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Memory Segmentation in 8086 Microprocessor Video Lecture - Computer Science Engineering (CSE)
FAQs on Memory Segmentation in 8086 Microprocessor Video Lecture - Computer Science Engineering (CSE)
| 1. What is memory segmentation in the 8086 microprocessor? |  |
Ans. Memory segmentation in the 8086 microprocessor refers to the division of memory into segments or sections. This allows the microprocessor to access a large memory space beyond the limitations of a 16-bit address bus.
| 2. How does memory segmentation work in the 8086 microprocessor? | |
Ans. In the 8086 microprocessor, memory segmentation works by dividing the memory space into multiple segments of up to 64KB each. Each segment is identified by a segment register, which holds the starting address of that segment. The effective address is calculated by combining the value in the segment register with the offset address.
| 3. What are the advantages of memory segmentation in the 8086 microprocessor? | |
Ans. Memory segmentation in the 8086 microprocessor offers several advantages. It allows for efficient memory management by dividing the memory space into smaller segments. It enables the use of larger memory beyond the 64KB limit of a 16-bit address bus. It also allows for the sharing of memory among different programs or processes, improving overall system performance.
| 4. Are there any limitations or challenges with memory segmentation in the 8086 microprocessor? | |
Ans. Yes, there are limitations and challenges with memory segmentation in the 8086 microprocessor. One limitation is the complexity of programming due to the need to calculate effective addresses using segment registers and offsets. Another challenge is the potential for memory fragmentation, where free memory is divided into small, non-contiguous segments, making it difficult to allocate larger blocks of memory.
| 5. How does memory segmentation differ from memory paging in the 8086 microprocessor? | |
Ans. Memory segmentation and memory paging are two different memory management techniques in the 8086 microprocessor. Memory segmentation divides the memory into segments of fixed sizes, while memory paging divides the memory into fixed-size pages. Segmentation uses segment registers to access memory, while paging uses page tables. Additionally, segmentation allows for a larger memory space beyond the 64KB limit, while paging provides a mechanism for virtual memory and memory protection.
Text Transcript from Video
in this session we'll discuss about
memory segmentation now what do you mean
by segmentation as we know what is a
memory and how much amount of memory
that can be interfaced with a gr8 6 is 1
MB that depends on the number of address
lines as it has 20 address lines so we
can interface up to 1 MB of memory now
what do you mean by segmentation
segmentation generally means dividing or
making it parts now here memory
segmentation is nothing but dividing or
dividing V entire 1 MB memory into some
parts or some logical segments ok so we
are dividing this entire 1 MB memory
into a number of logical segments ok and
the size of each and every logical
segment is 64 kilobytes why we are going
with only 64 kilobytes of memory
segmentation or 64 kilobytes of the size
of each and every segment because here
we have 16-bit registers for accessing
the memory so as we have 16-bit
registers we can only access up to 64
kilobytes with the help of 16-bit
registers so that is why here each and
every segment size is of 64 kilobytes ok
now so how many segments are present
there are totally this complete 1 MB
memory is divided into 16 logical
segments why because 64 kilobytes into
16 segments or 16 is nothing but 1 and B
ok so here the size of each and every
segment is 64 kilobytes and all these
segments are named as code segment data
segment extra segment and stack segment
so there are only 4 segments and there
are only 4 names given to these segments
like code segment data segment extra
segment as well our stack segment okay
and there can be any number of code
segments there can be any number of data
segments there can be any number of
extra segments as well as there can be
any number of stack segments and the
number of code data as well as extra as
well as stack segments depends on the
user program okay now here so the
complete physically available memory
be divided into number of logical
segments as I have told you now we have
the starting addresses five zeros and
the ending address is 5x okay now here
this address is represented in
hexadecimal form okay now if you
represent the same address in binary
form now it will give you a 20-bit
address which is nothing but the exact
location where actually the operand is
recited okay now it is a 20-bit address
but by using 16-bit address or 16-bit
register we cannot access or we cannot
we cannot retract the data which is
present in this particular memory
locations so for that purpose we use one
of the register for holding one kind of
address and the other register for
holding the other kind of address okay
now let us see what is that now here as
we have four segments code segment data
segment extra segment as well as data
say stack segment code segment is used
for holding the instructions so whatever
the instructions written by the user now
that are stored inside this code segment
and coming to the stack segment which is
mainly used for holding the return
addresses okay and this is used in case
of function calls or as interrupt
service routines okay and data in extra
segments are mainly used for holding the
data that is user data okay and as I
told you that the size of each segment
is 64 kilobytes and there are 16 logical
segments a segment may be located
anywhere in the memory so any of these
segments are located anywhere in the
memory and at a time only four segments
are active and those four signals have
code segment data segment extra segment
and then stacks it the four segments are
code okay and a segment is a 64
kilobytes block of memory 16 bit
contents of segment registers in the bi
you actually point to the starting
location of a particular segment now we
have some segment registers inside the
VIU now all these segment registers will
be pointing to the actual point or
starting location of the particular
segment we'll see that okay and what is
the advantage of going with segmentation
now here as
1 MB memory but we can access this 1 MB
memory with the help of 16-bit registers
ok and then the code or data will be
placed anywhere in the program that is
will be placed anywhere in the memory ok
and the segment registers are used to
allow the instruction data or stack
portion of a program to be more than 64
kilobytes long the above can be achieved
by using more than one code data or
stack signals if if assuming that we
have written a program now the code part
of the program is more than 64 kilobytes
then our two or three code segments are
assigned for that particular code that
depends on the size of the colon okay as
I told you that n number of code data
orders stack extra segments can be
assigned in the memory now let us see
what are these segments and sisters we
have totally four segment registers code
segment register data segment register
extra segment register and stack segment
register Nanyan confused with register
as well as a segment in the memory
those are segments and these are
registers now all these registers are
used for holding the starting address of
their respective segment it means code
segment register is used for holding the
starting address of code segment data
segment register used for holding the
starting address of data segment extra
segment register used for holding the
starting address of extra segment and
stack segment register is used for
coding the starting address of the stack
segment okay now all our 16-bit
registers but so as all these are
sixteen registers they would be holding
only the segment address all of these
starting headers now how to access each
and every individual location from the
memory okay now for accessing individual
memory locations inside the memory we
need offset address we need offset
headers and those offset addresses are
stored inside some other registers we
will see that now each of the segment
registers store the upper 16 bitterness
of the starting address of the
corresponding segments so they would be
holding only the upper 16 meters but not
the lower 16 bit address so we need to
hold districts here
sixteen others by using the offset
alistar's now he RCS is the register
holding these starting address of the
code segment and will get or we can
access a single location inside the code
segment by adding this segment address
that is a by adding the contents of this
code segment register along with
instruction pointer now this particular
register will be holding the offset
address of the code segment so by adding
the contents of IP and CS will get the
physical address inside the code segment
okay and if you go for stack segment now
stack segment register will be holding
the starting address of the stack
segment and will get the physical
address of a particular operand stored
inside the stack segment by adding this
particular register that is SS resistor
with this SP resistor and then we have
data segment register which will be
holding the starting address of data
segment and will get the exact location
of an operand which is resided in the
daesil data segment by adding da smith
and sign which is nothing but source
index register and in the same way extra
segment is added with di for getting the
physical address inside the extra
segment register now let us see an
example here this is the memory which is
started from the zero location as you
can see here code data extra as well as
tack assign some variants somewhere in
the program or somewhere in the memory
okay and the size of each every each and
every segment is 64 kilobytes okay now
here the code segment and it starts from
3/4 ba 0 and ends it double for b9 f now
this is just the distance between the
starting and the ending addresses 64
kilobytes okay now it is a 20 bitterness
but segment registers can all can hold
only 16 materials therefore it is going
to hold this upper 16-bit address which
is nothing but 3 4 P a 34b
a is stored inside this si si which is
member code segment register in the same
way b sr es r and s sr will be holding
the upper
address that is upper 16 bit
now the lowest 16-bit address will be
holding by some other registers those
are pointer and then index registers as
just now we have seen okay now
instruction pointer is the register
whose length is 16-bit and it is mainly
used for holding the offset address of
the code segment so when this register
is added with code segment register then
we'll get the 20 bit physical address
which indicates the address of the next
instruction that is to be fetched by the
processor from the memory
okay so CS along with IP will give us an
address that address indicates the
address of the next instruction to be
fetched from the memory now as you can
see here there is an offset address
whose lengthy 16 bits and our 20 bit
segment address okay now initially both
the addresses segment register as well
as offset register will be holding 16
16-bit addresses now segment register is
shifted towards left by four bits and
then added with this offset register
which will give you 20 bit physical
address let us see this with an example
now here you can see that offset value
is a 16-bit value okay which is 29 and
then segment address is again a 16-bit
value as you can see have four digits in
hexadecimal which is nothing but 16 bits
in binary so 1 triple 0 okay now we have
to get the 20 bit physical address from
these two addresses so how we are going
to get this initially this one triple 0
will be shifted towards left by 4 bits
by the address conversion mechanism
adder which is present inside the bus
interface unit okay now these address
calculations will be done by address
conversion mechanism adder which is
present in the BIE so this thousand will
be shifted towards left by 4 bits as you
can see here it is shifted towards left
by 4 bits okay so four zeros are
appended at the right side and then this
29 is added ok then you'll get the
required physical address that is 1 0 0
to 9 now this is the exact
location where actually the operand is
recited and here you can see how to form
the physical address in case of code
segment data segment extra segment stack
segment
okay so CS IP will be giving us the
physical address inside the code segment
SSSP or SS SP SS V P will be giving us a
physical address inside the stack
segment BSB X or B s si will be giving
us the physical address in our data
segment okay D is di for other than
string operations and then es di for
string operations will be giving us the
physical address inside the extra
segment as I told you that BX is also a
register which P which will be used for
holding the offset factors okay so this
is how the memory that is 1 MB memory is
divided into some of the segments and
that can be accessed with the help of
segment registers as well as offset
registers which will give us the
physical address which is nothing but
the exact location where actually the
operand is recited okay so in the next
session we'll discuss about addressing
modes thank you
you
memory segmentation now what do you mean
by segmentation as we know what is a
memory and how much amount of memory
that can be interfaced with a gr8 6 is 1
MB that depends on the number of address
lines as it has 20 address lines so we
can interface up to 1 MB of memory now
what do you mean by segmentation
segmentation generally means dividing or
making it parts now here memory
segmentation is nothing but dividing or
dividing V entire 1 MB memory into some
parts or some logical segments ok so we
are dividing this entire 1 MB memory
into a number of logical segments ok and
the size of each and every logical
segment is 64 kilobytes why we are going
with only 64 kilobytes of memory
segmentation or 64 kilobytes of the size
of each and every segment because here
we have 16-bit registers for accessing
the memory so as we have 16-bit
registers we can only access up to 64
kilobytes with the help of 16-bit
registers so that is why here each and
every segment size is of 64 kilobytes ok
now so how many segments are present
there are totally this complete 1 MB
memory is divided into 16 logical
segments why because 64 kilobytes into
16 segments or 16 is nothing but 1 and B
ok so here the size of each and every
segment is 64 kilobytes and all these
segments are named as code segment data
segment extra segment and stack segment
so there are only 4 segments and there
are only 4 names given to these segments
like code segment data segment extra
segment as well our stack segment okay
and there can be any number of code
segments there can be any number of data
segments there can be any number of
extra segments as well as there can be
any number of stack segments and the
number of code data as well as extra as
well as stack segments depends on the
user program okay now here so the
complete physically available memory
be divided into number of logical
segments as I have told you now we have
the starting addresses five zeros and
the ending address is 5x okay now here
this address is represented in
hexadecimal form okay now if you
represent the same address in binary
form now it will give you a 20-bit
address which is nothing but the exact
location where actually the operand is
recited okay now it is a 20-bit address
but by using 16-bit address or 16-bit
register we cannot access or we cannot
we cannot retract the data which is
present in this particular memory
locations so for that purpose we use one
of the register for holding one kind of
address and the other register for
holding the other kind of address okay
now let us see what is that now here as
we have four segments code segment data
segment extra segment as well as data
say stack segment code segment is used
for holding the instructions so whatever
the instructions written by the user now
that are stored inside this code segment
and coming to the stack segment which is
mainly used for holding the return
addresses okay and this is used in case
of function calls or as interrupt
service routines okay and data in extra
segments are mainly used for holding the
data that is user data okay and as I
told you that the size of each segment
is 64 kilobytes and there are 16 logical
segments a segment may be located
anywhere in the memory so any of these
segments are located anywhere in the
memory and at a time only four segments
are active and those four signals have
code segment data segment extra segment
and then stacks it the four segments are
code okay and a segment is a 64
kilobytes block of memory 16 bit
contents of segment registers in the bi
you actually point to the starting
location of a particular segment now we
have some segment registers inside the
VIU now all these segment registers will
be pointing to the actual point or
starting location of the particular
segment we'll see that okay and what is
the advantage of going with segmentation
now here as
1 MB memory but we can access this 1 MB
memory with the help of 16-bit registers
ok and then the code or data will be
placed anywhere in the program that is
will be placed anywhere in the memory ok
and the segment registers are used to
allow the instruction data or stack
portion of a program to be more than 64
kilobytes long the above can be achieved
by using more than one code data or
stack signals if if assuming that we
have written a program now the code part
of the program is more than 64 kilobytes
then our two or three code segments are
assigned for that particular code that
depends on the size of the colon okay as
I told you that n number of code data
orders stack extra segments can be
assigned in the memory now let us see
what are these segments and sisters we
have totally four segment registers code
segment register data segment register
extra segment register and stack segment
register Nanyan confused with register
as well as a segment in the memory
those are segments and these are
registers now all these registers are
used for holding the starting address of
their respective segment it means code
segment register is used for holding the
starting address of code segment data
segment register used for holding the
starting address of data segment extra
segment register used for holding the
starting address of extra segment and
stack segment register is used for
coding the starting address of the stack
segment okay now all our 16-bit
registers but so as all these are
sixteen registers they would be holding
only the segment address all of these
starting headers now how to access each
and every individual location from the
memory okay now for accessing individual
memory locations inside the memory we
need offset address we need offset
headers and those offset addresses are
stored inside some other registers we
will see that now each of the segment
registers store the upper 16 bitterness
of the starting address of the
corresponding segments so they would be
holding only the upper 16 meters but not
the lower 16 bit address so we need to
hold districts here
sixteen others by using the offset
alistar's now he RCS is the register
holding these starting address of the
code segment and will get or we can
access a single location inside the code
segment by adding this segment address
that is a by adding the contents of this
code segment register along with
instruction pointer now this particular
register will be holding the offset
address of the code segment so by adding
the contents of IP and CS will get the
physical address inside the code segment
okay and if you go for stack segment now
stack segment register will be holding
the starting address of the stack
segment and will get the physical
address of a particular operand stored
inside the stack segment by adding this
particular register that is SS resistor
with this SP resistor and then we have
data segment register which will be
holding the starting address of data
segment and will get the exact location
of an operand which is resided in the
daesil data segment by adding da smith
and sign which is nothing but source
index register and in the same way extra
segment is added with di for getting the
physical address inside the extra
segment register now let us see an
example here this is the memory which is
started from the zero location as you
can see here code data extra as well as
tack assign some variants somewhere in
the program or somewhere in the memory
okay and the size of each every each and
every segment is 64 kilobytes okay now
here the code segment and it starts from
3/4 ba 0 and ends it double for b9 f now
this is just the distance between the
starting and the ending addresses 64
kilobytes okay now it is a 20 bitterness
but segment registers can all can hold
only 16 materials therefore it is going
to hold this upper 16-bit address which
is nothing but 3 4 P a 34b
a is stored inside this si si which is
member code segment register in the same
way b sr es r and s sr will be holding
the upper
address that is upper 16 bit
now the lowest 16-bit address will be
holding by some other registers those
are pointer and then index registers as
just now we have seen okay now
instruction pointer is the register
whose length is 16-bit and it is mainly
used for holding the offset address of
the code segment so when this register
is added with code segment register then
we'll get the 20 bit physical address
which indicates the address of the next
instruction that is to be fetched by the
processor from the memory
okay so CS along with IP will give us an
address that address indicates the
address of the next instruction to be
fetched from the memory now as you can
see here there is an offset address
whose lengthy 16 bits and our 20 bit
segment address okay now initially both
the addresses segment register as well
as offset register will be holding 16
16-bit addresses now segment register is
shifted towards left by four bits and
then added with this offset register
which will give you 20 bit physical
address let us see this with an example
now here you can see that offset value
is a 16-bit value okay which is 29 and
then segment address is again a 16-bit
value as you can see have four digits in
hexadecimal which is nothing but 16 bits
in binary so 1 triple 0 okay now we have
to get the 20 bit physical address from
these two addresses so how we are going
to get this initially this one triple 0
will be shifted towards left by 4 bits
by the address conversion mechanism
adder which is present inside the bus
interface unit okay now these address
calculations will be done by address
conversion mechanism adder which is
present in the BIE so this thousand will
be shifted towards left by 4 bits as you
can see here it is shifted towards left
by 4 bits okay so four zeros are
appended at the right side and then this
29 is added ok then you'll get the
required physical address that is 1 0 0
to 9 now this is the exact
location where actually the operand is
recited and here you can see how to form
the physical address in case of code
segment data segment extra segment stack
segment
okay so CS IP will be giving us the
physical address inside the code segment
SSSP or SS SP SS V P will be giving us a
physical address inside the stack
segment BSB X or B s si will be giving
us the physical address in our data
segment okay D is di for other than
string operations and then es di for
string operations will be giving us the
physical address inside the extra
segment as I told you that BX is also a
register which P which will be used for
holding the offset factors okay so this
is how the memory that is 1 MB memory is
divided into some of the segments and
that can be accessed with the help of
segment registers as well as offset
registers which will give us the
physical address which is nothing but
the exact location where actually the
operand is recited okay so in the next
session we'll discuss about addressing
modes thank you
you
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