Basic Computer System Architecture Lecture PDF

Bachelor of Science (Honours) in Data Science and Artificial Intelligence DA 107 – Basic Computer System Architecture Introduction Outline First generation computers & John von Neumann architecture Basic Computing Concepts The Register File RAM A Closer Look At T...

Bachelor of Science (Honours) in Data Science and Artificial Intelligence DA 107 – Basic Computer System Architecture Introduction Outline First generation computers & John von Neumann architecture Basic Computing Concepts The Register File RAM A Closer Look At The Code Stream Register Vs Immediate 5 Outline Relative Addressing Mechanics of Program Execution Binary Encoding of Arithmetic Instructions Binary Encoding of Arithmetic Instructions With Immediate Value Binary Encoding of Memory Access Instructions The Store Instruction Example Programs 6 First Generation Computers & Architecture First generation computers & architecture The Electronic Numerical Integrator And Computer (ENIAC) designed & developed by Prof. John Mauchley and his student J. Presper Eckert John von Neumann worked on the ENIAC project It consisted of 18,000 vacuum tubes and 1500 relays It weighted 30 tons and consumed 140 KW of power It has 20 registers each capable of holding 10-digit decimal number ENIAC was programmed by setting up 6000 multiposition switches and connecting multitude of sockets This machine was built around 1946. 9 First generation computers & architecture Electronic Discrete Variable Automatic Computer (EDVAC) is the successor of ENIAC. John von Neumann built his own version of EDVAC. John von Neumann identified that programming computers with huge numbers of switches and cables was slow, tedious, and inflexible He came to realize that the program could be represented in digital form in the computer’s memory, along with the data He also saw that the clumsy serial decimal arithmetic used by the ENIAC, with each digit represented by 10 vacuum tubes (1 on and 9 off) could be replaced by using parallel binary arithmetic 10 First generation computers & architecture The basic design, which he first described, is now known as a von Neumann machine. It was used in the EDSAC (Electronic Delay Storage Automatic Calculator), the first stored-program computer, and even now, more than half a century later, is still the basis for nearly all digital computers. A sketch of the architecture is given 11 First generation computers & architecture Memory consisted of 4096 words (a word holds 40 bits each 0 or 1) Each word held either two 20-bit instructions or a 40-bit signed integer The instructions had 8 bits devoted to telling the instruction type and 12 bits for specifying one of the 4096 memory words. Together, the arithmetic logic unit and the control unit formed the ‘‘brain’’ of the computer. In modern computers they are combined onto a single chip called the CPU (Central Processing Unit). 12 Basic Computing Concepts Basic Computing Concepts At the heart of the modern computer is the microprocessor Commonly called the central processing unit (CPU) A computer takes a stream of instructions (code) and stream of data as input. Take the instruction +, take two numbers and perform addition Take the instruction -, take two numbers and perform subtraction Take the instruction *, take two numbers and perform multiplication Take the instruction /, take two numbers and perform division That is take stream of instructions and stream of data means, continuously perform these instructions on the data The results stream then is made up of results of these operations. 15 Basic Computing Concepts Fundamental functions a computer performs: ○ Reading ○ Modification ○ Writing To achieve the above three fundamental functions, computer needs: ○ Storage ○ Arithmetic logic unit ○ Bus 16 Basic Computing Concepts Storage To say that a computer "read" and "writes" numbers implies that there is at least one number-holding structure that it reads from and writes to. All computers have a place to put numbers - a storage area that can be read from and written to 17 Basic Computing Concepts Arithmetic logic unit To say that computer "modifies" numbers implies that the computer contains a device for performing operations on numbers. This device is ALU. It's the part of the computer that performs arithmetic operations (+, -, *, /) Numbers are read from storage into the ALU's data input port. Once inside the ALU, they are modified by means of an arithmetic calculation Output is written back to storage via the ALU's output port. 18 Basic Computing Concepts Bus To move numbers between the ALU and storage, some means of transmitting numbers is required. The ALU reads from and writes to the data storage area by means of the data bus Data bus is a network of transmission lines for shuttling numbers around inside the computers. Instructions travel into the ALU via the instruction bus 19 Basic Computing Concepts Storage Area Instructio t Resul Data ns ALU 20 Basic Computing Concepts The ALU goes through the following sequence of steps Obtain the two numbers to be added from data storage Add the numbers Place the result back into data storage These three steps are carried out billions (10 9) of times per second on a modern CPU. 21 The Register File The Register File Most computers have a relatively small number of very fast data storage locations attached to the ALU These storage locations are called registers The first x86 computers only had eight of them to work with These registers, which are arrayed in a storage structure called a register file, store the data that the code stream needs. 24 The Register File Building on the previous three-step description of what goes on when a computer's ALU is commanded to add two numbers, we can modify it as follows Obtain two numbers to be added from two source registers Add the numbers Place the result back in a destination register 25 The Register File - Example Addition on a simple computer with 4 registers named A, B, C and D Suppose each of these registers contains a number Task: Add contents of two registers and overwrite the contents of third register with the resulting sum Code Comments A+B=C Add the contents of registers A and B. Place the result in C. Overwrite whatever was there in C 26 The Register File - Example Three steps are to be performed Read the contents of registers A and B Add the contents of A and B Write the result to register C 27 RAM RAM Small set of registers are not very useful to perform reasonable computation To make a viable computer that does useful work, large storage is required This is where computer's main memory comes in Main memory, which in modern computers is always some type of random access memory (RAM), stores data on which computer operates A small portion of that data set at a time is moved to the registers for easy access from the ALU 30 RAM Main Memory Main memory is situated quite a bit farther away from the ALU Registers are internal parts of the microprocessor Main memory is a separate component and is connected to the processor via memory bus Registers ALU 31 CPU Computation How a computer uses main memory, the registers and the ALU to add two numbers Load the two operands (numbers) from main memory into two source registers Add the contents of the source registers and place the result in the destination register using the ALU. To do so, the ALU must perform these steps: ○ Read the contents of registers A and B into the ALUs input ports ○ Add the contents of A & B in the ALU ○ Write the result to the register C via ALU's output port Store the contents of the destination register in main memory 32 Computation The existence of main memory means that the user must manage the flow of information between main memory and the CPU's registers This means that user must issue instructions to more than just the processor's ALU He or she must also issue instruction to the parts of the CPU that handle memory traffic. 33 A Closer Look At The Code Stream A Closer Look At The Code Stream A code stream is an ordered sequence of instructions. Instructions are commands that tell the whole computer (Not just ALU) exactly what actions to perform A computer's list of actions encompasses more than simple arithmetic operations General instruction types If a programmer wants to add two numbers that are in main memory and then store the result back in main memory, he or she must write a list of instructions to tell the computer exactly what to do. 36 A Closer Look At The Code Stream A load instruction to move the two numbers from memory into the registers An add instruction to tell the ALU to add the two numbers A store instruction to tell the computer to place the result of the addition back into memory, overwriting what was previously there. These operations fall into two main categories Arithmetic instructions: These tell the ALU to perform an arithmetic calculation (add, sub, mul, div) Memory-access instructions: These tell the parts of the processor that deal with main memory to move data from and to main memory (load and store). 37 A Closer Look At The Code Stream A detailed example is presented to show how memory access and arithmetic operations work together within the context of the code stream. These examples are based on a hypothetical computer DLW-1. Assume DLW-1 has four registers A, B, C and D. DLW-1 is attached to a bank of main memory that laid out as a line of 256 memory cells numbered #0, #1, #2 …., #255 38 DLW-1's Arithmetic Instruction Format The DLW-1's arithmetic instructions are in the following instruction format Instruction source1, source 2, destination There are four parts to this instruction format each of which is called a field The instruction field specifies the type of operation being performed (+, -, /, *) The two source fields tell the computer which registers hold the two number being operated on The destination field tells the computer which register to place the result in. Instruction: add A, B, C 39 DLW-1's Memory Instruction Format To get the processor to move two operands from main memory into source registers, a memory-access instruction load is used. The load instruction loads the appropriate data from main memory into the appropriate registers. The store instruction takes data from a register and stores it in a location in main memory. In DLW-1's this is represented as: instruction source, destination 40 An Example DLW-1 Program Consider the piece of code in DLW-1 Line Code Comment 1 load #12, A Read the contents of memory cell #12 into register A 2 load #13, B Read the contents of memory cell #13 into register B 3 add A, B, C Add the numbers in registers A, B and store result in register C 4 store C, #14 Write the result of addition from register c into memory cell #14 Main #11 #12 #13 #14 Memory Before add 12 6 2 3 41 After add 12 6 2 8 Register Vs Immediate Immediate Values Programmer needs to know the exact memory location to load and store For small programs this may be feasible. But real computers have billions of possible locations Programmer needs a flexible way to access memory Modern computers allow the contents of a register to be used as memory address The arithmetic instructions required two source registers as input. It is possible to replace one or both source registers with explicit numerical value called immediate value 44 Immediate Values Example, increase contents of register A by 2. For this no need to store 2 in a register and call add A, B, C Instead, write: add, A, 2, A Memory addresses are numeric values. They can also be though of immediate values (a number prefixed by #) Memory addresses can be stored in registers Thus contents of a register D could be constructed as memory address 45 Immediate Value Assume number 12 is stored in register D then the add program is written as Line Code Comment 1 load #D, A Read the contents of memory cell whose location is in register D 2 load #13, B Read the contents of memory cell #13 into register B 3 add A, B, C Add the numbers in registers A, B and store result in register C 4 store C, Write the result of addition from register c into #14 memory cell #14 Main #11 #12 #13 #14 Memory Before add 12 6 2 3 46 Immediate Value Assume number 14 is stored in register D then the add program is written as Line Code Comment 1 load #11, Read the contents of memory cell whose location is in D register D 2 load #D, A Read the contents of memory whose location is in register D 3 load #13, B Read the contents of memory cell #13 into register B 4 add A, B, C Add the numbers in registers A, B and store result in register C 5 store C, Write the result of addition from register c into #14 Main memory #11 cell #14 #13 #12 #14 Memory Before add 12 6 2 3 47 Relative Addressing Relative Addressing A data segment is a block of contiguous memory cells If the starting address of the data segment is known, one can access all other memory locations in the segment using the formula base address + offset Where offset is the distance of bytes of the desired memory location from the data segment's base address 11 12 13 14 15 16 Assume base address is: 11 To access memory address 12: specify 11 + 1 (base address + offset) To access memory address 13: specify 11 + 2 (base address + offset) To access memory address 14: specify 11 + 3 (base address + offset) 50 DLW-1 Program With Use Of Relative Addressing The processor takes the number in D and add 108 Use the result as load's memory address Store also works in the same way as load instruction Load and store units on modern processor contain very fast integer addition hardware. Code Comments load #(D + Read the contents of memory cell at 108), A location #(D + 108) into register A store B, #(D + 108) Write the contents of register B into the memory cell at location #(D + 108) 51 Mechanics of Program Execution Mechanics of Program Execution To run a program, all instructions must be represented in binary notation. To represent instructions in binary notation, instructions are mapped to strings of binary numbers called opcodes. Each opcode designates a different operation 54 Mechanics of Program Execution Register name to 2-bit binary The 3-bit opcode for the code mapping is given as: hypothetical DLW-1 microprocessor is given as: Register Binary code Mnemonic opcode A 00 add 000 B 01 sub 001 C 10 load 010 D 11 store 011 55 Mechanics of Program Execution The binary values representing both the opcodes and the register codes are arranged in one number of a 16-bit (2-byte) format to get a complete machine language instruction. This is a binary number which can be stored in RAM and used by the processor 56 Binary Encoding of Arithmetic Instructions Binary Encoding of Arithmetic Instructions Arithmetic instructions have the simplest machine language instruction formats Table below shows the format for the machine language encoding of a register-type arithmetic instruction mode bit 0 denotes the instruction is register-type mode bit 1 denotes the instruction is immediate type 0 1 2 3 4 5 6 7 mode opcode source1 source2 8 9 10 11 12 13 14 15 destinatio 0 0 0 0 0 0 n 59 Binary Encoding of Arithmetic Instructions – Example add A, B, C 0 1 2 3 4 5 6 7 0 0 0 0 0 0 1 0 8 9 10 11 12 13 14 15 1 0 0 0 0 0 0 0 The machine instruction is: 0000000110000000 60 Binary Encoding of Arithmetic Instructions – Example add C, D, A 0 1 2 3 4 5 6 7 0 0 0 1 0 1 1 0 8 9 10 11 12 13 14 15 0 0 0 0 0 0 0 0 The machine instruction is: 0000101100000000 61 Binary Encoding of Arithmetic Instructions – Example sub C, D, A 0 1 2 3 4 5 6 7 0 0 0 1 0 1 1 1 8 9 10 11 12 13 14 15 0 0 0 0 0 0 0 0 The machine instruction is: 0001101100000000 62 Binary Encoding of Arithmetic Instructions With Immediate Value Binary Encoding of Arithmetic Instructions With Immediate Value Arithmetic instructions with immediate value 0 1 2 3 4 5 6 7 Mode Opcode Source1 destinatio n 8 9 10 11 12 13 14 15 8-bit immediate value 65 Binary Encoding of Arithmetic Instructions With Immediate Value Example: add 5, A, C 0 1 2 3 4 5 6 7 1 0 0 0 0 0 1 0 8 9 10 11 12 13 14 15 0 0 0 0 0 1 0 1 The machine instruction is: 1000001000000101 66 Binary Encoding of Arithmetic Instructions With Immediate Value Example: add 25, A, C 0 1 2 3 4 5 6 7 1 0 0 0 0 0 1 0 8 9 10 11 12 13 14 15 0 0 0 1 1 0 0 1 The machine instruction is: 1000001000011001 67 Binary Encoding of Memory Access Instructions Binary Encoding of Memory Access Instructions Memory access instructions use both register & immediate-type instruction formats Load instruction: immediate-type 0 1 2 3 4 5 6 7 Mode Opcode 0 0 destinatio n 8 9 10 11 12 13 14 15 8-bit immediate source address As load's source is immediate and not register, bits 4 & 5 takes value 0. 70 Binary Encoding of Memory Access Instructions Example immediate load instruction: load #12, A Load instruction: immediate-type; its machine representation is 0 1 2 3 4 5 6 7 1 0 1 0 0 0 0 0 8 9 10 11 12 13 14 15 0 0 0 0 1 1 0 0 As load's source is immediate and not register, bits 4 & 5 takes value 0. The machine instruction is: 1010000000001100 71 Binary Encoding of Memory Access Instructions Load instruction: register-type; its machine representation is 0 1 2 3 4 5 6 7 Mode Opcode Source1 0 0 8 9 10 11 12 13 14 15 Destinatio 0 0 0 0 0 0 n As load's instruction is register-type, bits 6 & 7 takes value 0. 72 Binary Encoding of Memory Access Instructions Load instruction: register-relative addressing; is similar to immediate- type format with base address and the offset stored in the second byte of the instruction 0 1 2 3 4 5 6 7 Mode Opcode Base destinatio n 8 9 10 11 12 13 14 15 8-bit immediate offset 73 The Store Instruction The Store Instruction The register-type binary format for a store instruction is the same as it is for a load, except that the destination field specifies a register containing a destination memory address Source1 field specifies the register containing the data to be stored to memory Immediate-type machine language format for a store is shown below 0 1 2 3 4 5 6 7 Mode Opcode source destinatio n 8 9 10 11 12 13 14 15 8-bit immediate destination address 76 Example Programs Examples Line Assembly Language Machine Language 1 load #12, A 10100000 00001100 2 load #13, B 10100001 00001101 3 add A, B, C 00000001 10000000 4 store C, #14 10111000 00001110 79 Thank You!