csce 121-200 fall 2013
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CSCE 121-200 Fall 2013. Prof. Jennifer L. Welch. Simplified Model of a Computer. How data is represented and stored How data is operated on How a program is stored How the program is executed on the data How programs are translated into machine language. How Data is Represented. - PowerPoint PPT PresentationTRANSCRIPT
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CSCE 121-200Fall 2013
Prof. Jennifer L. Welch
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Simplified Model of a Computer
• How data is represented and stored• How data is operated on• How a program is stored• How the program is executed on the data• How programs are translated into machine
language
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How Data is Represented• Fundamental unit of computer storage is a bit, data
entity that can take on two values, 0 and 1• Given a sequence of bits (01010001), what does it
represent?• Depends on the code
– Base 10 integer: 1,010,001– Base 2 integer: 1*20 + 1*24 + 1*26 = 81 (in base 10)– ASCII: The letter ‘Q’
• Must know the code to decode a sequence• Other codes:
– Negative numbers (e.g., 2’s complement)– Numbers with fractional parts (floating point)
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How Data is Stored• Byte: a group of 8 bits; 28 = 256 possibilities
– 00000000, 00000001, …, 11111110, 11111111
• Memory: long sequence of locations, each big enough to hold one byte, numbered 0, 1, 2, 3,…
• Address: the number of the location• Contents of a location can change• Use consecutive locations to store longer sequences
– e.g., 4 bytes = 1 word
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1 1 0 0 1 1 0 0 0 0 1 1 0 0 1 1 0 1 0 0 1 1 0 0 1 1 0
byte 0 byte 1 byte 2
…..
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Limitations of Finite Data Encodings• Overflow: number is too large
– suppose 1 byte stores integers in base 2, from 0 (00000000) to 255 (11111111). If the byte holds 255, then adding 1 to it results in 0, not 256
• Roundoff error:– insufficient precision (size of word): try to store
1/8, which is .001 in base 2, with only two bits– nonterminating expansions in current base: try to
store 1/3 in base 10, which is .333…– nonterminating expansions in every base:
irrational numbers such as πCSCE 121-200 5
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Kinds of Storage
The computer reads and writes the data stored in it. From fastest to slowest:
• cache: super-fast• main memory: random access, equally fast to
access any address• disk: random access, but significantly slower
than main memory• tape: sequential access, significantly slower
than diskCSCE 121-200 6
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How Data is Operated On
• Central processing unit (CPU) consists of:• arithmetic/logic unit (ALU):
– performs operations on data (e.g., add, multiply)
• registers: special memory cells, hold data used by ALU, even faster than cache
• control unit:– figures out what the ALU should do next– transfers data between main memory and registers
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Machine InstructionsGoal: add the number stored in address 3 and the number stored in
address 6; put the result in address 10.Control unit does the following:1. copies data from main memory address 3 into some register, say 1:
LOAD 3,1
2. copies data in main memory address 6 into some register, say 4: LOAD 6,4
3. tells ALU to add the contents of registers 1 and 4, and put result in some register, say 3:
ADD 1,4,3
4. copies data in register 3 into main memory address 10: STORE 3,10
LOAD, ADD and STORE are machine instructions.How does the control unit know which instruction is next? The program!
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How a Program is Stored
• Program: list of machine instructions using some agreed upon coding convention.
• Example:
• Program is stored same way data is stored!CSCE 121-200 9
opcode 1st operand 2nd operand 3rd operand
ADD 1 4 3
0 0 1 0 0 0 0 1 0 1 0 0 0 0 1 1
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How a Program is Executed• The control unit has
– instruction register: holds current instruction to be executed– program counter: holds address of next instruction in the
program to be fetched from memory
• Program counter tells where the computer is in the program. Usually the next instruction to execute is the next instruction in memory
• Sometimes we want to JUMP to another instruction (e.g., if or while)– unconditional JUMP: always jump to address given– conditional JUMP: only jump if a certain condition is true
(e.g., some register contains 0)CSCE 121-200 10
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Machine Cycle• fetch next instruction, as indicated by the
program counter (PC), and increment PC• decode the bit pattern in the instruction
register – figure out which circuitry needs to be activated to perform the specified instruction
• execute the specified instruction by copying data into registers and activating the ALU to do the right thing– a JUMP may cause the PC to be altered
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Diagram of Architecture
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PC:
IR:
R1:
R2:
R3:
R4:ALUcontrolunit
CPU:
…
… …
0 1 2 3 4 5
95 96 97 98 99 100
first instr. second instr. third instr.
program
datamain
memory:
bus
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Evolution of Programming Languages• Machine languages: all in binary
– machine dependent– painful for people
• Assembly languages: allow symbols for operators and addresses– still machine dependent– slightly less painful– assembler translates assembly language programs into machine
language
• High-level languages: such as Fortran, C, Java, C++– machine independent– easier for people– compiler translates high-level language programs into machine
language
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CompilationChallenges faced by a compiler:• one high-level instruction can correspond to
several machine instructions– Ex: x = (y+3)/z;
• fancier data structures, such as– arrays: must calculate addresses for references to
array elements– structs: similar to arrays, but less regular
• fancier control structures than JUMP– while, repeat, if-then-else, case/switch
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CompilationChallenges faced by a compiler:• functions (a.k.a. procedures, subroutines,
methods): must generate machine language code to:– copy input parameter values– save return address– start executing at beginning of function code– copy output parameter values at the end– set PC to return address
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Compilation Process• Lexical analysis: break up strings of characters into
logical components, called tokens, and discard comments, spaces, etc.– Ex: “total = sum + 55.32” contains 5 tokens
• Parsing: decide how the tokens are related– Ex: “sum + 55.32” is an arithmetic expression, “total = sum
+ 55.32” is an assignment statement
• Code generation: generate machine instructions for each high-level instruction.
• The resulting machine language program, called object code, is written to disk
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Linking• Linker: combines results of compiling different
pieces of the program separately. If pieces refer to each other, these references cannot be resolved during independent compilation.
• Combined code is written to disk.
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function p
…refers to x…
main…declares x…invokes p…
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Loading
• Loader: when it’s time to run the program, the loader copies the object code from disk into main memory– location in main memory is determined by the
operating system, not the programmer
• Loader initializes PC to starting location of program and adjusts JUMP addresses
• Result is an executable
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