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A Discussion on Assemblers

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Title: A Discussion on Assemblers


1
A Discussion on Assemblers
  • Mnemonic instructions, such as LOAD 104, are easy
    for humans to write and understand. Also labels
    can be used to identify particular memory
    locations.
  • They are impossible for computers to understand.
  • Assemblers translate instructions that are
    comprehensible to humans into the machine
    language that is comprehensible to computers
  • We note the distinction between an assembler and
    a compiler In assembly language, there is a
    one-to-one correspondence between a mnemonic
    instruction and its machine code. With compilers,
    this is not usually the case.
  • Assemblers create an object program file from
    mnemonic source code (assembly program) 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.

2
A Discussion on Assemblers
Mnemonic instructions or alphanumeric name Label
or Memory location name
  • Consider our example program at the right.
  • Note that we have included two directives HEX and
    DEC that specify the radix of the constants.
  • The first pass, creates a symbol table and the
    partially-assembled instructions as shown (ie.
    doesnt know X is located at address 104).
  • Also after the first pass, the translated
    instructions are incomplete

3
A Discussion on Assemblers
  • After the second pass, the assembler uses the
    symbol table to fill in the addresses and create
    the corresponding machine language instructions
  • After the second pass, it knows X is located at
    address 104 and that is totally translated to
    machine code

4
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 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.

5
Extending Our Instruction Set
  • We have included three indirect addressing mode
    instructions in the MARIE instruction set.
  • The first two are LOADI X and STOREI X, where X
    specifies the address of the operand to be loaded
    or stored.
  • In RTL
  • It would be the same conceptually for AddI, SubI,
    JumpI and JnS

MAR ? X MBR ? MMAR MAR ? MBR MBR ? MMAR AC ?
MBR
MAR ? X MBR ? MMAR MAR ? MBR MBR ? AC MMAR ?
MBR
STOREI X
LOADI X
6
Extending Our Instruction Set
  • Our first new instruction is the CLEAR
    instruction.
  • All it does is set the contents of the
    accumulator to all zeroes.
  • This is the RTL for CLEAR

AC ? 0
7
A Discussion on Decoding
  • As mentioned earlier, the control unit causes the
    CPU to execute a sequence of steps correctly
  • There are control signals asserted on various
    components in making the components active
  • A computers control unit keeps things
    synchronized, making sure that bits flow to the
    correct components as the components are needed.
  • There are two general ways in which a control
    unit can be implemented hardwired control and
    microprogrammed control.
  • With microprogrammed control, a small program is
    placed into read-only memory in the
    microcontroller.
  • Hardwired controllers implement this program
    using digital logic components. There is a direct
    connection between the control lines and the
    machine instructions.

8
A Discussion on Decoding
  • Your text provides a complete list of the
    register transfer language (or RTN) for each of
    MARIEs instructions.
  • The RTL or RTN actually defines the
    microoperations of the control unit.
  • Each microoperation consists of a distinctive
    signal pattern that is interpreted by the control
    unit and results in the execution of an
    instruction.
  • The signals are fed to combinational circuits
    within the control unit that carry out the
    logical operations for the instruction
  • Recall, the RTL for the Add instruction is

MAR ? X MBR ? MMAR AC ? AC MBR
9
A Discussion on Decoding
  • Each of MARIEs registers and main memory have a
    unique address along the datapath (0 through 7).
  • The addresses take the form of signals issued by
    the control unit.
  • Let us define two sets of three signals.
  • One set, P2, P1, P0, controls reading from memory
    or a register,
  • and the other set consisting of P5, P4, P3,
    controls writing to memory or a register.
  • Lets examine MARIEs MBR (with address 3)
  • Keep in mind from Ch 2 how registers are
    configure using flip-flops

10
A Discussion on Decoding - MBR
The MBR register is enabled for reading when P0
and P1 are high
The MBR register is enabled for writing when P3
and P4 are high
11

A Discussion on Decoding
  • We note that the signal pattern just described is
    the same whether our machine used hardwired or
    microprogrammed control.
  • In hardwired control, the bit pattern of machine
    instruction in the IR is decoded by combinational
    logic.
  • The decoder output works with the control signals
    of the current system state to produce a new set
    of control signals.

Unique output signal corresponding to the opcode
in the IR
Produce the series of signals that result in the
execution of the microoperations
Produces the timing signal for each tick of the
clock (sequential logic used here because the
series of timing signals is repeated) for tick,
a different group of logic can be activated
12
A Discussion on Decoding - ADD
Timing signal added with instruction bits produce
required behavior
  • Bit pattern for the Add 0011 instruction in the
    IR.

Result Here
Control lines and bits controlling the register
functions and the ALU
13
A Discussion on Decoding
  • The hardwired approach is FAST, however, the
    control logic are tied together via circuits and
    complex to modify
  • In microprogrammed control, the control can be
    easier modified
  • In microprogrammed control, instruction microcode
    produces control signal changes.
  • Machine instructions are the input for a
    microprogram that converts the 1s and 0s of an
    instruction into control signals.
  • The microprogram is stored in firmware, which is
    also called the control store.
  • A microcode instruction is retrieved during each
    clock cycle.

14
A Discussion on Decoding
  • All machine instructions are input into the
    microprogram. The microprograms job is to
    convert the machine instructions into control
    signals.
  • The hardwired approach, timing signals from the
    clock are ANDed using combinational logic
    circuits to invoke signals
  • In the microprogram approach, the instruction
    microcode produces changes in the data-path
    signals

15
A Closer Look at Instruction Set
ArchitecturesChapter 5
15
16
Introduction
  • In Ch 4, we learned that machine instructions
    consist of opcodes and operands. Opcodes specify
    the operations to be executed operands specify
    register or memory locations of data.
  • Sections 5.1 and 5.2 builds upon the ideas in
    Chapter 4 and looks more closer at Instruction
    Set Architecture (ISA) specifically, the
    instruction format
  • We will look at different instruction formats
  • We will see the interrelation between machine
    organization and instruction formats.
  • By understanding a high-level languages
    low-level instruction set architecture and
    format, this leads to a deeper understanding of
    the computers architecture in general.
  • As a computer scientist, in understanding the
    computers architecture in more detail, you can
    build more efficient and reliable programsm
  • When a computer architecture is in the design
    phase, the instruction set format must be
    determined first it must match the architecture
    and last for years

17
Instruction Formats
  • Instruction sets are differentiated by the
    following
  • Number of bits per instruction (16, 32, 64).
  • How the data is stored (Stack-based or
    register-based).
  • Number of explicit operands per instruction
    (0,1,2 or 3).
  • Operand location (instructions can be classified
    as register-to-register, register-to-memory or
    memory-to-memory).
  • Types of operations (or instructions) and which
    instructions can access memory or not.
  • Type and size of operands (operands can be
    addresses, numbers or characters).

18
Instruction Formats
  • As mentioned earlier, when a computer
    architecture is in the design phase, the
    instruction set format must be determined first
    it must match the architecture and last for
    years.
  • Instruction set architectures are measured
    several factors
  • the amount of space a program requires.
  • Instruction complexity (ie. decoding required).
  • Instruction length (in bits).
  • total number of instructions in the instruction
    set.

19
Instruction Formats
  • Issues to consider when designing an instruction
    set
  • Instruction length.
  • Whether short, long, or variable.
  • Short uses less space but is limited
  • Fixed size is easier to decode but wastes space
  • Memory organization.
  • Whether byte- or word addressable.
  • If memory has words and not byte-addressable, it
    could be difficult to access a single character
    (4 bits)
  • Number of operands.
  • How should operands be stored in the CPU
  • Number of addressable registers.
  • How many registers ?
  • How should the registers be organized ?
  • Addressing modes.
  • Choose any or all direct, indirect or indexed.
    For example, MARIE used the direct and indirect
    addressing modes
  • Given a byte-addressable machine, should the
    least significant byte be stored at the highest
    or lowest byte address ? Little Endian Vs Big
    Endian debate)

20
Instruction Formats
  • The term Endian refers to a computers Byte
    order, or the way the computer stores the bytes
  • If we have a two-byte integer, the integer may be
    stored so that the least significant byte is
    followed by the most significant byte or vice
    versa.
  • In Little endian machines, the least significant
    byte is followed by the most significant byte.
    (ie MSB-LSB) (most PCs, Intel)
  • Big endian machines store the most significant
    byte first (at the lower address). (ie. LSB-MSB)
    (most UNIX machines, Computer networks, Motorola)
  • As an example, suppose we have the hexadecimal
    number 12345678.
  • The big endian and small endian arrangements of
    the bytes are shown below.

21
Instruction Formats
  • Big endian
  • Is more natural.
  • The sign of the number can be determined by
    looking at the byte at address offset 0.
  • Strings and integers are stored in the same
    order.
  • Doesnt allow values on non-word boundaries (ie
    odd-numbered byte addresses). (ie. if a word is 2
    bytes, must start on 0, 2, 4, 6, 8, etc)
  • Conversion from a 16-bit integer address to a
    32-bit integer address requires arithmetic.
  • Little endian
  • Makes it easier to place values on non-word
    boundaries (ie odd-numbered byte addresses).
  • Conversion from a 16-bit integer address to a
    32-bit integer address does not require any
    arithmetic.

22
Instruction Formats
  • Once the designer decides on how the bytes should
    be ordered in memory,
  • the next consideration for the architecture
    design is how the CPU will store data.
  • We have three choices
  • 1. A stack architecture
  • 2. An accumulator architecture
  • 3. A general purpose register architecture.
  • In choosing one over the other, the tradeoffs are
    simplicity (and cost) of hardware design with
    execution speed and ease of use.

23
Instruction Formats
  • In a stack architecture, instructions and
    operands are implicitly taken from the stack.
  • A stack cannot be accessed randomly.
  • In an accumulator architecture, one operand of a
    binary operation is implicitly in the
    accumulator.
  • One operand is in memory, creating lots of bus
    traffic.
  • In a general purpose register (GPR) architecture,
    registers can be used instead of memory.
  • Faster than accumulator architecture.
  • Efficient implementation for compilers.
  • Results in longer instructions.

24
Instruction Formats
  • Most systems today are GPR systems.
  • There are three types
  • Memory-memory where two or three operands may be
    in memory.
  • Register-memory where at least one operand must
    be in a register.
  • Load-store where no operands may be in memory.
  • The number of operands and the number of
    available registers has a direct affect on
    instruction length.

25
Instruction Formats
  • Machine instructions that have no operands must
    use a stack (last in, first out (LIFO))
  • Stack architectures make use of push and pop
    instructions.
  • Push X places the data in memory location X onto
    the stack
  • Pop X removes the top element in the stack and
    stores it at location X.
  • Stack architectures require us to think about
    arithmetic expressions a little differently.
  • We are accustomed to writing expressions using
    infix notation, such as Z X Y.
  • Stack arithmetic requires that we use postfix
    notation Z XY.
  • This is also called reverse Polish notation,
    (somewhat) in honor of its Polish inventor, Jan
    Lukasiewicz (1878 - 1956).
  • places the operator after the operands

26
Instruction Formats
  • Example 1 Adding use a stack
  • CPU adds the top two elements of the stack,
    popping them both
  • And then push the sum onto the top of the stack
  • Example 2 Subtracting use a stack
  • The top stack element is subtracted from the
    next-to-the top element, both are popped
  • And the result is pushed onto the top of the stack

27
Instruction Formats
  • The principal advantage of postfix notation is
    that parentheses are not used.
  • The infix expression 3 4 is the postfix
    equivalent of 3 4
  • For example, the infix expression,
  • Z (X ? Y) (W ? U),
  • becomes
  • Z X Y ? W U ?
  • in postfix notation.

28
Instruction Formats
  • Example Convert the infix expression (23) - 6/3
    to postfix

The division operator takes next precedence we
replace 6/3 with 6 3 /.
2 3 - 6 3/
The quotient 6/3 is subtracted from the sum of 2
3, so we move the - operator to the end.
2 3 6 3/ -
29
Instruction Formats
  • We have seen how instruction length is affected
    by the number of operands supported by the ISA.
  • In any instruction set, not all instructions
    require the same number of operands.
  • Operations that require no operands, such as
    HALT, necessarily waste some space when
    fixed-length instructions are used.
  • One way to recover some of this space is to use
    expanding opcodes.

30
Instruction Formats
  • A system has 16 registers and 4K of memory.
  • We need 4 bits to access one of the registers. We
    also need 12 bits for a memory address.
  • If the system is to have 16-bit instructions, we
    have two choices for our instructions

31
Instruction Formats
  • If we allow the length of the opcode to vary, we
    could create a very rich instruction set

Is there something missing from this instruction
set?
32
Instruction Formats
  • Example Given 8-bit instructions, is it possible
    to allow the following to be encoded?
  • 3 instructions with two 3-bit operands.
  • 2 instructions with one 4-bit operand.
  • 4 instructions with one 3-bit operand.

33
Instruction Formats
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