1 00:00:00,002 --> 00:00:04,160 What we've viewed so far in the previous examples for memory addressing was the 2 00:00:04,160 --> 00:00:09,100 simplest form, which we have an address for it in our registry. 3 00:00:09,100 --> 00:00:16,268 And we just tell memory to return us the value of the location stored in, in the 4 00:00:16,268 --> 00:00:21,464 registry. Now we're going to see in, that so 5 00:00:21,464 --> 00:00:25,551 because in addressing mode. Now we're going to see the complete 6 00:00:25,551 --> 00:00:28,240 addressing modes supported by x86. Okay? 7 00:00:28,240 --> 00:00:32,595 So in these are the add, these are how addresses are used for accessing memory 8 00:00:32,595 --> 00:00:38,538 in a moving structure, them we saw a lot of them in the previous example. 9 00:00:38,538 --> 00:00:43,620 So that means the most general form of addressing more than a[INAUDIBLE] six is 10 00:00:43,620 --> 00:00:49,087 of the following form, okay you have D is a constant displacement that's added to, 11 00:00:49,087 --> 00:00:57,105 to, to the final address, okay, in this is a constant that has to be one. 12 00:00:57,105 --> 00:01:01,470 Two or four bytes. That's a constant displacement. 13 00:01:01,470 --> 00:01:03,770 That doesn't mean that it can be only off by four bytes. 14 00:01:03,770 --> 00:01:09,402 It means that the displacement distance itself has to be a value of one, two or 15 00:01:09,402 --> 00:01:13,120 four bytes. Okay. 16 00:01:13,120 --> 00:01:17,928 So now. The other primary here, our b, is a base 17 00:01:17,928 --> 00:01:22,326 register. It's called a base register because we 18 00:01:22,326 --> 00:01:27,040 adding, we're adding to that reference point. 19 00:01:27,040 --> 00:01:30,892 that's the base register where our, our address starts to be completed from. 20 00:01:30,892 --> 00:01:36,200 Ri, also called the index register, is used for things like arrays. 21 00:01:36,200 --> 00:01:42,640 That's why we have a multiplication here by the contents of Ri. 22 00:01:42,640 --> 00:01:47,544 And that's multiplied by a constant called S, which is the scale. 23 00:01:47,544 --> 00:01:50,200 Kay. And this value can be one, two, four or 24 00:01:50,200 --> 00:01:52,320 eight. Why those numbers? 25 00:01:52,320 --> 00:01:56,020 Well those numbers because we want to officially support. 26 00:01:56,020 --> 00:02:00,927 Arrays are value of these size, of these sizes, 'kay. 27 00:02:00,927 --> 00:02:04,939 So, but they're many special cases, on how to use this, complete addressing 28 00:02:04,939 --> 00:02:07,726 mode. For example, if you just use, in 29 00:02:07,726 --> 00:02:11,691 parenthesis, Rb, Ri, the final memory address is going to be the contents of Rb 30 00:02:11,691 --> 00:02:16,590 added. With the contents of Ri. 31 00:02:16,590 --> 00:02:20,230 And that's because in this case, S, variable S is just set to the, the 32 00:02:20,230 --> 00:02:24,680 constant S is set to one. The scale is set to one. 33 00:02:24,680 --> 00:02:34,280 Now we can also use D Rb, Ri, and that's just going to do Rb plus Ri, plus D. 34 00:02:34,280 --> 00:02:36,100 And the, the scale here is just set to one. 35 00:02:36,100 --> 00:02:38,850 That's the special case. Okay? 36 00:02:38,850 --> 00:02:42,390 And also, you could have no displacement at all, we're not adding anything, right? 37 00:02:42,390 --> 00:02:44,752 So it's in this displacing is in fact will be zero. 38 00:02:44,752 --> 00:02:50,140 Okay? So it adds rb, with the scale, multiplied 39 00:02:50,140 --> 00:02:55,540 by register Ri. So let's see some quick examples here. 40 00:02:55,540 --> 00:02:58,309 We have two registers in our example. EDX and ECX. 41 00:02:58,309 --> 00:03:05,740 EDX is set to F000 hex and ECX is set to 100. 42 00:03:05,740 --> 00:03:09,559 So now we're going to be exercising all these addressing modes here, let's start 43 00:03:09,559 --> 00:03:14,447 with the first one. this first one here our. 44 00:03:14,447 --> 00:03:22,788 So our d, the displacement is, 0x8. and we're adding this to the contents of 45 00:03:22,788 --> 00:03:25,122 edx. So, are we going to say that well, that's 46 00:03:25,122 --> 00:03:27,710 the value of edx? That's the constant the final addition is 47 00:03:27,710 --> 00:03:29,920 just an addition of these two. Okay? 48 00:03:29,920 --> 00:03:34,020 Now, the next one, is using this fall, this form here. 49 00:03:34,020 --> 00:03:36,950 Right? Which is just going to be rb plus ri, in 50 00:03:36,950 --> 00:03:43,188 this case just edx plus ecx, that's what we get here, f100. 51 00:03:43,188 --> 00:03:49,606 Now, the third example here, is using a a scale factor, okay. 52 00:03:49,606 --> 00:03:54,304 So, and what it's going to do is, it's going to get Edx, and add it to four 53 00:03:54,304 --> 00:04:00,544 times, because we have four here. Ecx, which is 100. 54 00:04:00,544 --> 00:04:07,405 So, in the end we get f400. So, now the final example here is d is 55 00:04:07,405 --> 00:04:12,869 set to 0x80. We do not have Rb, so Rb is empty. 56 00:04:14,130 --> 00:04:16,761 But we have our RI and we have a scale, too. 57 00:04:16,761 --> 00:04:19,923 So, what's this going to do? Well, it's going to do two times the 58 00:04:19,923 --> 00:04:22,658 contents of edx, okay? That's here. 59 00:04:22,658 --> 00:04:30,464 Added with the displacement, and we get the final address, 1e080. 60 00:04:30,464 --> 00:04:38,190 there's an instruction in X86 called leal that get a source. 61 00:04:38,190 --> 00:04:42,606 So it's source operand is an address mode expression is any of the expressions that 62 00:04:42,606 --> 00:04:46,244 I just showed you. And so what the this function does is it 63 00:04:46,244 --> 00:04:49,440 evaluates that address mode expression, and puts the resulting address into the 64 00:04:49,440 --> 00:04:52,989 destination operand, into the destination register. 65 00:04:52,989 --> 00:04:56,116 For example If I use this, what is it going to do? 66 00:04:56,116 --> 00:05:10,810 Well, eax is going to get, is going to give us eax equals edx + 4 times ecx. 67 00:05:10,810 --> 00:05:13,598 Isn't that cool? So, and you can use that to actually 68 00:05:13,598 --> 00:05:16,001 compute addresses. Right? 69 00:05:16,001 --> 00:05:18,481 So to go with that, this is without a memory reference, so if you have this in 70 00:05:18,481 --> 00:05:21,161 your c code,[INAUDIBLE] showed you this when he talked about pointers, and what 71 00:05:21,161 --> 00:05:24,980 do you want to do here? You want to get the address, of 72 00:05:24,980 --> 00:05:28,116 the[INAUDIBLE] element of the, of array x. 73 00:05:28,116 --> 00:05:30,117 Okay? See, so you just want to get the address, 74 00:05:30,117 --> 00:05:33,031 you're not doing a memory, you're not doing a memory access, so in this case 75 00:05:33,031 --> 00:05:36,628 you're going to be using this layout instruction. 76 00:05:36,628 --> 00:05:39,922 But this instruction is it, it can actually be used to compute any 77 00:05:39,922 --> 00:05:46,520 expression of the form, any arithmetic expression of the form x plus k times i. 78 00:05:46,520 --> 00:05:49,912 Now I'll tell a bit about some other arithmetic operations, then we're going 79 00:05:49,912 --> 00:05:54,670 to put everything together into a large example, okay, a larger example. 80 00:05:54,670 --> 00:05:55,615 So here we're doing, we have a bunch of instructions, the first one does a 81 00:05:55,615 --> 00:05:56,665 arithmetic addition okay, so it takes two operands as far from this destination and 82 00:05:56,665 --> 00:05:58,070 you get destination equals destination plus the source. 83 00:05:58,070 --> 00:06:06,966 Okay, the sub is very similar except that it does subtraction and imull, same 84 00:06:06,966 --> 00:06:14,762 thing, except that it does multiplication. 85 00:06:14,762 --> 00:06:20,861 So now we're going to see instructions that actually does shifts. 86 00:06:20,861 --> 00:06:27,125 So just shifts the shifts bit wise, okay. So shll is doing a shift arithmetic and 87 00:06:27,125 --> 00:06:31,589 it's called shift arithmetic because it preserves the sign bit, it does something 88 00:06:31,589 --> 00:06:39,173 special with the sign bit, okay. So this, L means left, that's why it's 89 00:06:39,173 --> 00:06:45,449 shifting to the left. So gets the destination value shifts left 90 00:06:45,449 --> 00:06:50,852 by the foruth operan and stores it back itno the destination register. 91 00:06:50,852 --> 00:06:57,282 Now SARL does the same thing but notice R because it shifts to the right. 92 00:06:57,282 --> 00:07:03,209 Okay. And this shrl, is shifting to, shifting 93 00:07:03,209 --> 00:07:08,900 to the right, okay. Except that this is shift logical 94 00:07:08,900 --> 00:07:12,342 community does not do anything with the sine bits. 95 00:07:12,342 --> 00:07:16,689 'Kay, so the xorl instruction it just does an xor of the destination with the 96 00:07:16,689 --> 00:07:21,170 source and stores it back to, to the destination. 97 00:07:21,170 --> 00:07:24,718 And does, a bitwise and, or does a bitwise or. 98 00:07:24,718 --> 00:07:30,890 'Kay, so then watch out for the argument order. 99 00:07:30,890 --> 00:07:34,884 That's actually very important. Yeah, because the, the, argument order, 100 00:07:34,884 --> 00:07:38,644 affects your results. Especially for subtraction. 101 00:07:38,644 --> 00:07:41,751 Hm? So it knows that there's no distinction 102 00:07:41,751 --> 00:07:45,603 here, between signed, and unsigned ints. Why is that? 103 00:07:46,810 --> 00:07:50,754 Well, it's because remember that if you use two complements you can do arithmetic 104 00:07:50,754 --> 00:07:55,046 without regards[INAUDIBLE] you simply can just[INAUDIBLE] or[INAUDIBLE] but most of 105 00:07:55,046 --> 00:07:58,526 these do have to worry about if you're doing shifts arithmetic or shifts 106 00:07:58,526 --> 00:08:06,500 logically because of the sign difference. Here's some, some other arithmetic 107 00:08:06,500 --> 00:08:10,072 operations. This is a unarlal unary instructions 108 00:08:10,072 --> 00:08:15,418 because it takes single operand. Incl just increments by 1, Decl, 109 00:08:15,418 --> 00:08:20,794 decrements by 1 and then neg-o Just changes the signment, okay just, just, it 110 00:08:20,794 --> 00:08:25,980 changes the sign. And, not just a bitwise not, of the 111 00:08:25,980 --> 00:08:30,024 contents of the register. And now you, you should look at the, the 112 00:08:30,024 --> 00:08:33,294 text book, for more instructions like mov, and so on. 113 00:08:33,294 --> 00:08:36,538 Divisions. Okay so now let's put everything together 114 00:08:36,538 --> 00:08:41,132 in a, in an example, that uses leal, and some other instructions as well. 115 00:08:41,132 --> 00:08:44,606 Okay? So recall that now we have the setup, 116 00:08:44,606 --> 00:08:47,770 that's just saving the, the stack pointer. 117 00:08:47,770 --> 00:08:51,220 And then finish is just restoring the stack pointer and returning. 118 00:08:51,220 --> 00:08:54,330 Okay? This is, this is, and this is the C code 119 00:08:54,330 --> 00:09:00,238 that we are implementing here. Let's see how this, this works. 120 00:09:00,238 --> 00:09:06,510 First of all, this is our stack so you, you know where our parameters are. 121 00:09:06,510 --> 00:09:08,050 We have three parameters. Kay. 122 00:09:08,050 --> 00:09:12,280 So this is IE 32 that means the parameters are stored onto the stack. 123 00:09:12,280 --> 00:09:17,230 Well these two instructions here are doing they're just reading X and Y and 124 00:09:17,230 --> 00:09:21,722 putting it into EAX and EDX. Okay? 125 00:09:21,722 --> 00:09:25,843 And that's what this one here is doing. This one is using the leal instruction, 126 00:09:25,843 --> 00:09:29,720 this is the leal instruction. What it's going to do, it's just going to 127 00:09:29,720 --> 00:09:35,446 add edx with eax and put it in ecx, okay? Recall that leal just evaluates 128 00:09:35,446 --> 00:09:41,074 expressions of any, any expression looks like an address. 129 00:09:41,074 --> 00:09:46,208 expression, okay. And now we use it again here, when we're 130 00:09:46,208 --> 00:09:49,989 multiplying. So, now we're executing this, this 131 00:09:49,989 --> 00:09:54,698 statement. And this layout, this instruction gets a, 132 00:09:54,698 --> 00:09:59,208 d, x. Okay, and it adds it with two times e to 133 00:09:59,208 --> 00:10:04,470 the x and is stored in e to the x. Okay? 134 00:10:04,470 --> 00:10:09,270 So that's why we get e dx equals y plus two times y which is the same thing as 135 00:10:09,270 --> 00:10:17,328 three times y. Now we're shifting arithmetic left here 136 00:10:17,328 --> 00:10:21,000 by 4. Hm, what does that mean to, to, to shift 137 00:10:21,000 --> 00:10:24,518 by 4. 4 bits is the same as multiplying by 2 to 138 00:10:24,518 --> 00:10:31,786 the 4 which is 16. That's why now we get 16 times 3. 139 00:10:31,786 --> 00:10:36,325 It equals 48. That's why we're now effectively storing 140 00:10:36,325 --> 00:10:41,266 48 times y into dx, because before we had, we had 3y, we multiply that by 16, 141 00:10:41,266 --> 00:10:47,470 now we have 48 multiplied by y, and storing into dx. 142 00:10:47,470 --> 00:10:51,160 And the next instruction, now is just, what is it doing. 143 00:10:51,160 --> 00:10:58,260 It's implementing this one here 'Kay? So, he does t5 equals t3 plus t4. 144 00:10:58,260 --> 00:11:01,650 But what he did was just inline this one. It's executing this one together as well, 145 00:11:01,650 --> 00:11:05,050 'kay? It's just putting it here, 'kay? 146 00:11:05,050 --> 00:11:08,495 That's why now what we're doing is four, that's the square number here, adding 147 00:11:08,495 --> 00:11:12,380 four. OK, which was a four here. 148 00:11:12,380 --> 00:11:20,540 With T4, there was a, computed by the previous instruction. 149 00:11:20,540 --> 00:11:24,882 And adding it with X, it just happens to be stored in EAX. 150 00:11:24,882 --> 00:11:30,269 And then storing back in EAX. And then finally. 151 00:11:30,269 --> 00:11:39,609 We are in this instruction here. This instruction is just implementing the 152 00:11:39,609 --> 00:11:43,244 final stack in our arith, arith function, okay? 153 00:11:43,244 --> 00:11:46,949 So things to note here is that, note that the instructions are in the different 154 00:11:46,949 --> 00:11:52,106 order than what appeared in the C code. Okay, so we see like at this, this orange 155 00:11:52,106 --> 00:11:56,322 instruction happens to implement that relates to these two instructions in c 156 00:11:56,322 --> 00:12:00,636 code, that these two statements in the c code. 157 00:12:00,636 --> 00:12:06,444 And also, green comes before blue in c, but blue comes before green in, in there, 158 00:12:06,444 --> 00:12:12,267 in the assembly code. Okay, and also it's interesting to note 159 00:12:12,267 --> 00:12:17,870 that some expressions require multiple instructions, and some instructions. 160 00:12:17,870 --> 00:12:24,610 Like this one that I just showed here, covers multiple expressions, okay? 161 00:12:24,610 --> 00:12:28,930 And you get the same exact code when you compile this. 162 00:12:28,930 --> 00:12:31,060 Why is that? Because the compiler is going to have to 163 00:12:31,060 --> 00:12:34,723 create temporary variables. Anyways. 164 00:12:34,723 --> 00:12:39,584 So, let me show you another example now. This example is implementing this logical 165 00:12:39,584 --> 00:12:43,874 function that evaluates all of these, that executes all of these expressions 166 00:12:43,874 --> 00:12:47,830 here. And again, here what this is like the, 167 00:12:47,830 --> 00:12:51,055 the previous example. We're just reading a parameter from the 168 00:12:51,055 --> 00:12:53,045 stack. Now, this function has two parameters, x, 169 00:12:53,045 --> 00:12:57,230 and y. this instruction is just loading the x 170 00:12:57,230 --> 00:13:02,774 parameter, which is 8 from EVP into EAX. Okay? 171 00:13:02,774 --> 00:13:05,060 And that's what the first instruction's doing now. 172 00:13:05,060 --> 00:13:09,040 This first instruction here, this x or is implementing this. 173 00:13:09,040 --> 00:13:11,122 A statement here. And look how interesting. 174 00:13:11,122 --> 00:13:14,832 This, this instruction is an interesting example because it has a register operand 175 00:13:14,832 --> 00:13:19,446 as well as a memory operand, okay? And this memory operand happens to point 176 00:13:19,446 --> 00:13:25,594 to er, where Y was located. So that's why this instruction implements 177 00:13:25,594 --> 00:13:30,284 X, X or Y. Now this instruction here is just 178 00:13:30,284 --> 00:13:35,892 shifting EAX right by 17 bits. So it's implementing this expression 179 00:13:35,892 --> 00:13:40,090 here, this, this statement here. And and we're shifting right so it 180 00:13:40,090 --> 00:13:44,840 chooses, it chooses the compiler chose the shift right instruction. 181 00:13:44,840 --> 00:13:50,620 And finally this instruction here is implementing these two statements, but 182 00:13:50,620 --> 00:13:56,060 there's something interesting here going on, right? 183 00:13:56,060 --> 00:14:00,190 What the compiler did is note that this expression here is a constant expression. 184 00:14:00,190 --> 00:14:03,318 But instead of inserting instructions to evaluate that, the compiler evaluates 185 00:14:03,318 --> 00:14:06,400 that at compile time, but uses a single number that represents that instruction 186 00:14:06,400 --> 00:14:11,370 using, that, that represents that statement and uses it in the instruction. 187 00:14:11,370 --> 00:14:14,702 So you don't have to do this in during execution time, which is what our compile 188 00:14:14,702 --> 00:14:18,040 time wants, because we know it's not going to change. 189 00:14:18,040 --> 00:14:22,313 Then that code, that's, that's a, that's going to be much faster.