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Hello, my name is Typhoon and this is the last lecture of our section, and many arithmetic instructions

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are available, but we are going to show a selection of them.

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And the others are similar to what you learned here.

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So before we investigate arithmetic instructions and keep in mind that we use the printf with more than

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two arguments, so we need an additional register.

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The first one.

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The first argument goes into RTI.

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The second one into RSI and the third one into RDX.

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So this is how Printf expects us to provide the arguments in Linux and you will learn more about that

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later when we talk about the calling conventions, which you will learn that in next sections.

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And here there is a instructions here.

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The first instruction is Add, which can be used to add signed or unsigned integers.

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The second operand source is added to the first operand destination and the result is placed in the

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first operand destination here.

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And the destination operand can be a register or a memory location.

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So the source can be an immediate value, a register or a memory location.

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The source and destination cannot be a memory location in the same instruction, and when the resulting

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sum is too large to fit in the destination, the F flag is set for signed integers.

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So for unsigned integer, the o f flag is then set.

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So.

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When the result is zero, the z f flag is set to one and when the result is negative, the F flag is

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set.

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And the subtraction with sub is similar to the Add instruction.

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Uh, we can see it right here, adding and subtracting.

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So here the.

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To increment a register or a value in a memory location with one use the I and C instruction.

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Similarly, the a C can be used to decrement a register or value in a memory.

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Location with one.

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So the arithmetic shift instructions are a special breed.

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So here you will see that at the end of the code here somewhere here.

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So here if you shift left here.

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So the shift left is Sal is in fact multiplying.

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So if you shift left one position, you are multiplying by two.

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Every bit is shifted.

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One place to the left and zero is added to the right.

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So take the binary number one shift left one place and you obtain binary 10 or 2 in decimal representation.

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So you are getting from 0 to 2 and shift one shift left, one place again and you have a binary one

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zero 0 or 4 in decimal representation.

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So if you shift left two position, you multiply by four.

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But what if you want to multiply by six?

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So you shift left two times and then add two times the original source in that order.

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So shift right as our R here.

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Uh, is similar to a shift left, but it means dividing by two.

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So every bit is shifted one place to the right and an additional bit is added to the left.

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So.

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Here.

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There is a complication.

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However, if the original value was negative, the leftmost bit would be one, and if the shift instruction

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added two added a zero, the bit at the left, the value would become positive and the result would

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be wrong.

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So in the case of negative value as a R will add a one bit to the left and in the case of positive value,

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zero bits will be added to the left and this is called sign extension.

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So by the way, a quick way to see if a hexadecimal number is negative is to look at byte seven.

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This is the leftmost byte continuing from byte zero, which is the rightmost byte.

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So the number is negative.

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If byte zero starts with an eight, nine, A, B, C, D, E, or F, but you need to take into account

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all eight bytes.

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For example, the byte zero x.

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Uh, the device.

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The 12 is still a positive number because the leftmost byte, which is not shown here, is zero.

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And there's also no arithmetic shift instructions.

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They will be discussed in next lecture.

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And next here we multiply integers for multiplying the unsigned integers you can use mul here.

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Um, now we will hear, uh, mul for unsigned multiplication.

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So I think, uh, so the in the product shift.

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Right.

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Increment number.

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Total integer quotient and product.

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Yes.

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And here but we use the I'm u l.

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This is assigned multiple action which offers more flexibility so I can take one, 2 or 3 operands.

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And in our example we use one operand, as you can see here.

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And here.

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And the as I said in our example, we use this one operand and the operand following the I mul instruction.

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I mul instruction is multiplied with the value in rax right.

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So you may expect that the resulting product is stored in Rax, but that is not entirely correct.

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So let's illustrate an example here.

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So you can verify that when you multiply, for example, a two digit number with a three digit number,

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the product has 4 or 5 digits.

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So when you multiply 4 to 8 bit digit with 30 bit digit, you will obtain 77, 77 bit digit or 78 bit

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digit.

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And that value does not fit in 64 bit register.

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Right.

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So to cope with this, the instruction I moved here Am will store the lower 64 bits of the resulting

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product in Rax and upper 64 bits in RDX.

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And this can be very deceptive.

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Okay, so now let's experiment a little bit here.

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So go back to the source code here, modify the number one so that it contains here.

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Number one.

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One, two.

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Three, four.

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One, 234567890.

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And again.

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One, two, three, four, five, six, seven.

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That's it yeah seven and now and modify the number 2 to 100 and the product will just fit in racks so

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you can check the here you would.

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As you can see, we got an error, but missing node Unix like implies executable stack.

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Yeah, this is just a warning and not an error here.

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And as you can see here, we got very different result here.

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And if it started the budget.

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And in order to debug this, we need to we will put a break before the I mole instruction.

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The started here.

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And we will, uh, here we will put a break before the I instruction here, which is for the multiplying.

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That's it.

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Now.

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Let's go.

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And here we stopped it now.

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Ugh.

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Now.

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The result of the multiplication here will be.

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That's actually and that's the result of the multiple action here will be this number.

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And as you can see in racks after the instruction is executed.

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Right.

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Uh, the new modified.

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Now, uh, actually, let's actually modify number two into 10,000.

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Obsession.

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Let's stop the debugging.

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It still have that breakpoint here?

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One, two, three.

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Now it's in 10,000.

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The bugging is started.

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Let's go to email here.

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And with this, um, after, uh, of course, after setting the restart, after setting the 10,000 in

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number to stop the debugging and then start it again.

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And after look after the racks after executing moved here like this.

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And here.

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Um.

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So you can see that the product is a large negative value here, right?

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So this is because the most significant bit in Rax is a one.

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And CSM concludes that this must be a negative number.

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Also, Printf thinks that Rax contains a negative number because Rax contains a one bit in the leftmost

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position, so it's assumed to be negative.

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So be careful with printf.