Transcript Addressing Modes

Course code: 10CS46
UNIT-2
Machine Instructions and
Programs contd:
Engineered for Tomorrow
Prepared by : Alpana Dahiya and Akshatha P S
Department :Computer Science
Date :
Outline
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Addressing Modes
Assembly Language
Basic Input Output Operations
Stacks and Queues
Subroutines
Additional Instructions
Encoding of Machine Instructions
7 Hours
ADDRESSING MODES
The different ways in which the location of an operand is specified in an
instruction is called as Addressing mode.
Generic Addressing Modes:
• Immediate mode
• Register mode
• Absolute mode
• Indirect mode
• Index mode
• Base with index
• Base with index and offset
• Relative mode
• Auto-increment mode
• Auto-decrement mode
Implementation of Variables and
Constants
Variables:
The value can be changed as needed using the appropriate
instructions. There are 2 accessing modes to access the
variables. They are
• Register Mode
• Absolute Mode
Register Mode:
– The operand is the contents of the processor register. The name
(address) of the register is given in the instruction.
Absolute Mode (Direct Mode):
– The operand is in a memory location. The address of this
location is given explicitly in the instruction.
Constants
Address and data constants can be represented using:
Immediate Addressing Mode:
The operand is specified in the instruction
itself
Move 200immediate, R0
Move #200, R0
Indirection and pointers
• Indirect mode – the Effective address of the operand
is the contents of a register or memory location
whose address appears in the instruction
• The register of memory location that contains the
address of an operand is called a pointer
• Analogy of treasure hunt
– Instead of finding the treasure, we find the address
where we find the treasure
– By changing the contents of register R1 or location A in
the following figure, the same add instruction fetches
different operands to add to register R0.
Add
(R1),R0
Add
(A),R0
Main
memory
B
R1
Operand
B
Register
(a) Through a general-purpose register
A
B
B
Operand
(b) Through a memory location
Figure 2.11. Indirect addressing.
Indirect addressing logic for the program of adding
n numbers using loop
Address
LOOP
Contents
Move
Move
Clear
Add
Add
Decrement
Branch>0
Move
N,R1
#NUM1,R2
R0
(R2),R0
#4,R2
R1
LOOP
R0,SUM
Initialization
Figure 2.12. Use of indirect addressing in the program of Figure
2.10.
Indexing and Arrays
• Index mode – the effective address of the operand is
generated by adding a constant value to the contents of a
register.
• Index register
• X(Ri): EA = X + [Ri]
• The constant X may be given either as an explicit number
or as a symbolic name representing a numerical value.
• If X is shorter than a word, sign-extension is needed.
Indexing and Arrays
• In general, the Index mode facilitates access to
an operand whose location is defined relative
to a reference point within the data structure
in which the operand appears.
• Several variations:
(Ri, Rj): EA = [Ri] + [Rj]
X(Ri, Rj): EA = X + [Ri] + [Rj]
Indexed addressing
Add
20(R1),R2
1000
1000
R1
20
R1
20 = offset
1020
Operand
(a) Offset is given as a constant
Add
1000(R1),R2
1000
20 = offset
1020
Operand
(b) Offset is in the index register
Figure 2.13. Indexed addressing.
n
Student ID
Test 1
Test 2
N
LIST
LIST + 4
LIST + 8
LIST + 12
LIST + 16
Test 3
Student ID
Test 1
Test 2
Test 3
•
•
•
Figure 2.14. A list of students' marks.
Student 1
Student 2
LOOP
Move
#LIST,R0
Clear
R1
Clear
R2
Clear
R3
Move
N,R4
Add
4(R0),R1
Add
8(R0),R2
A dd
12(R0),R3
Add
#16,R0
Decrement
R4
Branch>0
LOOP
Move
R1,SUM1
Move
Move
R2,SUM2
Note: contents of R0,
which is used as
indexed register, are not
changed when it is used
in indexed addressing
mode to access test
scores.
Contents of R0 change
only in last Add
instruction, to move
from one student
record to the next
R3,SUM3
Figure 2.15. Indexed addressing used in accessing test scores in the list in Figure
2.14.
Indexed addressing contd…
• Several variations of the basic form of indexed
addressing provide efficient access to memory
operands
– A second register may be used to contain the offset X,
called as based indexed addressing mode
– Denoted as ( Ri, Rj )
– Effective address is the sum of contents of Ri and Rj
– Second register is called base register.
– Provides more flexibility to the user
– Eg., suppose, in the previous example, instead of only 3
items, each student record contain a large no of items, say
k
– We can replace the three Add instructions by a single
instruction inside a second loop
Indexed addressing contd…
• Several variations of the basic form of indexed
addressing provide efficient access to memory
operands
– A second register may be used to contain the offset X,
called as based indexed addressing mode
– Denoted as ( Ri, Rj )
– Effective address is the sum of contents of Ri and Rj
– Second register is called base register.
– Provides more flexibility to the user
– Eg., suppose, in the previous example, instead of only 3
items, each student record contain a large no of items, say
k
– We can replace the three Add instructions by a single
instruction inside a second loop
Solution
Indexed addressing mode contd..
• based indexed addressing mode with offset
– Uses two registers plus a constant
– X( Ri, Rj ) - Effective address is the sum of the
constant X and the contents of registers Ri and Rj
– This mode implements 3-Dimensional array
•
•
•
•
•
Relative addressing
Till now for index mode, registers we used are general purpose registers
A useful variation of this is to use Program Counter PC instead of a general
purpose register
X(PC) – to address a memory location that is X bytes away from the location
presently pointed to by the program counter
Relative mode- the effective address is determined by the index mode using
the program counter in place of the general-purpose register Ri.
MOST common use is to specify the target address in branch instructions
– Eg., Branch>0 LOOP
– If branch condition is true, the program execution goes to
branch target location identified by name LOOP
– Can compute this location by specifying it as an offset from the
current value of the program counter.
– Since branch target address can be either before or after the
branch instruction, the offset is given as a signed number
Example for relative addressing
mode
Address
LOOP


Contents
Move
Move
Clear
Add
Add
Decrement
Branch>0
Move
N,R1
#NUM1,R2
R0
(R2),R0
#4,R2
R1
LOOP
R0,SUM
Initialization
 Assume 4 instructions starting from LOOP are located at
memory locations 1000,1004,1008 & 1012.
 Updated contents at the time branch target address is
generated is 1016. to branch to location LOOP(1000),
the offset value needed is X = -16
Auto increment addressing logic for the
program of adding n numbers using loop
LOOP
Move
Move
Clear
Add
Decrement
Branch>0
Move
N,R1
#NUM1,R2
R0
(R2)+,R0
R1
LOOP
R0,SUM
Initialization
Figure 2.16. The Autoincrement addressing mode used in the program of Figure 2.12.
Autodecrement mode
• The contents of a register specified in the instruction
are first automatically decremented and are then used
as the effective address of the operand
• Denoted by –(Ri)
• Operands are accessed in descending address order.
• The way the autoincrement and autodecrement modes
are specified in very useful when implementing an
important data structure called a stack.
– Always we can use two instructions to perform the
functionality of autoincrement and autodecrement mode
ASSEMBLY LANGUAGE
• Machine instructions are represented by patterns of 0s
and 1s
– Such patterns are awkward to deal when discussing or
preparing programs
– Hence, usage of symbolic names to represent the patterns
– Such symbolic acronyms( short forms) is called as
mnemonics
• A complete set of symbolic names and rules for their
use constitute a programming language, referred to as
assembly language
• The set of rules for using the symbolic names is called
as syntax of the language
Assembly language
• Programs written in an assembly language can
be automatically translated into a sequence of
machine instructions by a program called an
assembler.
– The user program in its original form is called as
source program
– The assembled machine language program is
called object program
• The assembly language for a given computer
may or may not be case sensitive
Assembly language
 A move instruction is written as
 MOVE R0,SUM
 The mnemonic MOVE represents the binary pattern, or OP
code, for the operation performed by the instruction
 The assembler translates this mnemonic into the binary OP code
that the computer understands
 OP-code followed by at least one blank space character
 Then the information that specifies the operands is given
 In above example , source operand is in register R0
 Followed by specification of the destination operand – separated
from source operand by a comma, with no intervening blanks
 In above example destination operand is in memory location that
has its binary address represented by name SUM
Assembly language
• Since there are several possible addressing modes for
specifying operand locations, the assembly language
must indicate which mode is being used.
– # sign indicates immediate operand ( eg., ADD #5,R3)
– In some assembly languages the immediate addr mode is
intended in the OP-code mnemonic( eg., ADDI 5,R3)
• Indirect addressing mode is usually specified by putting
parentheses around the name denoting pointed
operand.
– If number 5 is to be placed in a memory location whose
address is held in register R2, it is specified as MOVE
#5,(R2)
Assembler directives
 In addition to providing mechanisms for representing instructions in a
program, the assembly language allows programmer to specify other
information needed to translate the source program to object program
 Eg., if name SUM is used to represent value 200, it can be conveyed to
assembler as
 SUM EQU 200
 This statement will not appear in the object program.. Rather informs the
assembler that the name SUM is to be replaced by the value 200
 Such statements are called as assembler directives
 Eg., ORIGIN – where in the memory should the data/instruction block that
follows be placed
 DATAWORD – load the location with specified value
 RETURN – identifies the point at which execution of the program must be
terminated. Assembler returns control to OS by inserting an appropriate
machine instruction
LOOP
100
104
Move
Move
N,R1
#NUM1,R2
108
112
116
Clear
Add
Add
R0
(R2),R0
#4,R2
120
124
128
132
Decrement
Branch>0
Move
R1
LOOP
R0,SUM
SUM
N
200
204
NUM1
NUM2
208
212
NUM n
Memory
address
label
Assembler
directives
SUM
N
NUM1
Statements
that generate
machine
instructions
100
Assembler
directives
START
LOOP
Operation
Addressing
or data
information
EQU
200
ORIGIN
204
DATAWORD 100
RESERVE
400
ORIGIN
100
MOVE
N,R1
MOVE
#NUM1,R2
CLR
R0
ADD
(R2),R0
ADD
#4,R2
DEC
R1
BGTZ
LOOP
MOVE
R0,SUM
RETURN
END
START
604
Fig2.17.Memory arrangement for program in Fig 2.12.
Figure 2.18. Assembly language representation
Any statement
that results in
instructions or
data being placed
in memory
location may be
given a memory
address label.
Label is assigned
the value equal
to the address of
that location
Memory
address
label
Assembler directives
SUM
N
NUM1
Statements that
generate
machine
instructions
Assemblerdirectives
START
LOOP
Operation
EQU
ORIGIN
DATAWORD
RESERVE
ORIGIN
MOVE
MOVE
CLR
ADD
ADD
DEC
BGTZ
MOVE
RETURN
END
Addressing
or data
information
200
204
100
400
100
N,R1
#NUM1,R2
R0
(R2),R0
#4,R2
R1
LOOP
R0,SUM
START
Figure 2.18. Assembly language representation for the program in Figure 2.17.
contd….
• Most assembly languages require statements in a
source program to be written in the form
– Label Operation Operand(s) Comment
– Label – is optional and is associated with memory
address where the machine language instruction or
the data items are stored
– Operation field contains the OP-code mnemonic of
the desired instruction or assembler directive
– Operand field contains addressing information for
accessing one or more operands, depending on the
type of instruction
– Comment field is ignored by the assembler program
• Used for documentation purposes
Assembly and execution of programs
 ASSEMBLER replaces
 All symbols denoting operations and addressing modes with
binary codes
 All names and labels with their actual values.
 In a branch instruction, branch target is not replaced by actual
address.
 Usually it uses relative addressing mode. Assembler computes branch
offset, and puts into the machine instruction
 The assembler keeps track of all names and the numerical values
that correspond to them in a symbol table
 When a name appears a second time, it is replaced with its value from
the table.
 A problem arises if some name appears before it is given a value. Eg.,
forward branching.
 Solution is to scan through the program twice – once to collect the values
of all names and second time to substitute values for all names from
symbol table. Such an assembler is called as TWO-PASS ASSEMBLER
Assembly and execution of programs
 Assembler stores object program on a magnetic disk
 The object program must be loaded into memory
before it is executed
 A utility program called loader does this. For this loader
must already be in memory
 Loader must know the length of the program and the
address in the memory where it will be stored
 Assembler usually places this info in a header preceding the object
code
 When the object program begins execution, it proceeds to
completion unless there are errors.
 To help debugging, system software usually includes a debugger
program
Number notations
• Can use either decimal, or binary or
hexadecimal notation
• Eg., ADD #93, R1
• This is in decimal
• In binary – ADD #%01011101,R1
– Here % tells that we are using a binary number
• In hexadecimal – ADD #$5D, R1
– Here $ denotes the usage of hexadecimal
BASIC INPUT/OUTPUT OPERATIONS
 Till now we have assumed that the data operated in stored in memory
already.
 Now, let’s see how data transfer takes place between computer and
outside world
 The I/O operations are very important and the way they are done can
significantly affect the performance of the computer
 The rate of data transfer from the keyboard to a computer is limited by the
typing speed of the user
 Few characters per second
 The rate of output transfers from the computer is determined by the rate
at which characters can be transmitted over the link between the
computer and display device.
 Typically several thousand characters per second
 Both are much slower than the speed of processor
 Can execute many millions of instructions per second
 Hence need for the mechanisms to synchronize the transfer of data
between them.
Solution to speed disparity
• On output, the processor sends the first
character and then waits for a signal from the
display that character has been received
– Then it sends the second character and so on
• For input, the processor waits for a signal
indicating that a character key has been struck
and that its code is available in some buffer
register associated with keyboard
– Then only will processor proceed to read the code
SIN
Key board
SOUT
Display
Figure 2.19 Bus connection for processor
, keyboard, and display
.
DATAIN/DATAOUT – buffer
registers
SIN/SOUT – status control flags
The striking of a key on keyboard does not automatically cause the
corresponding character to be displayed on the screen!
One block of instructions in I/O program causes the character to be
transferred to processor and another block of instructions will cause the
character to be displayed on the screen
How data is read?
• The technique used is called program – controlled I/O
• Striking a key stores the corresponding character code
in a 8-bit buffer register called DATAIN.
• To inform the processor that a valid character is in
DATAIN, a status control flag SIN is set to 1.
• A program monitors SIN, and when SIN is set to 1, the
processor reads the contents of DATAIN.
• When the character is transferred to the processor, SIN
is automatically cleared to 0.
• When a second character is read, SIN is again set to 1
and the process repeats
How data is displayed?
• A buffer register DATAOUT and status control flag SOUT
is used.
• When SOUT is 1, display is ready to receive a character
• Under the control of a program, ( hence program –
controlled I/O), the processor monitors SOUT and when
SOUT is set to 1, the processor transfers a character
code to DATAOUT
• The transfer of character to DATAOUT clears SOUT to 0.
–When the display is ready to receive a second character,
again SOUT is set to 1.
–SIN , SOUT , DATAIN, DATAOUT are part of circuitry called
as device interface, connected to processor through bus
Programs for I/O
• Assume that SIN is set to 0 initially and SOUT is set to 1.
–This initialization is performed by device control circuits
• How to refer to the buffers DATAIN and DATAOUT?
–Many computers use arrangement called as memory-mapped
I/O
–Some memory address values are used to refer to peripheral
device buffer registers such as DATAIN and DATAOUT
–Thus no special instructions required to access the contents of
these registers
–Eg., Movebyte DATAIN, R1 moves the contents of keyboard
character buffer DATAIN to R1
–Similarly, Movebyte R1, DATAOUT
–The status flags SIN and SOUT are automatically cleared when
the buffer registers are referenced
–Movebyte signifies that the operand is a byte, not a word
Programs for I/O
• It is common to include SIN and SOUT in device status
registers, which have distinct memory addresses
–Assume that the status registers are INSTATUS and OUTSTATUS
respectively for input and output
–Also assume that SIN and SOUT are referred by bit3 in the
registers.
• Read operation
–READWAIT
–
–
Testbit
Branch=0
Movebyte
#3, INSTATUS
READWAIT
DATAIN,R1
Testbit
Branch=0
Movebyte
#3,OUTSTATUS
WRITEWAIT
R1,DATAOUT
• Write operation
–WRITEWAIT
–
–
Move
#LOC,R0
READ
TestBit
Branch=0
MoveByte
#3,INSTATUS
READ
DATAIN,(R0)
ECHO
TestBit
Branch=0
MoveByte
#3,OUTST ATUS
ECHO
(R0),DATAOUT
Compare
#CR,(R0)+
Branch 0
READ
Initialize pointer register R0 to pointto the
address of the firstlocation in memory
wherethe characters are tobe stored.
Wait for a character tobe entered
in the keyboardbuffer DATAIN.
Transferthe character fromDATAIN into
the memory(this clears SIN to 0).
Wait for the display to become ready.
Move thecharacterjust read to the display
buffer register(this clearsSOUT to 0).
Checkif the character just read is CR
(carriage return). If it is not CR, then
branch bac k and read another character.
Also, increment the pointer to store the
next character.
Figure 2.20. A program that reads a line of characters and displays it.
STACKS and QUEUES
• Stacks mainly used to organize control and information
linkage between main program and the subroutine
• A stack is a list of elements, usually words or bytes, with
the accessing restriction that elements can be added or
removed at one end of the list only.
–This end is called as the top of stack
–The other end is called as bottom
–Hence also called as pushdown stack
–Eg., a pile of trays in a cafetaria
–Storage mechanism is also referred by LIFO ( Last-in-FirstOut)
–Push – put an element on top of stack
–Pop – remove an element from top of stack
0
Stack
pointer
register
SP
•
•
•
- 28
Current
top element
17
739
Stack
BOTTOM
Normal practice
is stack grows in
the direction of
decreasing
memory
addresses
k
2 - 1
•
•
•
43
•
•
•
Figure 2.21. A stack of words in the memory.
Bottom
element
Stack
pointer(SP) – is
a processor
register used to
keep track of
the address of
the top
element of
stack
Push and Pop operations on Stack
• Stack pointer keeps track of the address of the top of
stack
• To Push an element
–Subtract #4,SP
–Move NEWITEM,(SP)
• To Pop an element
–Move (SP),ITEM
–Add #4,SP
• If processor support autoincrement and autodecrement
• Push  Move NEWITEM,-(SP)
• Pop  Move (SP)+,ITEM
Stacks contd…
• Care has to be taken that stack is not popped
when empty and pushed when full
• Usage of Compare instruction
• Compare src,dst
• [dst] – [src] and affect flags accordingly
• eg/., BOTTOM – at addres 2000
• Till location 1500
SAFEPOP
Compare
Branch> 0
#2000,SP
EMPTYERROR
Move
(SP)+,ITEM
Check to seeif the stack pointer contains
an addressvaluegreaterthan 2000. If it
does,the stack is empty . Branch to the
routine EMPTYERROR for appropriate
action.
Otherwise,pop the top of the stack into
memory location ITEM.
(a) Routine for a safe pop operation
SAFEPUSH
Compare
Branch 0
#1500,SP
FULLERROR
Move
NEWITEM,– (SP)
Check to seeif the stack pointer
contains an addressvalueequal
to or less than1500. If it does, the
stack is full. Branch to the routine
FULLERROR for appropriateaction.
Otherwise,push the element in memory
location NEWITEM onto the stack.
(b) Routine for a safe push operation
Figure 2.23. Checking for empty and full errors in pop and push operations.
Queue – First In First Out List
 Data elements are inserted in memory at the increasing order of
memory addresses i.e., at the back of Queue.
 Data elements are retrieved from memory at the decreasing order
of memory addresses i.e. from the front end of Queue.
 Operations such as QINSERT and QDELETE can be performed on a
queue to insert and delete & element respectively.
 A Queue consisting of 6 elements may appear as follows:
Difference between implementation of
stacks and queues
• Stacks
–One end of a stack is fixed ( bottom ), while the other end rises
and falls as data are pushed and popped. Hence a single pointer
is needed to point to the top of stack at any given time.
–A stack is assigned a fixed amount of space in memory.
• Queues
–Both ends of queue move to higher addresses as data are added
at the back and removed from the front. So two pointers are
needed to keep track of the two ends of the queue.
–Without further control, a queue would continuously move
through the memory of a computer in the direction of higher
addresses.
• One way to limit the queue to a fixed region in memory is to use a circular
buffer.
SUBROUTINES
• In a given program, it is many times needed to perform
a particular subtask on different data values. Such a
sub task is called as subroutine
– Eg., a subroutine to calculate sine function, or to sort a list
of values in increasing or decreasing order
• One copy of the instructions that constitute subroutine
is placed in the memory, and any program that requires
the use of subroutine branches to its starting location.
– This branching to a subroutine is termed as calling the
subroutine
– The instruction which performs this is Call instruction
–
Subroutines - Linkage
• After a subroutine has been executed, the calling program must resume
execution continuing immediately after the instruction that called the
subroutine.
– This is performed by a Return instruction
– Since a subroutine may be called from different places, we need to make
provision to return to appropriate location
– The location where the calling program resumes execution is the location
pointed to by the updated PC while the Call instruction is being executed.
– Hence, need to save the contents of PC by Call instruction to enable correct
return.
• The method in which computer calls and returns from subroutines is
referred to as subroutine linkage method
– Simplest way is to save the return address in a specific location, which may be
a register dedicated to perform this function
– Such a register is called as link register
– When the subroutine completes its task, the Return instruction returns to the
calling program by branching indirectly through the link register
Memory
location
Memory
location
Calling program
200
204
Call
SUB
next instruction
Subroutine SUB
1000
first instruction
Return
1000
PC
204
Link
204
Call
Call instruction performs the
Figure 2.24.
following operations
i. Store the contents of PC in
link register
ii. Branch to the target address
specified by the instruction
Subroutine linkage using a link register.
Return
Return instruction
performs the following
operation
i. Branch to the address
contained in link
register
Subroutines - nesting and processor
stack
• Subroutine nesting – it’s a practice where in one subroutine calls another
subroutine.
– In this case, the return address of the second call is also stored in link register, destroying its
previous contents.
– Hence, need to save the contents of link register in some other location before calling another
subroutine.
– Else the return address of first subroutine is lost
• Subroutine nesting can be carried out to any depth.
– Last subroutine that is called completes computations and returns to the subroutine which
called it.
– Hence return addresses are generated and used in a last-in-first-out order.
– Hence return addresses can be pushed onto a stack.
– Many processors do this automatically as one of the operations performed by call instruction
– A special pointer called stack pointer, points to a stack called processor stack.
– the call instruction pushes the contents of the PC onto the processor stack and loads the
subroutine address into the PC.
– The return instruction pops the return address from the processor stack into the PC
Subroutines - parameter passing
• When calling a subroutine, a program must provide to
the subroutine the parameters, (operands or their
addresses), to be used in the computation
• The subroutine returns other parameters ( the results of
computation) back.
• This exchange of information between calling program
and subroutine is called as parameter passing
• Can be done in several ways
–Can place the parameters in registers or memory locations
• Placing in processor registers in highly efficient and straight forward
• No of parameters that can be passed is limited by the number of
general purpose registers available
–The parameters may also be placed on processor stack.
• Highly flexible and can handle a large no of parameters
Addition of N numbers in a loop using subroutines. Parameters are passed through
registers
Calling program
Move
Move
Call
Move
..
.
N,R1
#NUM1,R2
LISTADD
R0,SUM
R1 servesas a counter.
R2 pointsto thelist.
Call subroutine.
Save result.
Clear
Add
Decrement
Branch>0
Return
R0
(R2)+,R0
R1
LOOP
Initialize sumto 0.
Add entry from list.
Subroutine
LISTADD
LOOP
Return to calling program.
Figure 2.25. Program of Figure 2.16 written as a subroutine; parameters passed through registers.
Program to add a list
on n numbers written
as a subroutine.
Parameters are
passed through
processor stack
NOTE : parameter passing
either by pass by value or
pass by reference
n – is passed by value (actual
value is passed)
NUM1 – passed by
reference. address is passed
( not the actual entries of
the list but the address of
first element is passed
Figure 2.26. Program of Figure 2.16 written as a subroutine; parameters passed on the stack.
Subroutines – Stack Frame
• The locations which will be used by the
subroutine on stack for private work,
constituting a private work space for the
subroutine, which is created at the time the
subroutine is entered and freed up when the
subroutine returns control to the calling
program, is called as stack frame.
• Can also allocate space for local memory
variables
Stack frames
• In addition to stack pointer, it is useful to have
another additional pointer register, called
frame pointer (FP)
• The local variables are local to the subroutine,
hence can be allocated space in stack frame
associated with the subroutine
• FP points to the location just above the return
address
• Contents of FP remain fixed throughout the
execution of subroutine
Move FP,-(SP)
Move SP,FP
SP
(stack pointer)
saved [R1]
saved [R0]
localvar3
localvar2
localvar1
FP
(frame pointer)
saved [FP]
Return address
Subtract #12,SP
Stack
frame
for
called
subroutine
param1
Access parameters
using index addr
mode.
8(FP), 12(FP)
Local variables can be Figure 2.27.
accessed using
addresses
-4(FP), -8(FP)
param2
param3
param4
Old TOS
(top-of-stack)
A subroutine stack frame example.
The calling program is
responsible for removing
parameters from stack
frame, some of which may
be results passed back by
the subroutine
Stack frames for nested subroutines
• Stack is a proper data structure for holding
return addresses when subroutines are nested
• Complete stack frames for nested subroutines
build up on the processor stack
• Saved contents of FP in the current stack
frame are the frame pointer contents for the
stack frame of the subroutine that called the
current subroutine
Memory
lo cation
Main
Figure shows
only flow of
control and
data between
subroutines.
Not the actual
computations
Instructions
program
2000
2004
2008
2012
2016
2020
First
2100
2104
2108
2112
3000
P ARAM2, – (SP)
P ARAM1, – (SP)
SUB1
(SP),RESUL T
#8,SP
Place parameters
FP ,– (SP)
SP ,FP
R0 – R3, – (SP)
8(FP),R0
12(FP),R1
Sav e frame poin ter register.
Load the frame poin ter.
Sav e registers.
Get first parameter.
Get second parameter.
P ARAM3,
SUB2
(SP)+,R2
Place a parameter
on stack.
Pop SUB2
in to R2.
Store result.
Restore stack
on stack.
lev el.
subroutine
SUB1
2160
2164
Second
..
.
Mo v e
Mo v e
Call
Mo v e
Add
next instruction
..
.
Commen ts
Mo v e
Mo v e
Mo v eMultiple
Mo v e
Mo v e
..
.
Mo v e
Call
Mo v e
..
.
Mo v e
Mo v eMultiple
Mo v e
Return
R3,8(FP)
(SP)+,R0
(SP)+,FP
– (SP)
– R3
result
Place answ er on stack.
Restore registers.
Restore frame p oin ter register.
Return to Main program.
subroutine
SUB2
Mo v e
Mo v e
Mo v eMultiple
Mo v e
..
.
Mo v e
Mo v eMultiple
Mo v e
Return
FP ,– (SP)
SP ,FP
R0 – R1, – (SP)
8(FP),R0
Sav e frame p oin ter register.
Load the frame p oin ter.
Sav e registers R0 and R1.
Get the parameter.
R1,8(FP)
(SP)+,R0
(SP)+,FP
Place SUB2 result on stack.
Restore registers R0 and R1.
Restore frame p oin ter register.
Return to Subroutine
1.
– R1
Figure 2.28. Nested subroutines.
FP
[R1] from SUB1
[R0] from SUB1
[FP] from SUB1
2164
param3
[R3] from Main
FP
[R2] from Main
[R1] from Main
[R0] from Main
[FP] from Main
2012
Stack
frame
for
second
subroutine
Stack
frame
for
first
subroutine
param1
param2
Old TOS
Figure 2.29. Stack frames for Figure 2.28.
ADDITIONAL INSTRUCTIONS
• Instructions so far
–Move
–Load
–Store
–Clear
–Add
–Subtract
–Increment
–Decrement
–Branch
–Testbit
–Compare
–Call
–Return
Even this small set of instructions has a
number of redundancies
Eg., Load and Store instructions can be
replaced by a single Move instruction
Increment by Add and Decrement by
Subtract instruction
Clear by a Move ( moving zero )
Additional instructions
–Logic instructions
–Not
• Complement all the bits in destination operand
• Not R0 ( complement all bits of R0)
• Can get 2’s complement representation of a negative number
– Not R0 ( R0 contains, say 3)
– Add #1,R0
• Can get 2’s complement of a number by a single instruction
– Negate R0
–Bitwise And , Or
• Performs and & or operations bitwise
• And #$FF000000,R0
–Problem : write a program which checks whether an
entered 4letter word contains Z in the beginning.
–
–
–
–
Additional instructions
Answer
And $#FF000000, R0
Compare #$5A000000,R
Branch=0 YES
– And instruction is often used in practical programming
tasks where all bits of an operand except for some
specified field are to be cleared to zero
– Problem: Write program to clear the bits b10 to b31 in a
4 byte word
Additional instructions – shift and
rotate
instructions
– Shift instructions
• For general operands we use logical shifts
• For signed numbers, we use arithmetic shifts ( to preserve
the sign of the number)
– Logical Shifts
•
•
•
•
2 instructions – for left shift and for right shift
LShiftL count, destination
LShiftR count, destination
Count can be given either as immediate operand or given
inside a processor register
• Vacated positions are filled with zeros
• The bits shifted out are passed through the Carry flag, C, and
then dropped
C
R0
0
before:
0
0
1
1
1
0
.
.
.
0
1
1
after:
1
1
1
0
.
.
.
0
1
1
0
0
(a) Logical shift left
LShiftL
0
#2,R0
R0
C
before:
0
1
1
1
0
.
.
.
0
1
1
0
after:
0
0
0
1
1
1
0
.
.
.
0
1
(b) Logical shift r
ight
LShiftR #2,R0
R0
C
before:
1
0
0
1
1
.
.
.
0
1
0
0
after:
1
1
1
0
0
1
1
.
.
.
0
1
(c) Ar ithmetic shift r
ight
AShiftR #2,R0
Figure 2.30. Logical and arithmetic shift instructions.
Additional instructions – Shift and
Rotate instructions
• Digit packing
• 2 decimal digits represented in ASCII are
located in memory at byte locations LOC and
LOC + 1
• Represent each of these digits in 4-bit BCD
and store in a single byte location PACKED (the
result is said to be in packed-BCD format)
• Hint: rightmost 4 bits of the ASCII code for a
decimal digit correspond to the BCD code for
the digit
Move
MoveByte
LShiftL
MoveByte
And
Or
MoveByte
#LOC,R0
(R0)+,R1
#4,R1
(R0),R2
#$F,R2
R1,R2
R2,PACKED
R0 points to data.
Load first byte into R1.
Shift left by 4 bit positions.
Load secondbyte into R2.
Eliminate high-order bits.
ConcatenatetheBCD digits.
Store the result.
Figure 2.31. A routine that packs two BCD digits.
Additional instructions – Shift and
Rotate instructions
• Arithmetic Shifts
• Shifting a number one bit left is same as
multiplying the number by 2
– Overflow may occur
• Shifting right is equivalent to dividing the
number by 2
– Reminder is lost
– On a right shift sign-bit must be repeated as the
fill in bit for the vacated position, main difference
between Arithmetic shift and logical shift
Additional instructions – Shift and
Rotate instructions
•
•
•
•
•
•
Rotate instructions
4 in number
Rotate left without Carry – RotateL
Rotate left with Carry – RotateLC
Rotate right without Carry – RotateR
Rotate right with Carry - RotateRC
C
R0
before:
0
0
1
1
1
0
.
.
.
0
1
1
after:
1
1
1
0
.
.
.
0
1
1
0
1
(a) Rotate left without carry
RotateL
C
#2,R0
R0
before:
0
0
1
1
1
0
.
.
.
0
1
1
after:
1
1
1
0
.
.
.
0
1
1
0
0
(b) Rotate left with carry
RotateLC #2,R0
R0
C
before:
0
1
1
1
0
.
.
.
0
1
1
0
after:
1
1
0
1
1
1
0
.
.
.
0
1
(c) Rotate r
ight without carr
y
RotateR #2,R0
R0
C
before:
0
1
1
1
0
.
.
.
0
1
1
0
after:
1
0
0
1
1
1
0
.
.
.
0
1
ight with carr
y
(d) Rotate r
RotateRC #2,R0
Figure 2.32. Rotate instructions.
Additional instructions – Shift and
Rotate instructions
Multiplication and Division
• Multiply Ri,Rj
• Product can be as large as 2n bits for 2 n-bit numbers
• Answer will not necessarily fit in Rj
• Lower order bits in Rj, higher order bits in Rj+1
• Some systems provide Divide Ri ,Rj
• Rj/Ri and quotient in Rj
• May place remainder in Rj+1
• Computers which do not have divide and multiply
instructions can implement same using basic
instructions like add, subtract, shift and rotate
ENCODING OF MACHINE
INSTRUCTIONS
• Till now we have not used acronyms for
instructions because they are specific to a
particular processor in market ( eg., we used
Move instead of MOV )
• To be executed, an instruction must be encoded
in a compact binary pattern. Such encoded
instructions are properly referred to as machine
instructions
– The instructions that use symbolic names and
acronyms are called as assembly language
instructions, which are converted to machine
instructions using assembler program
Encoding of machine instructions
• The type of operation to be performed and the type of
operands used is specified using an encoded binary
pattern referred to as the OP code for the given
instruction
–Suppose 8 bits are assigned for OP code to specify upto
256 different instructions
–If we assume that word length is 32-bits and if we want to
fit an entire instruction into a single memory word, then
we are left with 24 bits to specify rest of the information
• Eg., Add R1,R2 has to specify the registers R1 and R2, in
addition to the OP code
• If system has 16 registers, then 4 bits are required to
identify each register and some additional bits to specify
that we are using register addressing mode
Encoding of machine instructions
• Move 24(R0) ,R5
– 16 bits to denote OP code, 2 registers, some bits to
express that source operand uses index addressing
mode and that index value is 24
• Suppose 3 bits are used to specify addressing
mode
– We need 6 bits – to denote the chosen addressing
mode for both source and destination operands
– 32 – ( 16+6) = 10 bits left to give the index value (
signed number )
– Eg., LShiftR #2,R0
– Move #$3A, R1 how many bits left to specify
immediate value?
Encoding of machine instructions
• Branch>0 LOOP
– Here 8 bits for OP code. How many bits for Branch
OFFSET?
– What is the limit for the branch target address
with respect to branch insrtuction?
– Jump instruction - uses Absolute or Register
Indirect mode to specify branch target
Encoding of machine instructions
• CISC processors use instructions that are complex and
require more than a single word to specify the
instruction
• Consider instruction
• Add LOC,R2
– Here 8 bits for OP code, 10 bits for addr mode and
specifying the register, remaining 14 bits cannot hold a 32bit address of the absolute addressing mode
– Hence we need to go either for multiple word instruction
– Or
– Register Indirect
• Load address of LOC to R3
• Add (R3),R2
8
7
7
10
OP code
Source
Dest
Other info
(a) One-word instruction
OP code
Source
Dest
Other info
Memory address/Immediate operand
(b) Two-word instruction
OP code
Ri
Rj
Rk
Other info
(c) Three-operand instruction
Figure 2.39. Encoding instructions into 32-bit words.
Encoding of machine instructions
• How to load 32-bit address to register?
• One way is to use indexed addressing mode to load
the address into register ( for this need to place the
address in a memory location close to the program)
• Other option is to use Logical and Shift instructions –
passing the address by parts
• The restriction that the instruction must fit in a single
word ( 32 bit here ) is the style of RISC machines
– Places a restriction that all manipulation of data
contained only in processor registers
– Move (R3) ,R1
– Add R1,R2
Thank You