Using IAS instructions, Slides of Assembly Language Programming

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Cosc 2150:

Computer Organization

Simplified IAS language

The Computer Level Hierarchy

• (^) Each virtual machine layer is an abstraction of the level below it.
• (^) The machines at each level execute their own particular instructions, calling upon machines at lower levels to perform tasks as required.
• (^) Computer circuits ultimately carry out the work.

The Computer Level Hierarchy

• Level 4: Assembly Language Level

—Acts upon assembly language produced from Level 5,

as well as instructions programmed directly at this

level.

• Level 3: System Software Level

—Controls executing processes on the system.

—Protects system resources.

—Assembly language instructions often pass through

Level 3 without modification.

The Computer Level Hierarchy

• Level 2: Machine Level

—Also known as the Instruction Set Architecture (ISA)

Level.

—Consists of instructions that are particular to the

architecture of the machine.

—Programs written in machine language need no

compilers, interpreters, or assemblers.

The Computer Level Hierarchy

• Level 0: Digital Logic Level

—This level is where we find digital circuits (the chips).

—Digital circuits consist of gates and wires.

—These components implement the mathematical logic

of all other levels.

IAS Hardware

• There are 4 registers

• (^) PC = Program Counter
• (^) IR = Instruction Register
• (^) AC = Accumulator Register
• (^) MQ = Multiplier Quotient Register
  • (^) Used in Multiplication and division.

Language

• See Handout

Example 1

main () { int a=15; int b=5; int c; c = a + b; } 0 15 a 1 5 b 2 0 c 3 begin 4 .c = a +b 5 load M(0) 6 add M(1) 7 stor M(2) 8 halt

Example 2 (continued)

0 15 1 5 2 a 3 b 4 c 5 begin

6. a = 15 7 load M(0) 8 stor M(2)
7. b = 5 10 load M(1) 11 stor M(3)
8. c = a + b 13 add M(2) 14 stor M(4) 15 halt

Or you could

optimize the code a

little.

NOTE FOR

HOMEWORK.

You do not need to

optimize, but it

must be at least as

efficient as the

“original” code.

Example 3

main () { int a=15, b=5, c; if (a >= b) c = a – b; else c = a + b; } 0 15 a 1 5 b 2 c 3 begin

4. If (a >=b) 4 load M(0) 5 sub M(1) 6 jump+ M(8) 7 jump M(12) 8 .true, c=a-b 8 load M(0) 9 sub M(1) 10 stor M(2) 11 jump M(15) 12 .false c = a+b 12 load M(0) 13 add M(1) 14 stor M(2) 15 halt

Example3 ( with a > b)

main () { int a=15, b=5, c; if (a > b) c = a – b; else c = a + b; } 0 15 a 1 5 b 2 c 3 1 4 begin

5. a > b 5 load M(0) 6 sub M(1) 7 sub M(3) 8 jump+ M(10) 9 jump M(14)
6. True, c = a- b 10 load M(0) 11 sub M(1) 12 stor M(2) 13 jump M(17)
7. False, c = a + b 14 load M(0) 15 add M(1) 16 stor M(2) 17 halt

Example 4

main () { int x=0; while (x < 5) { x=x+1;//or ++x; } } A NOTE FOR HOMEWORK You CAN NOT change a top testing loop to a bottom testing loop! 0 5 1 1 2 0 x 3 begin

4. While (x<5) 4 load M(2) 5 sub M(0) 6 jump+ M(11)
5. X=x+ 7 load M(2) 8 add M(1) 9 stor M(2) 10 jump M(4) 11 halt

Example 5

main () { int I , x = 0; for (I =1;I<=6; ++I) { if (I % 2==0) { // % is mod, so I mod 2 x = x +I; } } x = x * 2; }

Code on next side.

Example 5 (continued)

0 7 1 1 2 2 3 i 4 0 x 5 begin

6. i = 7 load M(1) 8 stor M(3)
7. I<= 10 load M(3) 11 sub M(0) 12 jump+ M(27)
8. If (I%2) 14 load M(3) 15 div M(2) 16 sub M(1) 17 jump+ M(23)
9. x +=I; 19 load M(4) 20 add M(3) 21 stor M(4)
10. ++I 23 load M(3) 24 add M(1) 25 stor M(3) 26 jump M(10)
11. x = x * 27 load MQ,M(4) 28 mul M(2) 29 stor M(4) 30 halt