + d 1  B 1 + d 0  B 0 N-digit base B  B N unique values in N digits of base B

16 Example: Base 2 (Binary) Digits: 0, 1 (2 of them) “Binary digit” = “Bit” Example: –11010 two = (1  2 4 ) + (1  2 3 ) + (0  2 2 ) + (1  2 1 ) + (0  2 0 ) = = 26 ten Choice for machine implementation! –1 = on, 0 = off

17 Base Conversion Let’s do decimal-to-binary conversion: A ten = d n-1 d n-2 …d 1 d 0two Given a base-10 number A, come up with n-digit binary number that has the same value! –X = the number –Let N be the largest power of 2 that fits into X –Put a 1 in that position –X = X – 2^N –Repeat until you are done!

18 Base Conversion, cont’d From binary to decimal From decimal to binary From binary to hexadecimal From hexadecimal to binary From decimal to hexadecimal? (more complicated; later)

19 Base Conversion, cont’d Binary to hex (base 16), or hex to binary base conversion: –Take 4 bits in binary and convert them into one hex digit and vice versa –For binary  hex: 4-bit groups, starting from the right –For hex  binary: translate each hex digit into 4 bits, starting from the right Since binary notation tends to be long, hex notation is frequently used in assembly language (and in C programs). More on binary number representation will be discussed when we study arithmetic

20 Before moving on to chapter 2…. We’ll mention some concepts in program performance, so you have ideas in mind We’ll return to this material later in the course.

21 Program Performance Program performance is measured in terms of time! Program execution time depends on: Number of instructions executed to complete a job How many clock cycles are needed to execute a single instruction The length of the clock cycle (clock cycle time)

22 Clock, Clock Cycle Time Circuits in computers are “clocked” At each clock rising (or falling) edge, some specified actions are done, usually within the next rising (or falling) edge Instructions typically require more than one cycle to execute Function block (made of circuits) clock clock cycle time

23 Program Performance time = (# of clock cycles)  (clock cycle time) # of clock cycles = (# of instructions executed)  (average cycles per instruction) We’ll do specific calculations later But now, let’s move on to Chapter 2