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i CHAPTER 0.1 - NUMBERS AND ARITHMETIC You don't habitually use the base two system to balance your checkbook, so it would be counterproductive to teach you machine arithmetic on a base two system. What number systems have you had a lot of experience with? The base 10 system springs to mind. I'm going to show you what happens on a base 10 system so you will understand the structure of what happens with computer arithmetic. BASE 10 MACHINE Each place inside the microprocessor that can hold a number is called a REGISTER. Normally there are a dozen or so of these. Our base 10 machine has 4 digit registers. They can represent any number from 0000 to 9999. They are exactly like an industrial counters or the counters on your tape machines.{1} If you add 27 to a register, the microprocessor counts forward 27; if you subtract 153 from a register, the microprocessor counts backwards 153. Every time you add 1 to a register, it increments by 1 - that is 0245, 0246, 0247, 0248. Every time you subtract 1 from a register, it decrements by 1 - that is 3480, 3479, 3478, 3477. Let's do some more incrementing. 9997, 9998, 9999, 0000, 0001, 0002. Whoops! That's a problem. When the register reaches 9999 and we add 1, it changes to 0000, not 10,000. How can we tell the difference between 0000 and 10,000? We can't without a little help from the CPU.{2} Immediately after an arithmetical operation, the CPU knows whether you have gone through 10,000 (9999->0000). The CPU has something called a carry flag. It is internal to the CPU and can have the value 0 or 1. After each arithmetical operation, the CPU sets the CARRY FLAG to 1 if you went through the 9999/0000 boundary, and sets the carry flag to 0 if you didn't.{3} Here are some examples, showing addition, the result, and the carry flag. The carry flag is normally abbreviated by CF. number 1 number 2 result CF 0289 4782 5071 0 4398 2964 7382 0 8177 5826 4003 1 ____________________ 1. Exactly like industrial counters that have several hundred thousand parts, that is. 2. The CPU (central processing unit) is the chip(s) that does all the arithmetic. In the case of the PC, it is the 8086. 3. When you set a flag to 0, it is called CLEARING the flag. ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson The PC Assembler Tutor ii ______________________ 6744 4208 0952 1 Note that you must check the carry flag immediately after the arithmetical operation. If you wait, the CPU will reset it after the next arithmetical operation. Now let's do some decrementing. 0003, 0002, 0001, 0000, 9999, 9998. Golly gosh! Another problem. When we got to 0000, rather than getting -1, -2, we got 9999, 9998. Apparently 9999 stands for -1, 9998 stands for -2. Yes, that's the system on this, on the 8086, and on all computers. (Back to that in a moment.) How do we tell that the number went through 0 ; i.e. 0000->9999? The carry flag comes to the rescue again. If the number goes through the 9999/0000 boundary in either direction, the CPU sets the CF to 1; if it doesn't, the CPU sets the CF to 0. Here's some subtraction, with the result and the carry flag. number 1 number 2 result CF 8473 2752 5721 0 2836 4583 1747 1 0654 9281 8627 1 9281 0654 8627 0 Look at examples 3 and 4. The numbers are reversed. The results are the same but they have different signs. But that is as it should be. When you reverse the order in a subtraction, you get the same absolute value, only a different sign (15 - 7 = 8 but 7 - 15 = -8). Remember, the CF is reliable only immediately after the operation. NEGATIVE NUMBERS The negative numbers go 9999=-1, 9998=-2, 9997=-3, 9996=-4, 9995=-5 etc. A more negative number is denoted by a smaller number in the register; -5 = 10,000 -5 = 9995; -498 = 10,000 -498 = 9502, and in general, -x = 10,000 -x. Here are some negative numbers and their representations on our machine. number machine no number machine no -27 9973 -4652 5348 -8916 1084 -6155 3845 As you will notice, these numbers look exactly the same as the unsigned numbers. They ARE exactly the same as the unsigned numbers. The machine has no way of knowing whether a number in a register is signed or unsigned. Unlike BASIC or PASCAL which will complain whenever you try to use a number in an incorrect way, the machine will let you do it. This is the power and the curse of machine language. You are in complete control. It is your responsibility to keep track of whether a number is signed or unsigned. Which signed numbers should be positive and which negative? This has already been decided for you by the computer, but let's think Chapter 0.1 - Numbers and Arithmetic iii ____________________________________ out what a reasonable solution might be. We could have from 0000 to 8000 positive and from 9999 to 8001 negative, but that would give us 8001 positive numbers and 1999 negative numbers. That seems unbalanced. More importantly, if we take -(3279) the machine will give us 6721, which is a POSITIVE number. We don't want that. For reasons of symmetry, the positive numbers are 0000-4999 and the negative numbers are 9999-5000.{4} Our most negative number is -5000 = 10,000 -5000 = 5000. 10'S COMPLEMENT It's time for a digression. If we are going to be using negative numbers like -(473), changing from an external number to an internal number is going to be a bother: i.e. -473 -> 9527. Going the other way is going to be a pain too: i.e. 9527 -> -473. Well, it would be a problem except that we have some help. 0000 = 10,000 = 9999 +1 - 473 result 9526 +1 = 9527 Let's work this through carefully. On our machine, 0000 and 10000 (9999+1) are the same thing, so 0 - 473 is the same as 9999+1-473 which is the same as 9999-473+1. But when we have all 9s, this is a cinch. We never have to borrow - all we have to do is subtract each digit from 9 and then add 1 to the total. We may have to carry at the end, but that is a lot better than all those borrows. We'll do a few examples: (-4276) 0000 = 10,000 = 9999 +1 -4276 result 5723 +1 = 5724 (-3982) 0000 = 10,000 = 9999 +1 -3982 result 6017 +1 = 6018 (-2400) 0000 = 10,000 = 9999 +1 -2400 result 7599 +1 = 7600 (-1989) 0000 = 10,000 = 9999 +1 ____________________ 4. That way, if we tell the machine that we are working with signed numbers, all it has to do is look at the left digit. If the digit is 5-9, we have a negative number, if it is 0-4, we have a positive number. Note that 0000 is considered to be positive. This is true on all computers. The PC Assembler Tutor iv ______________________ -1989 result 8010 +1 = 8011 This is called 10s complement. Subtract each digit from 9, then add 1 to the total. One thing we should check is whether we get the same number back if we negate the negative result; i.e. does -(-1989)) = 1989? From the last example, we see that -1989 = 8011, so: (-8011) 0000 = 10,000 = 9999 +1 -8011 result 1988 +1 = 1989 It seems to work. In fact, it always works. See the footnote for the proof.{5} You are going to use this from time to time, so you might as well practice some. Here are 10 numbers to put into 10s complement form. The answers are in the footnote. (1) -628, (2) -4194, (3) -9983, (4) -1288, (5) -4058, (6) -6952, (7) -162, (8) -9, (9) -2744, (10) -5000.{6} The computer keeps track of whether a number is positive or negative. After an arithmetical operation, it sets a flag to tell whether the result is positive or negative. This flag has no meaning if you are using unsigned numbers. The computer is saying, "If the last arithmetical operation was with signed numbers, then this is the sign of the result." The flag is called the sign flag (SF). It is 0 if the number is positive and 1 if the number is negative. Let's decrement again and look at both the sign flag and carry flag. NUMBER SIGN CARRY 3 0 0 2 0 0 1 0 0 0 0 0 9999 1 1 ____________________ 5. Let x be any number. Then: -x = ( 10,000 - x) = ( 9999 + 1 - x ) ; -(-x) = ( 10,000 - (-x) ) = ( 9999 + 1 - (-x) ) = ( 9999 + 1 - ( 9999 + 1 - x ) ) = ( 9999 + 1 - 9999 - 1 + x ) = x 6. (1) -628 = 9372 , (2) -4194 = 5806 , (3) -9983 = 0017, (4) -1288 = 8712 , (5) -4058 = 5942 , (6) -6952 = 3048 (7) -162 = 9838 , (8) -9 = 9991 , (9) -2744 = 7256, (10) -5000 = 5000. This last one is a little strange. It changes 5000 into itself. In our system, 5000 is a negative number and it winds up as a negative number. This happens on all computers. If you take the maximum negative number and take its negative, you get the same number back. Chapter 0.1 - Numbers and Arithmetic v ____________________________________ 9998 1 0 9997 1 0 9996 1 0 That worked pretty well. The sign flag changed from 0 to 1 when we went from 0 to 9999 and the carry flag was set to 1 for that one operation so we could see that we had gone through the 9999/0000 boundary. Let's do some more decrementing. NUMBER SIGN CARRY 5003 1 0 5002 1 0 5001 1 0 5000 1 0 4999 0 0 4998 0 0 4997 0 0 4996 0 0 This one didn't work too well. 5000 is our most negative number (-5000) and 4999 is our most positive number; when we crossed the 4999/5000 boundary, the sign changed but there was nothing to tell us that the sign had changed. We need to make another flag. This one is called the overflow flag. We check the carry flag (CF) for the 0000/9999 boundary and we check the overflow flag for the 5000/4999 boundary. The last decrementing example with the overflow flag: NUMBER SIGN CARRY OVERFLOW 5003 1 0 0 5002 1 0 0 5001 1 0 0 5000 1 0 0 4999 0 0 1 4998 0 0 0 4997 0 0 0 4996 0 0 0 This time we can find out that we have gone through the boundary. We'll come back to how the computer sets the overflow flag later, but let's do some addition and subtraction now. UNSIGNED ADDITION AND SUBTRACTION Unsigned addition is done the same way as normally. The computer adds the two numbers. If the result is over 9999, it sets the carry flag and drops the left digit (i.e. 14625 -> 4625, CF = 1, 19137 -> 9137 CF = 1, 10000 -> 0000 CF = 1). The largest possible addition is 9999 + 9999 = 19998. This still has a 1 in the left digit. If the carry flag is set after an addition, the result must be between 10000 and 19998. The PC Assembler Tutor vi ______________________ Since this is unsigned addition, we won't worry about the sign flag or the overflow flag for the moment. Here are some examples of unsigned addition. NUMBER 1 NUMBER 2 RESULT CF 5147 2834 7981 0 6421 8888 5309 1 2910 6544 9454 0 6200 6321 2521 1 Directly after the addition, the computer has complete information about the number. If the carry flag is set, that means that there is an extra 10,000, so the result of the second example is 15309 and the result of the fourth example is 12521. There is no way to store all that information in 4 digits in memory so that extra information will be lost if it is not used immediately. Subtraction is similar. The machine subtracts, and if the answer is below 0000, it sets the carry flag, borrows 10000 and adds it to the result. -3158 -> -3135 + 10000 -> 6842 CF = 1 ; -8197 -> -8197 + 10000 -> 1803 CF = 1. After a subtraction, if the carry flag is set, you know the number is 10000 too big. Once again, the carry flag information must be used immediately or it will be lost. Here are some examples: NUMBER 1 NUMBER 2 RESULT CF 3872 2655 1217 0 9826 5967 3859 0 4561 7143 7418 1 2341 4907 7434 1 If the carry flag is set, the computer borrowed 10000, so example 3 is 7418 - 10000 = -2582 and example 4 is 7434 - 10000 = -2566. MODULAR ARITHMETIC What the computer is doing is modular arithmetic. Modular arithmetic is like a clock. If it is 11 o'clock and you go forward 1 hour it's now 12 o'clock; if it's 11 and you go backwards 1 hour it's now 10. If it's 11 and you go forward 4 hours it's not 15, it's 3. If it's 11 and you go backward 15 hours it's not -4, it's 8. The clock is doing mod 12 arithmetic.{7} (A+B) mod 12 (A-B) mod 12 From the clock's viewpoint, 11 o'clock today, 11 o'clock yesterday and 11 o'clock, June 8, 1754 are all the same thing. If ____________________ 7. To be a perfect analogy 12 o'clock should be 0 o'clock. Chapter 0.1 - Numbers and Arithmetic vii ____________________________________ you go forward 200 hours (that's 12X16 + 8) you will have the same result as going forward 8 hours. If you go backwards 200 hours (that's -(12X16 + 8) = -(12X16) -8) you get the same result as going backwards 8 hours. If you go forward 4 hours from 11 (11+4) mod 12 = 3 you get the same result as going backwards 8 hours (11-8) mod 12 = 3. In fact, these come in pairs. If A + B = 12, then going forward A hours gives the same result as going backwards B hours. Forwards 9 = backwards 3; forwards 7 = backwards 5; forwards 11 = backwards 1. In the mod 12 system, the following things are equivalent: (+72 + 4) (+72 - 8) (+60 + 4) (+60 - 8) (+48 + 4) (+48 - 8) (+36 + 4) (+36 - 8) (+24 + 4) (+24 - 8) (+12 + 4) (+12 - 8) ( 0 + 4) ( 0 - 8) (-12 + 4) (-12 - 8) (-24 + 4) (-24 - 8) (-36 + 4) (-36 - 8) (-48 + 4) (-48 - 8) (-60 + 4) (-60 - 8) They form what is known as an equivalence class mod 12. If you use any one of them for addition or subtraction, you will get the same result (mod 12) as with any other one. Here's some addition:{8} (+48 + 4) + 7 = (48 + 11) mod 12 = 11 (-48 - 8) + 7 = (48 - 1 ) mod 12 = 11 ( 0 - 8) + 7 = ( 0 - 1 ) mod 12 = 11 (-60 + 4) + 7 = (-60 +11) mod 12 = 11 And some subtraction: (+48 + 4) - 2 = (48 + 2 ) mod 12 = 2 (-48 - 8) - 2 = (48 - 10) mod 12 = 2 ( 0 - 8) - 2 = ( 0 - 10) mod 12 = 2 (-60 + 4) - 2 = (-60 + 2) mod 12 = 2 Our pretend computer doesn't cycle every 12 numbers, it cycles every 10,000 numbers - it is a mod 10,000 machine. On our machine, the number 6453 has the following equivalence class: (+30000 + 6453) (+30000 - 3547) (+20000 + 6453) (+20000 - 3547) (+10000 + 6453) (+10000 - 3547) ( 0 + 6453) ( 0 - 3547) (-10000 + 6453) (-10000 - 3547) (-20000 + 6453) (-20000 - 3547) (-30000 + 6453) (-30000 - 3547) ____________________ 8. (-10) mod 12 = 2 ; (-11) mod 12 = 1 The PC Assembler Tutor viii ______________________ Any one of these will act the same as any other one. Notice that 10000 - 3547 is the subtraction that we did to get the representation of -3547 on the machine. -3547 = 9999 + 1 3547 6452 + 1 = 6453 6453 and -3547 act EXACTLY the same on this machine. What this means is that there is no difference in adding signed or unsigned numbers on the machine. The result will be correct if interpreted as an unsigned number; it will also be correct if interpreted as a signed number. 6821 + 3179 = 10000 so -3179 = 6821 and 3179 = -6821 5429 + 4571 = 10000 so -4571 = 5429 and 4571 = -5429 Since -3179 and 6821 act the same on our machine and since -4571 and 5429 act the same, let's do some addition. Take your time so you understand why the signed and unsigned numbers are giving the same results mod 10000: --------------------------------------------------------- 6821 + 497 = 7318 -3179 + 497 = (10000 - 3179) + 497 = 10000 -2682 = -2682 7318 + 2682 = 10000 so -2682 = 7318 ---------------------------------------------------------- 5429 + 876 = 6305 -4571 + 876 = (10000 - 4571) + 876 = 10000 - 3695 = -3695 6305 + 3695 = 10000 so -3695 = 6305 ---------------------------------------------------------- Here's some subtraction: ----------------------------------------------------------- 6821 - 507 = 6314 -3179 - 507 = (10000 - 3179) - 507 = 10000 - 3686 = -3686 6314 + 3686 = 10000 so -3686 = 6314 ---------------------------------------------------------- 5429 - 178 = 5251 -4571 - 178 = (10000 - 4571) - 178 = 10000 - 4749 = -4749 5251 + 4749 = 10000 so -4749 = 5251 ----------------------------------------------------------- It is the same addition or subtraction. Interpreted one way it is Chapter 0.1 - Numbers and Arithmetic ix ____________________________________ signed addition or subtraction; interpreted another way it is unsigned addition or subtraction. The machine could have one operation for signed addition and another operation for unsigned addition, but this would be a waste of computer resources. These operations are exactly the same. This machine, like all computers, has only one integer addition operation and one integer subtraction operation. For each operation, it sets the flags of importance for both signed and unsigned arithmetic. For unsigned addition and subtraction, CF, the carry flag tells whether the 0000/9999 boundary has been crossed. For signed addition and subtraction, SF, the sign flag tells the sign of the result and OF, the overflow flag tells whether the result was too negative or too positive. SIGN EXTENSION Although our base 10 machine is set up for 4 digit numbers, it is possible to use it for numbers of any size by writing the appropriate software. We'll use 12 digit numbers as an example, though they could be of any length. The first problem is converting 4 digit numbers into 12 digit numbers. If the number is an unsigned number, this is no problem (we'll write the number in groups of 4 digits to keep it readable): 4816 -> 0000 0000 4816 9842 -> 0000 0000 9842 127 -> 0000 0000 0127 what if it is a signed number? The first thing we need to know about signed numbers is, what is positive and what is negative? Once again, for reasons of symmetry, we choose positive to be 0000 0000 0000 to 4999 9999 9999 and negative to be 5000 0000 0000 to 9999 9999 9999.{9} This longer number system cycles from 9999 9999 9999 to 0000 0000 0000. Therefore, for longer numbers, 0000 0000 0000 = 1 0000 0000 0000. They are equivalent. 0000 0000 0000 = 9999 9999 9999 + 1. If it is a positive signed number, it is still no problem (recall that in our 4 digit system, a positive number is between 0000 and 4999, a negative signed number is between 5000 and 9999). Here are some positive signed numbers and their conversions: 1974 -> 0000 0000 1974 1 -> 0000 0000 0001 3909 -> 0000 0000 3909 ____________________ 9. Once again, the sign will be decided by the left hand digit. If it is 0-4 it is a positive number; if it is 5-9 it is a negative number. The PC Assembler Tutor x ______________________ If it is a negative number, where did its representation come from in our 4 digit system? -x -> 9999 + 1 -x = 9999 - x + 1. This time it won't be 9999 + 1 but 9999 9999 9999 + 1. Let's have some examples. 4 DIGIT SYSTEM 12 DIGIT SYSTEM -1964 9999 + 1 9999 9999 9999 + 1 -1964 -1964 8035 -> 8036 9999 9999 8035 + 1 -> 9999 9999 8036 -2867 9999 + 1 9999 9999 9999 + 1 -2867 -2867 7132 -> 7133 9999 9999 7132 + 1 -> 9999 9999 7133 -182 9999 + 1 9999 9999 9999 + 1 -182 -182 9817 -> 9818 9999 9999 9817 + 1 -> 9999 9999 9818 As you can see, all you need to do to sign extend a negative number is to put 9s to the left. Can't those 9s on the left become 0s when we add that 1 at the end? No. In order for that to happen, the right four digits must be 9999. But that can only happen if the number to be negated is 0000: 9999 9999 9999 + 1 -0000 9999 9999 9999 + 1 -> 0000 0000 0000 In all other cases, adding 1 does not carry anything out of the right four digits. It is impossible to truncate one of these 12 digit numbers to a 4 digit number without making the results unreliable. Here are two examples: (number) 0000 0168 7451 -> 7451 (now a negative number) (actual value) +168 7451 -2549 (number) 9999 9643 2170 -> 2170 (now a positive number) (actual value) -356 7830 +2170 We now have 12 digit numbers. Is it possible to add them and subtract them? Yes but only 4 digits at a time. When you add with pencil and paper you carry left from each digit. The computer can carry left from each group of 4 digits. We'll do the following addition: 0138 6715 6037 + 2514 2759 7784 Chapter 0.1 - Numbers and Arithmetic xi ____________________________________ Do this with pencil and paper and write down all the carries. The computer is going to do this in 3 parts: 1) 6037 + 7784 2) 6715 + 2759 + carry (if any) 3) 0138 + 2514 + carry (if any) The first addition is our regular addition. It will set the carry flag if the 0000/9999 boundary was crossed (i.e. the result was larger than 9999). In our case CF = 1 since the result is 13821. The register holds 3821. We store 3821. Next, we need to add three things: 6715 + 2759 + CF (=1). There is an instruction like this on all computers. It adds two numbers plus the value of the carry flag. Our first addition was ADD (add two numbers). This time the machine instruction is ADC (add two numbers and the carry). The result of our second addition is 9475. The register holds 9475 and CF = 0. We store 9475. Finally, we need to add three more things: 0138 + 2514 + CF (=0). Once again we use ADC. The result is 2652, CF = 0. We store the 2652. That is the whole result: 2652 9475 3821 If CF = 1 at this point, the number has crossed the 9999,9999,9999/0000,0000,0000 boundary. This will work for signed numbers also. The only difference is that at the very end we don't check CF, we check OF to see if the 4999,9999,9999/5000,0000,0000 boundary has been crossed. Just to give you one more example we'll do a subtraction using the same numbers: 0138 6715 6037 2514 2759 7784 Notice that in order for you to do this with pencil and paper you'll have to put the larger number on top before you subtract. With the machine this is unnecessary. Go ahead and do the subtraction with pencil and paper. The machine can do this 4 digits at a time, so this is a three step process: 1) 6037 - 7784 2) 6715 - 2759 - borrow (if any) 3) 0138 - 2514 - borrow (if any) The first one is a regular subtraction and since the bottom number is larger, the result is 8253, CF = 1. (Perhaps you are puzzled because that's not the result that you got. Don't worry, it all comes out in the wash). Step two subtracts but also subtracts any borrow (We had a borrow because CF = 1). There is a special instruction called SBB (subtract with borrow) that does just that. 6715 - 2759 - 1 = 3955, CF = 0. We store the 3955 and go on to the third part. This also is SBB, but since we had no The PC Assembler Tutor xii ______________________ borrow, we have 0138 - 2514 - 0 = 7624, CF = 1. We store 7624. This is the end result, and since CF = 1, we have crossed the 9999,9999,9999/0000,0000,0000 boundary. This is going to be the representation of a negative number mod 1,0000,0000,0000. With pencil and paper, your result was: -2375 6044 1747 The machine result was: 7624 3955 8253 But CF was 1 at the end, so this represents a negative number. What number does it represent? Let's take its negative to get a positive number with the same absolute value: 9999 9999 9999 + 1 7624 3955 8253 2375 6044 1746 + 1 = 2375 6044 1747 This is the same thing you got with pencil and paper. The reason it looked wierd is that a negative number is always stored as its modular equivalent. If you want to read a negative number, you need to take its negative to get a positive number with the same absolute value. If we had been working with signed numbers, we wouldn't have checked CF at the very end, we would have checked OF to see if the 4999,9999,9999/5000,0000,0000 boundary had been crossed. If OF = 1 at the end, then the result was either too negative or too positive. OVERFLOW How does the machine decide that overflow has occured? First, what exactly is overflow and when is it possible for overflow to occur? Overflow is when the result of a signed addition or subtraction is either larger than the largest positive number or more negative than the most negative number. In the case of the 4 digit machine, larger than +4999 or more negative than -5000. If one number is negative and the other is positive, it is not possible for overflow to occur. Take +32 and -4791 as examples. If we start with the positive number (+32) and add the negative number (-4791), the result can't possibly be too positive. Similarly, if we start with the negative number (-4791) and add the positive number (+32), the result can't be too negative. Therefore, the result can be neither too positive nor too negative. Make sure you understand this before going on. What if both are positive? Then overflow is possible. Here are some examples: Chapter 0.1 - Numbers and Arithmetic xiii ____________________________________ (+3500) + (+4500) = 8000 = -2000 (+2872) + (+2872) = 5744 = -4256 (+1799) + (+4157) = 5956 = -4044 In each case, two positive numbers give a negative result. How about two negative numbers? (7154) + (6000) = 3154 = +3154 (actual value) -2946 -4000 (5387) + (5826) = 1213 = +1213 (actual value) -4613 -4174 (8053) + (6191) = 4244 = +4244 (actual value) -1947 -3809 The numbers underneath are the negative numbers that the numbers above them represent. In these cases, adding two negative numbers gives a positive result. This is what the machine checks for. Before the addition, it checks the signs of the numbers. If the signs are the same, then the result must also be the same sign or overflow has occurred.{10} Thus + and + must have a + result; - and - must have a - result. If not, OF (the overflow flag) is set (OF = 1). Otherwise OF is cleared (OF = 0). MULTIPLICATION Unsigned multiplication is easy. The machine simply multiplies the two numbers. Since the result can be up to 8 digits (the maximum result is 9999 X 9999 = 9998 0001) the machine uses two registers to hold the result. We'll call them R1 and R2. 5436 X 174 R1 0094 R2 5864 2641 X 2003 R1 0528 R2 9923 You need to know which register holds which half of the result, but besides that, everything is straightforward. On this machine R1 holds the left four digits and R2 holds the right four digits. Notice that our machine has changed the modular base from N to N*N (from 1 0000 to 1 0000 0000). What this means is that two things which are modularly equivalent under addition and subtraction are not necessarily equivalent under multiplication and division. 6281 and -3719 will not work the same. ____________________ 10. The machine checks something considerably more obscure because it is easier to implement in semiconductor logic, but what it is actually doing is checking to see if the two numbers being added have the same sign. If they do, the result must be the same sign or overflow has occurred. The PC Assembler Tutor xiv ______________________ The machine can't do signed multiplication. What it actually does is convert the numbers to positive numbers (if necessary), perform unsigned multiplication, and then do sign adjustment of the results (if necessary). It uses 2 registers for the result. SIGNED MULTIPLICATION REGS RESULT (number) (5372) X (3195) R1 8521 = -1478 6460 (actual value) -4628 X +3195 R2 3540 (number) (9164) X (8746) R1 0104 = +104 8344 (actual value) -836 X -1254 R2 8344 (number) (9927) X (0013) R1 9999 = -949 (actual value) -73 X +13 R2 9051 Looking at the last example, if we performed unsigned multiplication on those two numbers, we would have 9927 X 0013 = 0012 9051, a completely different answer from the one we got. Therefore, whenever you do multiplication, you have to tell the machine whether you want unsigned or signed multiplication. DIVISION Unsigned division is easy too. The machine divides one number by the other, puts the quotient in one register and the remainder in another. Once again, the only problem is remembering which register has the quotient and which register has the remainder. For us, the quotient is R1 and the remainder is R2. 6190 / 372 R1 0016 16 remainder 238 R2 0238 9845 / 11 R1 0895 895 remainder 0 R2 0000 As with multiplication, signed division is handled by the machine changing all numbers to positive numbers, performing unsigned division, then putting back the appropriate signs. SIGNED DIVISION REGS RESULT (number) (7192) / (9164) R1 0003 +3 rem. -300 (actual value)-2808 / -836 R2 9700 (number) (3753) / (9115) R1 9996 -4 rem. +213 (actual value)+3753 / -885 R2 0213 Looking at the last example, 3753 / 9115, if that were unsigned multiplication the answer would be 0 remainder 3753, a completely different answer from the signed division. Every time you do a division, you have to state whether you want unsigned or signed division.