00:01
So, here we need to calculate and update the content of registers accordingly in the fill in the blank.
00:11
So, let's do it step by step by one by one fill in the blank.
00:16
So, for first we need to put lr 5 6 which is known as load register.
00:25
So, for this we need to understand r5 is loaded with the content of memory in location 6 which is 7 times 0 4.
00:35
So, on the basis of this we can say that r5 will be 0 0 0 0 0 0 0 4.
00:50
Okay.
00:50
Now, let's move towards the second thing which is la 1 2 in bracket 0 3.
00:59
So, this is we consider as load address.
01:03
So, the address is calculated as r3 plus 2 which is 0 0 0 0 0 3 6 c plus 2.
01:18
So, here it is 0 0 0 0 0 3 6 e to the value of memory location 0 0 0 0 0 3 6 e is in memory location the value of this well is 0 0 0 0 0 1 2 3.
01:45
So, we can say this r1 will be 1 2 3.
01:52
Now, let's move forward for the next thing l6 4 6 1.
02:01
So, we consider it as load.
02:03
So, it is calculated as r6 plus r1.
02:09
So, which is 2 7 times 0 4 plus 5 times 0 1 2 3.
02:18
So, we got it 1 2 3 4 5 1 2 7.
02:25
So, is its value is value is 8 times f 4 5 6 7 8.
02:45
So, the value located into r6.
02:51
So, this is the value of r6.
02:54
Now, let's move forward here we need to find br 4 8 which is add register.
03:08
So, for the add register r4 is added the content of r8.
03:15
Okay, which is 7 times 0 4.
03:21
So, as the result of addition is so here we need to do r4 plus r8.
03:28
So, when we add this 8 times f.
03:38
So, here we got it we got it 0 0 0 0 0 0 0 which is 7 times 0 3...