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Advanced videos So let us study that what instructions

come under this category in this category the instructions which

perform the arithmetic operations they are included arithmetic operations like

addition subtraction increment and decrement the first instruction under this

category is a D D addition the operand of this instruction can be a register or

it can be your memory so this instruction is add register so this instruction add it is adding the

contents of the operand operand can be a register if R is there then it will be a

register and if M is there then it will be a memory so the contents of either

the register or the memory they are added to the contents of the accumulator

so one of the operand will be in the accumulator and another operand will be

the register or a memory now if the operand is memory then the location of

that memory is specified by the contents of the actual registers how we are

getting the location of the memory the actual register pair will indicate the

location of the memory suppose we have here 8 0 4 0 so at this memory location

the second data will be stored suppose at 8 0 4 0 we have 2 so this 2 and

accumulator is having 3 so 3 & 2 they got added up and the result of the

operation will be stored in the accumulator itself so accumulator will

have 6 here now if we have a register here so in that register like be

register see register b e:h l any of the resistors can be used and that resistor

will be added up with the accumulator if it is a register then b c d e HL any of

the registers can be used and their content will be added up with the

accumulator now all the flags are modified to reflect the result of the

addition in the 0 8 5 we have five types of flags sign flag zero flag parity flag

auxilary carry flag and carry flag so while performing this operation addition

operation if the result obtained after the addition if the result is zero then

zero flag will be said if a carry is generated in adding of

two numbers then carry flag will be set to 1 and if auxilary carry is generated

then auxilary carry flag is set to 1 if the result is having even numbers of 1

then parity flag is set otherwise it is reset and if the result is 0

coming out to be 0 of it is a negative number

then sign flag in zero flags are modified so all the flags they are

modified to reflect the result of the addition all the five flags will be

changed so this was the add R or either M instructions next instruction we have

a DC R and in this is add register to accumulator with Daddy so this instruction is a DC R or M no

operand can either be a register or it can be a memory location so this is

adding the register to the accumulator with caddy caddy flag is used so we have

this see here the contents of the operand at the register or memory and

the carry flag they are added to the contents of the accumulator and the

result of the addition is stored in the accumulator itself here also all the

flags are modified and if the operand it is a memory then the location of that

memory will be specified by the HL register pair let’s take an example

suppose we have a DC B that is adding the register B with the accumulator and

the carry flag also suppose an accumulator we have the number 2 and in

the B register we have the number 3 and in carry flag suppose carry flag is set

to 1 so this instruction a DC it is going to add the be registered with the

accumulator register and also the carry flag and the result will be stored in

the accumulator so 2 plus 3 plus 1 it will be 7 so

accumulator will have 7 here now suppose if carry

flag is equal to 0 it is the set condition if it is 1 and reset condition

if its value is 0 so if it is 0 then 0 plus 3

plus two so the content of accumulator will be six so this is how the carry

flag and the register they are added up and the accumulator and result is stored

in the accumulator it’s same now if we are having a memory address suppose we

are having ADC M so M is the address in the HL register pair HL will have 8 0 5

0 suppose so 8 0 5 0 memory location it is having the number 7 and we have the

accumulator has 2 and carry flag has 1 so 7 plus 2 plus 1 it will be 10 so

accumulator contents will be a suppose if it is 0 then 7 plus 2 plus 0 so the

accumulator content will be 9 so this is how this instruction works

it adds the carry flag and the register and accumulator or it can add a memory

location plus the carry and the accumulator now the next instruction is

Adi this instruction is add immediate to

accumulator so in this instruction the operand is an 8-bit data the operand is

defined in the instruction itself so this instruction is going to add this

8-bit data with the accumulator and the result of the addition will be stored in

the accumulator itself so this instruction adds the 8-bit data

to the contents of the accumulator and the result is stored in the accumulator

itself and all the flags are modified in this instruction let’s take an example

we have ad I and 45 so this is an 8-bit data fortify this will be added with the

accumulator suppose accumulator has here five so this five and forty five they

are going to add up and the result will be stored in the accumulator itself so

it will be 50 so in this way this instruction works that it adds the

immediate data given in the instruction itself to the accumulator and the result

is stored in the accumulator itself our next instruction is ECI this instruction is at the in EEG data

which is given in the instruction along with the carry to the accumulator so this instruction it is adding this

8-bit data along with the caddy flag to the contents of the accumulator and the

result is the stored in the accumulator itself let’s take an example if we have

AC I again 45 then this 45 number will be added to the accumulator accumulator

is having five and carry flag is suppose set so it is equal to one so 45 plus

five plus one so the accumulator will get the result 51 if it is having 0 then

the accumulator will have the result 15 so this instruction ACA it is adding the

immediate data which is in the instruction to the contents of the

accumulator and the carry flag all three are added up and the result is

the stored in the accumulator flags are also modified if suppose after the

addition any carry is generated then carry flag is again modified and if 0 if

the result is 0 then 0 flag is said sign parity auxilary carry all the five flags

they are modified during this instruction the next instruction is AC this instruction is to add the

accumulator with the image a data plus the carry so this instruction is adding this image

a data which is given in the instruction in cells to the accumulator along with

the carry so this 8-bit data which is the operand

and the carry flag they are added to the contents of the accumulator and the

result of this arithmetic operation it is stored in the accumulator now we know

that whenever any automatic and logical operation is performed the floods like

sign flag zero flag parity auxilary caddy and caddy all these flags are

modified so as to reflect the result of the addition if the number after

obtained after the addition it is a negative number or it is a positive

number then sign flag will be modified if the result is zero then zero flag

will be modified if any caddy is generated then carry an auxilary carry

they are modified and if the resultant number it is having even number of ones

or odd number of ones then parity flaccus modified so all the flags are

modified due to this instruction to reflect the result of the addition now

if we take an example of this instruction then a CI and an 8-bit data

suppose we have 45 so this instruction can be written in the program to execute

this instruction now what it will do suppose in the accumulator we have 0 4

and this 8-bit date is 45 and suppose the carry flag is reset it is in the

reset condition that is its value is 0 then 4 + 0 + 45 all these 3 will be

added and the result will be stored in the accumulator so in the accumulator we

will get 4 & 4 5-it will be 49 if it is 1 then 4 plus 1

plus 45 then the accumulator will have 50 ok so this is how the caddy flag and

the mej data and accumulator they are added up now next instruction is dad now

in 8 0 8 5 there is an exception that no registers the contents of the register

pairs they cannot be added if you want to add any of the register then you have

to add with it to the accumulator like we have b c d e h l so no two resistors

pair they can be added if you have to add the contents of any of the register

then you have to move the contents of the other register into the accumulator

so we can say that addition can only be performed if one of the operand it is

stored in the accumulator no two other resistors they can be added but an 8 0 8

5 this instruction bag it is an exception this instruction add to

register pairs ok so this instruction is about and the registered pair to h&L registers so because b c d e HL

they cannot be added up and if you want to add then one of the operon should be

stored in the accumulator so this instruction is an exception it is adding

the contents of the register pair either b c or either de with hn so one of the

operand will be in the BC registers or de resistor and one of the operand will

be in the HL register so earlier instructions whatever we have

studied about the addition of two contents or two numbers then their

addition was about 8-bit addition like 8-bit data will be added to another

8-bit data and the result is also an 8-bit data now using this instruction

guide we can add a 16-bit number and the 16-bit number it is stored in the

register pair any of the register pair BC or either de and another operand will

be stored in the Etzel resistor so both the numbers they will be added up and

the result will be stored in the HL register let’s take an example we will

write bad H so what it is doing it is adding the contents of the HL register

pair with the HL register pair okay so suppose in HL register pair if we have 0

2 and 0 3 this is the content of HL register pair then this will be added

because here we have written that h-he on e HL will be added with a chill only

so here also we will write 0 2 0 3 both will be added up and the result of the

addition will be 0 2 0 4 it will be 0 2 is 0 4 and here we will have 6 now this

addition it was a 8 bit addition also you can say that HNH these are 8 bit

data and the result is also 8 bit you can have 16-bit data also here and then

the we’ll be the 16-bit content so this

instruction it is allowing the 16-bit addition of the data now here we can

also have suppose we have bad B then the a children sister pair will be added

with the BC register pair L will be added with C and H will be added with B

higher-order registers will be added and lower order registers will be added and

accordingly if the Cadi is generated then the Cadi flag will be modified and

the result of the addition it is stored in the HL register pair only so the contents of the souls are just a

pair they are not altered if in this case like that be if we have in BC

resistor we have a 16-bit data then this resistor and HL resistor pair they are

added and the result is restored in HL but the contents of the BC register they

are not changed so the contents of the source resistor pair they are not

altered and if the result which we are getting after the 16-bit addition if it

is greater than 16 bits sub that is it means caddy is generated then this caddy

flag is said and no other Flags they are modified so this was the that

instruction next instruction we have is su B R and in this instruction is for

the subtraction it subtracts the contents of the accumulator from the

register or the memory so here we can have the operand as a register also and

as a memory location also so what this instruction suv’s doing it is to

subtract so subtracting the register or memory

from the accumulator so this instruction is to sub the

contents of the operand if the operand is memory or it can be a register

they are subtracted from the accumulator so for subtraction the subtraction is

always done like from smaller number is subtracted from the larger number now

here we are subtracting the contents of the memory or the register from the

accumulator suppose an accumulator we have the number three and in the

register suppose we are having the register B here and in this register we

are having the number zero one then one will be subtracted from three and the

result will be stored in the accumulator itself so accumulator will now have the

number zero – now this abduction it is a binary subtraction here decimal

subtraction you have to perform so so if the accumulator in this case of

cumulated was having the larger number so easily the subtraction killing we

perform now if the accumulator it is having a lesser number suppose one is in

the accumulator and three is in the register then you have to subtract the

three from one so now you will get a negative number and this negative number

will modify the flags in the sign flag it will be set that means it values will

be one if the result of the subtraction is a negative number and if some Gaddy

is generated means borrow is generated you have to borrow your then carry flag

and all these flags are also modified do you do this instruction

now if the operand it is of memory then the address of this memory location it

is a specified in the HL register Pam suppose we have written here sub n 18

then the address of this memory location will be given by the HL register pair

suppose we have a 0 5 0 here then our second number will be stored at this

memory location suppose we have one here then this one will be subtracted from

accumulator and result will be stored in the

accumulator itself so this is how the this instruction will be executed it

subtracts the contents of the operand from the accumulator and also all the

flags they are modified to reflect the result of this subtraction next

instruction we have is SBB now in case of addition we were having carry now in

case of subtraction we will have borrow so this instruction is doing what it is

subtracting the source source is our operand it can be either a register or

it can be a memory location it is subtracting the source and the borrow

from the accumulator so this instruction what it is doing the

contents of the operand operand can be either the register or it can be a

memory location it can be a memory and the borrow flag they are subtracted from

the contents of the accumulator now if the operand it meant it is a

memory then its address is given by the HL register bear and all the flags are

modified just the differences there that in the SUV instruction we are

subtracting the contents of the register and memory from the accumulator

but in this SBB instruction we are having this borrow flag also so this

borrow flag is also considered and it is also subtracted from the accumulator now

example we can write here s BB we can write any of the registers we can use

like s BB C resistor we have used or we can write s BB in hip it is M then its

address is given by the it will register pair suppose initially we have two zero

three four then one of the operand it will be in the accumulator and we have

the borrow flag we have to consider the borrow flag also if it is zero or bun

and the second operand is stored at this memory location to 0 3 4 so this is how

this instructions execute now next instruction we have is it is again for

the subtraction now this instruction is sui this I means

we have an immediate data here and this immediate data it is given in the

instruction it’s same so what you what this instruction does it subtracts the

container this 8-bit data from the accumulator so this is subtracting the

immediate data with the accumulator no borrow flag you don’t have to consider

borrow black in this only this data and the accumulator so this instruction it is subtracting

the 8-bit data from the contents of the accumulator result is stored in the

accumulator and all the flags are modified due to the execution of this

instruction now example we have sui and 45 so this number 45 it will be

subtracted from the accumulator suppose an accumulator we have 50 then 45 will

be subtracted from 50 and the result will be stored in accumulator which is

zero five and if there is any borrow suppose in this number we have is 445

suppose we are having here fifty-five then we will subtract this 55 from 50

then we have to take a borrow then the flags will be modified to reflect the

result that a borrow has been taken to perform this subtraction so this was su

I that is subtracting the Maj data from the accumulator next we have s bi8

mu theta now this instruction it considers the borrow because here B is

for the borrow and I is for the immediate data and s is for the

subtraction so this is doing the subtracting the change it from public up

with vodka so this instruction it is subtracting

the 8-bit data along with the borrow flag from the contents of the

accumulator and the result is stored in the accumulator itself and also all the

flags are modified to reflect the result of the is for the subtraction and the

addition instructions you can notice that in all the addition and subtraction

instructions we have used though one of the our contents will be in the

accumulator we have add a bi a DC and AC n these are other addition and for

subtraction we have su b then Sui then we have is bi and we have SB B so these

four were photo addition and these work for the subtraction we were having two operands two numbers

we require and one of the number is in accumulator and the second number it is

in the either it is a stored in the register or it is a stored at a memory

location and also we can have an 8-bit data directly given in the instruction

itself that is immediate datum and one instruction boson exception that is the

guide instruction which is adding two resistor pairs it is adding a registered

payer with the Etzel register pair so these were for the addition and

subtraction now next instructions which come under the category of arithmetic

operations or arithmetic instructions arithmetic instructions are increment

and decrement increment is increasing the number by one and decrement means

decreasing the number by one so first we will study the increment instruction so

we have INR R and M so this instruction is to increment register or memory by one so this

resistor it can be any of the resistors a b c we have d e h n l so any of the

resistor it can be incremented by 1 or it can be a memory also now this memory

address of this memory location will be given by the HL resistor pair so this instruction INR it is increasing

the content of the register or the memory by one and the result is stored

in the same place suppose take an example we have INR B so in the B

register a number is stored suppose it is 3 so this I naught B it will

increment the instruction this register by 1 so the content of B will now become

0 4 so the result of this increment instruction it is stored in the same

place if we have written INR C so in the C register if we have 4 then this will

be incremented and it will become 5 it will be incremented by 1 just remember

that here only the register is incremented by 1 we are not taking the

register pair here only the register it can be any of the register it can be

accumulator b c d e HL any of these registers and they can be incremented by

1 now if the operand it is of memory then the address of this memory location

will be given by tau H will register pay so this was INR okay we have a

modification of this instruction we have i NX i NH r now this instruction is used

to increment the register pair by 1 so your register pair is incremented by

one we can have the resistance pair as BC we can have de and we can have a

chain so any of these register spills can be incremented by one using this IX

instruction now result of this increment it will be stored in the same place

let’s take an example if we have BC register we have eight zero five zero

then it will be plus one incremented by one and the result will be BC will have

eight zero five one because we will add this one in the lower order register C

is the lower order register so plus one will make it fifty one and eighty will

remain as it is so this complete register pair is incremented by one so

the sine X instruction we can use to increment the register pair and result

is restored at the same place similarly we can have an X this was an X B then we

can have an X D also and we can have INX it’s also so through this the BC

register pair will be implemented through this de will be incremented and

through this H L will be incremented so here register pairs are incremented by

one now these were for the increment and then we can have the same instructions

just difference of this will be that now we will have deployment in the register

or the register pair by one so just we were having iron ore for

incremented INR a denim here we have DC RR n m4 decrementing the register or

memory by one so the contents of the register register

can be BC accumulator can also be their de h L any of the register it can be

decremented by 1 and if it is immunity then the address of this memory location

it is a specified in it shall register pair the result of this instruction it

is stored in the same place suppose we have I this we have d CR and C and in C

register we are having 5 so this will be decremented by 1 and the new contents

will be 0 4 so the content will remain stored at this the result is stored in

the same place it is not stored in the accumulator it is stored at the same

place and the number is decremented by 1 similarly we can have d CR m and M the

setlist will be specified in HL register pair we have supposed to 0 3 4 then at 2

0 3 4 if we have a number 6 then it will be decremented by 1 and now the new data

will be 0 5 so this is of this instruction VCR works now next

instruction we have is d CX this instruction it decrements the

registered pair buy one just like I on expose for incrementing the registered

pair by one DC X is four decrementing the register pair by one so the contents of the register pet will

be decremented like b c d e and agent any of the register pair it can be

decremented by 1 and the result is stored at the same place now in de

suppose we are having 2 0 3 4 then it will be decremented by 1 so the new

contents of the de register pair will be 2 0 3 3 and we can have the instruction

for this d CX d we can have d CX be also we can have d

CX h also so this is decrementing the register pairs by 1 so these were the

increment and decrement instructions I not INEX we were having INR INEX for the

increment and we have DC r and d CX for the degree matter now next instruction

we have this be a a this is decimal a just accumulator B is for decimal a is for a just and

this a is for the accumulator this instruction is used in the BCD

operations when the microprocessor it is performing the BCD addition then this

instruction is used so this instruction it is used to bend

the microprocessor it is performing BCD operations BCD means battery coded

decimal now in the microprocessor we have all the numbers in the decimal

notation that is in the form of zero and once so the this instruction decimal a

just accumulator it changed the contents of the accumulator which are given in

the binary value to two 4-bit BCD digits because accumulator it is an 8-bit

transistor so we were having 8-bit data now this 8-bit data which is given in

the accumulator it has changed into two 4-bit data’s this is for bit and 4-bit

pitches in the BCD notation because this 8-bit data it is in the decimal notation

so it will be converted into the BCT dot notation and it will get converted into

two 4-bit data’s now this is the only instruction which is using the auxilary

carry flag to perform this BCD conversion so it can be asked that which

instruction uses the auxilary carry flag so this is the answer that the decimal

is just accumulator instruction it uses the auxilary carry flag to perform the

BCD conversion although other flags they are also modified to reflect the result

of the addition now how this BCD conversion is performed

let’s see so if I do the note of the four bits

suppose we are having the number this is our 8 bit data and the lower or the four

bits they are greater than 9 or if the auxilary caddy is said so if this

auxilary carry flag it is said that means its value is 1 then this

instruction it adds 6 to the lower order 4 bits so

this 6 will be added to the lower order 4 bits n we can have the 4 bits which

are in the BCD notation now if the value this was for the lower order 4 bits now

if we have the value of the higher order 4 bits it is greater than 9 or if this

auxilary carry flag it has said that means it is 1 then the accumulator then

this instruction it adds 6 to the higher order bits so if a higher order bits are

greater than 9 then the instruction it adds and if this caddie flag it is said

then this 6 is added to the higher order bits and the result will we have is the

4 bits in the BCD notation so this is how the BCD conversion is performed how

a decimal number is this converted into the BCD notation if lower order bits

they are greater than 9 then add 6 to the lower order bits and if the higher

order bits they are greater than 9 carry flag asset then add six to the

higher order bits to perform this BCD conversion in the microprocessor we will

just write D a a this instruction it does having implicit addressing mode

implicit addressing because the operand it is given in the instruction itself

operand is the data or it is something on which we are performing the operation

so we are here performing the operation on the accumulator and accumulator it is

specified in the instruction itself we need not to specified externally it is

implicit in the instruction so this type of addressing is implicit addressing so

this instruction da it is an example of implicit addressing mode so this was all

about the arithmetic instructions the arithmetic instructions they involve the

operations like addition subtraction increment decrement and the last

instruction we studied was the decimal adjust accumulator is the BCD conversion

so these instructions were for the fall in the category of arithmetic

instructions next we will study the branching instruction

its 3+2=5, and not 6, please check that…it confuses the beginner initially

7:50 …….its 2+3+1= 6 not 7….plz check it, I think you overlooked

Nice

Nice Explanation mam

Thanks so much

Thanks a lot ji

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