Machine Level Architecture: Machine code and processor instruction set
Machine code
[edit | edit source]As we should hopefully already know, computers can only understand binary, 1s and 0s. We are now going to look at the simplest instructions that we can give a computer. This is called machine code.
Machine code allows computers to perform the most basic, but essential tasks. For this section we are going to use the Accumulator (you met this register earlier) to store the intermediate results of all our calculations. Amongst others, the following instructions are important for all processors:
- LDA - Loads the contents of the memory address or integer into the accumulator
- ADD - Adds the contents of the memory address or integer to the accumulator
- STO - Stores the contents of the accumulator into the addressed location
Assembly code is the easy to read interpretation of machine code, there is a one to one matching, one line of assembly equals one line of machine code:
Machine code | Assembly code |
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000000110101 = | Store 53
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Let's take a look at a quick coding example using assembly code.
LDA #23 ;loads the number 23 into the accumulator
ADD #42 ;adds the number 42 to the contents of the accumulator = 65
STO 34 ;save the accumulator result to the memory address 34
The code above is the equivalent of saying x = 23 + 42
in VB.NET.
- the above example is not entirely factually correct as the variable "x" has never been created(defined)in the Assembly code example.
Instruction set
[edit | edit source]There are many different instructions that we can use in machine code, you have already met three (LDA, ADD, STO), but some processors will be capable of understanding many more. The selection of instructions that a machine can understand is called the instruction set. Below are a list of some other instructions that might be used:
ADD ;add one number to another number
SUB ;subtract one number from another number
INC ;increment a number by 1
DEC ;decrement a number by 1
MUL ;multiply numbers together
OR ;boolean algebra function
AND ;boolean algebra function
NOT ;boolean algebra function
XOR ;boolean algebra function
JNZ ;jump to another section of code if a number is not zero (used for loops and ifs)
JZ ;jump to another section of code if a number is zero (used for loops and ifs)
JMP ;jump to another section of code (used for loops and ifs)
Let us look at a more complex example of assembly code instructions:
LDA #12 ;loads the number 12 into the accumulator
MUL #2 ;multiplies the accumulator by 2 = 24
SUB #6 ;take 6 away from the accumulator = 18
JNZ 6 ;if the accumulator <> 0 then goto line 6
SUB #5 ;take 5 away from the accumulator (this line isn't executed!)
STO 34 ;saves the accumulator result (18) to the memory address 34
You'll notice that in general instructions have two main parts:
- opcode - instruction name
- operand - data or address
Depending on the word size, there will be different numbers of bits available for the opcode and for the operand. There are two different philosophies at play, with some processors choosing to have lots of different instructions and a smaller operand (Intel, AMD) and others choosing to have less instructions and more space for the operand (ARM).
- CISC - Complex Instruction Set Computer - more instructions allowing for complex tasks to be executed, but range and precision of the operand is reduced. Some instruction may be of variable length, for example taking extra words (or bytes) to address full memory addresses, load full data values or just expand the available instructions.
- RISC - Reduced Instruction Set Computer - less instructions allowing for larger and higher precision operands.
Exercise: Instruction sets What is the instruction set: Answer: the range of instructions that a CPU can execute Name and explain the two parts that make up an machine code instruction: Answer:
For a word with 4 bits for an opcode and 6 bits for an operand
Answer:
For a 16 bit word with 6 bits for an opcode
Answer:
Why might a manufacturer choose to increase the instruction set size? Answer: so that they can increase the number of discrete instructions that can be executed What might be the problem with increasing the space taken up by the opcode? Answer: less space for the operand, meaning reduced range and precision in data be processed in a single instruction Give two benefits for increasing the word size of a processor? Answer:
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Addressing modes
[edit | edit source]You might notice that some instructions use a # and others don't, you might even have an inkling as to what the difference is. Well here is the truth:
# = number [no hash] = address
Let's take a look at a quick example:
Assembly code | Main memory start | Main memory end | ||||||||||||||||||||||||
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LOAD #10
ADD #12
STORE 12
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This code loads the number 10 into the accumulator, then adds the number 12, it then stores the result 22 into memory location 12. |
Let's take a look at doing this without the hashes:
Assembly code | Main memory start | Main memory end | ||||||||||||||||||||||||
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LOAD 10
ADD 12
STORE 12
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This code loads the value stored in memory location 10 into the accumulator (9), then adds the value stored in memory location 12 (7), it then stores the result into memory location 12 (9 + 7 = 16). |
There are many types of addressing modes. But we only need to know 3, they are:
Addressing Mode | Symbol | Example | Description |
---|---|---|---|
Memory Location | LOAD 15 | 15 is treated as an address | |
Integer | # | LOAD #15 | 15 is treated as a number |
Nothing | HALT | Some instruction don't need operands such as halting a program |
Exercise: Assembly code and Addressing modes For the following memory space, what would it look like after executing the assembly code below:
LOAD 14
ADD #12
STORE 12
Answer:
For the following memory space, what would it look like after executing the assembly code below:
LOAD #100
STORE 213
LOAD 214
ADD 213
STORE 214
Answer:
For the following memory space, what would it look like after executing the assembly code below:
LOAD 100
ADD 101
DIV #7
STORE 102
Answer:
Write some assembly code to do the following:
Answer:
LOAD #34
ADD #35
STORE 100
Write some assembly code to do the following:
Answer:
LOAD #100
DIV #2
ADD #4
STORE 100
List and give examples of three addressing modes: Answer:
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Machine code and instruction sets
[edit | edit source]There is no set binary bit pattern for different opcodes in an instruction set. Different processors will use different patterns, but sometimes it might be the case that you are given certain bit patterns that represent different opcodes. You will then be asked to write machine code instructions using them. Below is an example of bit patterns that might represent certain instructions.
Machine code | Instruction | Addressing mode | Hexadecimal | Example |
---|---|---|---|---|
0000 | STORE | Address | 0 | STO 12 |
0001 | LOAD | Number | 1 | LDA #12 |
0010 | LOAD | Address | 2 | LDA 12 |
0100 | ADD | Number | 4 | ADD #12 |
1000 | ADD | Address | 8 | ADD 12 |
1111 | HALT | None | F | HALT |
Exercise: Machine Code Using the table above provide machine code to do the following: LOAD 12
ADD #6
Answer:
0010 00001100 0100 00000110
0001 00000111 0100 00001001 0000 00011110 Answer:
LOAD #7
ADD #9
STORE 30
Explain what the above code does: Answer: loads the integer 7 into the Accumulator, adds the integer 9 to the Accumulator, stores the result, 16, in memory location 30
0001 00111011 0100 00001001 0000 00011110 1111 00000000 Answer:
1 3 B 4 0 9 0 1 E F 0 0 If we were lacking Assembly code, why might we want to convert machine code into Hexadecimal? Answer: It makes it easier for humans to read and understand. |
Extension: Little Man Computer If you would like to play around with Assembly language a great place to start is the Little man computer. You can find a Java applet and some examples at the York University website or a javascript version created by Peter Higginson |