Instruction set of 8086 Microprocessor I Video Lecture - Computer Science Engineering (CSE)

Okay welcome to the topic number seven till now we have covered the architecture the pain diagram
The addressing modes and now we come to the instruction set of 8086
It's a big long list of instructions which 8086 gives
But that is what it makes it powerful and easy to use
Why study the instruction set
As you know that instruction sets are required so that you can write codes and programs so that you can tell what
8086 has to do so we there are a lot of
data manipulations lot of
Data transfer, which has to be done and all these instructions will give you that power to manipulate
data
To calculate some sense to values manipulate some sense of values and display something
So for this reason a deep understanding of all this instructions how they work
and how they fetch data from the memory or a
Register, and how they are manipulated how the flags are affected that understanding will give you a very
head start to use or to start programming 8086
In this section we are going to study
Different instruction sets there are there are different instruction types
And I know we will see all the examples and how each instruction works, and how the data flows in an instruction
So there are lot of types there are nine types listed here. There are data. Copy transfer instructions arithmetic instructions logical instructions
shift and rotate instructions branch instructions loop instructions machine control instructions
flag manipulation instruction and string instructions
Let us jump into that the first is a data copy and transfer instructions as the name suggests
Basically, what will happen in this instructions is that data is getting we big getting copied to a certain location
It will be copied from a certain location so what the format
Given to us is move destination comma source so what is going to happen is that data from source will be copied to the destination
So there will be a transfer of data from source to destination let us see the example
Which is given on the right of the screen?
that is
move
that
Is move ax comma B X so what is going to happen is that data VX is getting copied to X
What will happen the ax has some data?
And that will be overwritten with the data of BX so that is what is going to happen
So you see that BX the data is getting copied to the X register
So ax has FF 3 3 and the date
BX has the data 1 0 a B and what should happen is that this data FF 3 3 should be
overwritten by 1 0 a B
so you see that and now the
Instruction has been executed and the data has been copied to X register
Next is instruction is the push instruction push instruction basically does is that it pushes the?
Source which is there in the instruction or the 16-bit data in the instruction or the 16-bit data in the register and?
pushes that data into the stack memory
So you know that stack memory is a temporary memory?
Which is used during the execution of a program?
Or if you want to cut some kind of store some kind of temporary data some dead
Flag values some register values you can keep it in the stack memory
So what is going to happen is that if you?
See the example push
BX what will basically do is that it will push the value of BX into the stack memory
So you can see that
Right now the this is the stack memory and there is FF. And 3 3 which is stored in the stack memory and
FF is the topmost data there definitely there will be data is here and here
But then FF is the topmost data and this memory location and this memory location is empty
stack pointer is a register which will point to a 16-bit value which is and
16-bit value, which is the address of the data?
Which is the at the top of the stack so?
right now you see that FF is at our topmost data in the stack memory and the address of FF is 5 0 0 2
so that is the
Data, which will be there in the stack pointer
so after the execution of data
What should what will happen is that this data will be copied to the stack memory so stack memory?
address or stack memory content will be decremented by 2
So what will happen is that first data goes and the higher data goes in?
it is there in the
Location above FF so now if you see that the stack pointer has in
decremented with by 1
So now when the lower byte goes again the stack pointer
Will be decremented by 1 and the value becomes 5 thousand so 16-bit data
Definitely take two places in the stack memory and now the stack memory is pointing at five thousand earlier. It was five thousand two
Similarly exactly opposite will take place in pop
instruction basically pop instruction will take out the last
put data in the stack memory to the
Any register which is given along with the instructions so in the example given on the screen pop D is?
What will basically do is that it will take a 16-bit data from the stack memory?
It will basically take the data
16-bit data from Sacre that is 1 0 and a B and
Put that data into D. S right now des has two two three three and this data will be lost when this
Instruction is executed
So what is happening? Is that the first the lower byte has been copied and then the higher byte will be copied and
In this case exactly opposite to push instruction the stack pointer is
incremented and
The it is pointing at the top most memory location that is
Data with the top most memory that is FF
Exchange destination and source is a very powerful instruction and which is required a lot of times
When you want to swap details between two registers, and if you see that
In example exchange B. X comma a export is going to happen
Is that the data of B X will be copied to a X and data of X will be copied to B X so?
definitely it uses a temporary register to do this and
You can see the data getting transferred after the execution of program
Next instruction is in instruction and the out instruction in instruction is basically used
To take data from the i/o ports available of i/o ports of 8086
So what is going to do is that it is going to with its address lines select a port?
Basically it will select a port and from that port it will take a 8-bit data
And it will store it in accumulator. It could be 8-bit
Port or a 16-bit port it depends upon
What kind of peripheral you are connecting if it's in IE 2 phi phi for an example will give you an 8-bit?
put address for each ports of
8 to 5 5 that is port a port B port see they are all 8 bits so all the
Addresses are it puts so from that
It to 5 you can take an 8-bit data, which will be stored in the accumulator, so ill
Accumulator it will be storing the data it can be if you have a peripheral which can operate 16-bit data
It can also say 16-bit data into a X register
So in this example what we see is that in Al comma 8 P Hanks? What is going to happen?
Is that 18 X is the port address 80 X is not the data?
80 X is the port address and the data from Port will be transferred to
accumulator so data from
Port the data from port 80 is
Twelve which has come to L register
similarly if you see the second example it is a in X comma D X
so in ax comma D X DX is the
Default location if you want to have a 16-bit port address you cannot give along with the instructions
What you have to do is that you have to put that value into B X register in a step?
Before in instruction, and then do in ax comma D. X now what will happen is that from port?
8000 DX s 8,000 and the port 8000 address a
Data will come to ax register the data will come to ax register
So in this case one two three four has come to ax register