Lecture 15 Todays lecture MARIE programming Assembler Extending

Lecture 15 • Today’s lecture – MARIE programming – Assembler – Extending the instruction set 1

A Simple Program • Consider the simple MARIE program given below: add two numbers together and the sum in memory. • The instructions stored at addresses 0 x 100 – 0 x 106 (hex): 2

A Simple Program • What happens inside the computer when the program runs. • This is the LOAD 104 instruction: 3

A Simple Program • Our second instruction is ADD 105: 4

A Simple Program • The third instruction is Store 106. • After the program is done, the binary content of location 0 x 106 change to 0 x 000 C (12 in decimal). 5

A Discussion on Assemblers • Mnemonic instructions, such as LOAD 104, are easy for humans to write and understand. • They are impossible for computers to understand. • Assemblers translate assembly language that are comprehensible to humans into the machine language that is comprehensible to computers. • The assembler reads a source file (assembly program) and produces an object file ( the machine code). 6

Creating an Object Program Files • Assemblers create an object program file from mnemonic source code in two passes. • During the first pass, the assembler assembles as much of the program as it can, while it builds a symbol table that contains memory references for all symbols in the program. • During the second pass, the instructions are completed using the values from the symbol table. 7

Creating an Object Program Files - Example • The first pass, creates a symbol table and the partiallyassembles instructions. • After the second pass, the assembly is complete. 8

Extending Our Instruction Set • So far, all of the MARIE instructions that we have discussed use a direct addressing mode. • This means that the address of the operand is explicitly stated in the instruction. • It is often useful to employ a indirect addressing, where the address of the operand is given in the instruction. – If you have ever used pointers in a program, you are already familiar with indirect addressing. • Six new instructions using indirect addressing. 9

LOADI and STOREI • LOADI X (load indirect): go to address X, use the value at X as the actual address of the operand to load into the AC. • STOREI X (Store indirect): go to address X, use the value at X as the destination address for storing the value in the AC. • In RTL: MAR X MBR M[MAR] MAR MBR M[MAR] AC MBR LOADI X MAR MBR M[MAR] X M[MAR] MBR AC MBR STOREI X 10

ADDI and JUMPI • The ADDI instruction is a combination of LOADI X and ADD X. • JUMPI instruction: go to address X, use the value at X as the actual address of the location to jump to. • In RTL: MAR X MBR M[MAR] MAR MBR M[MAR] AC + MBR ADDI X MAR X MBR M[MAR] PC MBR JUMPI X 11

JNS and CLEAR • JNS: The jump-and-store instruction gives us limited subroutine functionality: store the PC at address X and jump to X+1 • CLEAR: set AC to zero. • In RTL: MBR PC MAR X M[MAR] MBR X AC 1 AC + MBR PC AC JNS X AC 0 CLEAR 12

MARIE Programming Examples • 400 14

MARIE Programming Examples Example: a simple subroutine to double the value stored at X. 15

Exercise 1 Consider the MARIE program below. Hex Address 100 101 102 103 104 105 106 107 108 109 10 A 10 B Label Start, A, B, C, D, Instruction LOAD A ADD B STORE D CLEAR OUTPUT ADDI D STORE B HALT HEX 00 FC DEC 14 HEX 0108 HEX 0000 a) List the hexadecimal code for each instruction. b) Draw the symbol table. c) What is the value stored in the AC when the program terminates? 16

Exercise 1 - Solution Consider the MARIE program below. Hex Address 100 101 102 103 104 105 106 107 108 109 10 A 10 B Label Start, A, B, C, D, Instruction LOAD A ADD B STORE D CLEAR OUTPUT ADDI D STORE B HALT HEX 00 FC DEC 14 HEX 0108 HEX 0000 1108 3109 210 B A 000 6000 B 10 B 2109 7000 00 FC 000 E 0108 0000 AC = 0108 upon termination a) List the hexadecimal code for each instruction. b) Draw the symbol table. c) What is the value stored in the AC when the program terminates? 17