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In this lesson we're going to look at how the compiler acts on conditions.

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Let's go over what we said about the P.C. in the previous lesson.

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As you remember we said a piece these times for the program counter and that it stores the address of

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the next instruction to be executed.

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Yeah we are going to look at how that happens.

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Also remember our previous program worked in an incremental linear waterfall manner.

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Here we are going to implement conditions and see what branch and happiness in the disassembly view

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and see how the compiler works with that show.

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We're going to start by modifying our current code which looks like this currently.

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Remember we started by just increment and with our decrement and just with a plus plus contour and here

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we have eight of them.

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Meaning we add in one to counter eight times if it comes eight.

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After that as it's implied.

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So instead of typing this out eight times let's just put it in a loop and see how the compiler would

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deal with this loop although the results would be the same.

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Let's observe in the disassembly view how the compiler works with that we can start by just wiping all

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of this off.

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This is eight of them and we could just use a while loop like we have here and see while the counter

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is less than eight and then we can put what we want which is plus plus counter

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so this should give us the same resource and let's see.

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So let's just exit the debugger and run this code by compiling it first and then back to the debug view.

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Okay.

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So let's step in and see first of all let's just check whether the Locos view gives us the value eight

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at the end.

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OK.

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It doesn't give us eight.

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It ends at seven.

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This is because as it says here not in scope the variable is out of scope.

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We can just cut this from here and put it here and then it should be scope.

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Now let's exit this and then come back and then go back to debug.

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Okay.

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Now let's try this step.

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See now we have the value eight which means it gives us the same resource over here.

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You can see that.

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Let's change it to decimal form here.

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We have eight here.

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Now let's go back to the top and see how it works its way to the answer eight in the disassembly view.

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So let's observe what goes on in the disassembly view.

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First of all we need to get it I wrote back to the beginning of the main function and we have to do

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that by exiting the debugger and coming back and I will explain why we have to do that in a second.

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Let's exit with this button here and then click the same button.

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Okay to come in here.

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So as we can see currently it's pointing here.

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This the instruction that's going to be executed.

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We already dealt with this.

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So we're going to step.

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Line by line and see what goes on into disassembly.

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You would click to step with step.

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It's executed here.

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The next instruction to be executed will be to branch into this while fun this while loop.

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Remember this phase while loop is the program's loop.

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This is what keeps the program running all the time.

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In fact the while loop we are interested in is the second one here.

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That's the one that keeps adding the value one to the variable counter the next instruction will be

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to branch to this while loop and the B here stands for branch.

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And then this address is where the wire loop is.

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So let's step in now with enter the programs while loop the next instruction will be to branch into

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the while loop of our variable counter.

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So let's go.

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Yes.

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Now we are in our while loop and we can see so far counter is there was 0 0.

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We are here because we've executed this already counts here.

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In fact like we did previously this just changed it to a decimal so few.

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So let's step again which step counter is 1.

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What happened was like we had in the previous lesson RS zero of course are zero plus one does Holcomb

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counter became one.

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Remember in the previous lesson anytime we stepped we had our share of plus one.

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So after you had kept incrementing mean in the first one gave us one the second one gave us to the third

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one gave us three.

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That's what went on.

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It was always our zero answer a one after you answer one.

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Now let's take a look at this one here.

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Let's step and see rotted and having eight different add s up codes to perform the additions over here

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we have just one at s up code and this one address code works in conjunction with a C and P and b L2

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up codes to perform all the eight additions when we increment the value is first added as we see here.

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And then to see MP which is to compare instruction this new forward and how the campaign structure works

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is that it uses the means of subtraction to make its decision as we see here.

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Counter has the value to and over here our loop says the counter should keep increasing as long as the

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value is less than eight.

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So what compares to acids.

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It takes two and then minus eight.

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And at this point the answer compare gets it's minus six because two minus eight is a negative six.

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And when this happens the AP S.R. register we spoke about it's updated.

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Remember we said the AP is our house the N C C and C bit the N stand for the negative bit dizzy for

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the zero bit C for Curry bit and V for overflow bit.

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So the negative bit of the AP s our register is updated because the results from the C MP is a negative

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number.

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So after that branch which is BLT this BLT here you see is a variant of the branch instruction.

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This BLT is moving forward and the way this P L 2 works is that it goes to the AP s register and it

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checks the various bits and it makes its decision based on that and what happens is when in the AP s

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I register D The and it is set or enabled the BLT modifies the P.C. the program counter to branch to

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a different location rather than allowing the P.C. to work in its own orderly manner.

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And as we said the result of this C MP modified the end bit.

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Therefore this keeps happening until we hit the number eight and when we hit the number eight will return

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to the beginning of the loop.