1 00:00:00,1 --> 00:00:01,924 Okay, now we are ready to start a new section. 2 00:00:01,924 --> 00:00:07,976 We just saw in the previous section Introduction to Architecture and Machine 3 00:00:07,976 --> 00:00:13,850 Code and how it relates to C, and now we are ready to start assembly programming, 4 00:00:13,850 --> 00:00:20,10 okay? So this actually is going to teach you 5 00:00:20,10 --> 00:00:26,240 about movie movie instructions. Not movie instructions, move instructions 6 00:00:26,240 --> 00:00:30,2 that move data between registers and memory and between registers themselves 7 00:00:30,2 --> 00:00:34,568 and what their operands. we are also going to see the various ways 8 00:00:34,568 --> 00:00:39,610 we have available to address memory to specify memory locations. 9 00:00:39,610 --> 00:00:43,108 then we can look at an example called swap, that just exchanges two pieces of 10 00:00:43,108 --> 00:00:46,530 data in memory, and both in 32 and 64-bits. 11 00:00:46,530 --> 00:00:50,818 So that's going to be an opportunity for you to see the difference between 32-bit 12 00:00:50,818 --> 00:00:55,443 assembly, and x86, and x86-64. Then we're going to look at arithmetic 13 00:00:55,443 --> 00:00:59,80 operations, how to use condition codes, how to do branches. 14 00:00:59,80 --> 00:01:02,618 how to write loops with assembly, and then we're going to end with switch 15 00:01:02,618 --> 00:01:06,12 statements. So if we recall from last section, we 16 00:01:06,12 --> 00:01:08,730 there's three basic kinds of instructions. 17 00:01:08,730 --> 00:01:13,353 Instructions that move data between memory and registers, instructions that 18 00:01:13,353 --> 00:01:18,450 perform arithmetic operations, like addition for example. 19 00:01:18,450 --> 00:01:22,232 And it could have as operands both a memory or or registers, and then the 20 00:01:22,232 --> 00:01:27,94 instructions that transfer control. Instructions that change the flow of 21 00:01:27,94 --> 00:01:30,88 execution of your program in the processor, okay? 22 00:01:30,88 --> 00:01:33,41 So, now let's start. We're going to see now is instructions 23 00:01:33,41 --> 00:01:35,990 that move data between registers and memory. 24 00:01:35,990 --> 00:01:39,830 There's two basic types of instructions. One is called the of that type. 25 00:01:39,830 --> 00:01:43,286 One is called load. A load takes an address as a parameter, 26 00:01:43,286 --> 00:01:47,434 goes to memory, gets that, the data stored in that address and puts it into 27 00:01:47,434 --> 00:01:51,830 the register, okay? A store does the opposite. 28 00:01:51,830 --> 00:01:55,275 A store takes a register as a parameter as well and also an address and takes the 29 00:01:55,275 --> 00:01:59,809 data off the register. And stores in the specified memory 30 00:01:59,809 --> 00:02:05,520 address and recall that memory is just, is index just like an array, okay? 31 00:02:05,520 --> 00:02:09,90 You have an array or a table, and it has multiple entries. 32 00:02:09,90 --> 00:02:11,550 So you're going to access that using an index. 33 00:02:11,550 --> 00:02:15,390 And the point of an index in an array or in a table, the memory is an address, an, 34 00:02:15,390 --> 00:02:21,3 a memory location. Okay, so we saw briefly that the last 35 00:02:21,3 --> 00:02:27,936 section, that there are eight general purpose registers in IA32. 36 00:02:27,936 --> 00:02:34,130 Of the eight registers, six of them, these six here are general purpose. 37 00:02:34,130 --> 00:02:38,260 And these two here, the esp and ebp have spectial meaning, and we're going to see 38 00:02:38,260 --> 00:02:42,790 how they're special in a, in a little bit. 39 00:02:42,790 --> 00:02:46,330 So there are multiple types of instructions to move data. 40 00:02:46,330 --> 00:02:48,890 They all start with these three letters here mov. 41 00:02:48,890 --> 00:02:53,44 Then this x here which in fact could be one of these three b, w, or l depending 42 00:02:53,44 --> 00:02:58,938 on the amount of data that's being moved. If you use movl, l says for long words so 43 00:02:58,938 --> 00:03:02,550 you're going to be moving 4-bytes worth of data, okay? 44 00:03:02,550 --> 00:03:06,970 If you use movw you're going to be moving two bytes worth of data, and if you use 45 00:03:06,970 --> 00:03:11,50 movb you're going to be using, you're going to be moving 1-byte worth of data, 46 00:03:11,50 --> 00:03:16,41 okay? now in move instructions takes two 47 00:03:16,41 --> 00:03:21,404 operands, where the data comes from and where the data goes to. 48 00:03:21,404 --> 00:03:27,346 So now let's, let's look at the specific instructions. 49 00:03:27,346 --> 00:03:31,254 Movl source, and destination, so move data from the source to the destination. 50 00:03:31,254 --> 00:03:32,718 And these are operands, these are the source, this is a source operand and 51 00:03:32,718 --> 00:03:37,36 destination operand. And here's the type of operands, that you 52 00:03:37,36 --> 00:03:44,283 could have in a move instruction. The first one's called Immediate, think 53 00:03:44,283 --> 00:03:49,5 of it as a constant. This is a constant data, constant integer 54 00:03:49,5 --> 00:03:55,150 data imbedded in the instructions itself. For example, you could have x 400 as one 55 00:03:55,150 --> 00:04:00,58 of, as one of your operands and it only makes sense to be a source. 56 00:04:00,58 --> 00:04:02,830 This is doesn't make to have a constant, it doesn't make sense to have a constant 57 00:04:02,830 --> 00:04:06,618 as a destination. So a constant can only be a source of 58 00:04:06,618 --> 00:04:11,504 your move operation, okay? So now the other type of operand is a 59 00:04:11,504 --> 00:04:14,722 register, okay? It could be any of the eight registers 60 00:04:14,722 --> 00:04:19,733 here that we have, okay? In for example, if we do mov eax comma 61 00:04:19,733 --> 00:04:26,558 edx, what are we going to be doing? This'll eax, and edx are are operands, so 62 00:04:26,558 --> 00:04:30,284 eax in this case is going to be the source, and edx is going to be the 63 00:04:30,284 --> 00:04:35,106 destination. And we're going to get the contents in 64 00:04:35,106 --> 00:04:40,234 eax, in stored in, in edx, okay? So as I said before, esp and ebp, are 65 00:04:40,234 --> 00:04:45,968 reserved for special use, but they can still be used as operands. 66 00:04:45,968 --> 00:04:51,682 Now, the third and final type of operant in moving structures is memory. 67 00:04:51,682 --> 00:04:56,236 So, since we're talking about movl and we move 4-bytes of data that means that an 68 00:04:56,236 --> 00:05:01,135 address there is, which is be specifying the beginning of this four wide, 4-bytes 69 00:05:01,135 --> 00:05:06,726 word, that's going to be used as an operand. 70 00:05:06,726 --> 00:05:10,970 Okay, in the one way it is an operand, is to use to specify memory operand. 71 00:05:10,970 --> 00:05:14,6 Here's an example. It's to put, for example, a register IN 72 00:05:14,6 --> 00:05:20,110 parenthesis, when you're doing that you see that eax contains and address. 73 00:05:20,110 --> 00:05:25,240 And now the operant for your move abrasion is the data containing the 74 00:05:25,240 --> 00:05:31,130 address that is stored in riser eax. But as we'll see later there are many 75 00:05:31,130 --> 00:05:34,370 many types of address modes many ways of specifying addresses. 76 00:05:34,370 --> 00:05:40,242 This is just one of them and so, the many combinations of moving structures. 77 00:05:40,242 --> 00:05:47,382 So for example, continuing with our movl instruction, so the source can either be 78 00:05:47,382 --> 00:05:55,530 an immediate value, a registered value or a memory value, okay? 79 00:05:55,530 --> 00:05:59,373 Suppose that you have an immediate as a, as a source operand, and then you can 80 00:05:59,373 --> 00:06:04,340 have a register as a destination operand. Here's what we have, we have four is a 81 00:06:04,340 --> 00:06:07,352 con, that's an example here. Have an example here. 82 00:06:07,352 --> 00:06:12,472 four is a constant, so it is a source operand eax is the register and what it 83 00:06:12,472 --> 00:06:19,205 is going to do, it is going to store value x4 in register eax. 84 00:06:19,205 --> 00:06:23,605 Now, we can also have memory as a destination, so if we do movl, this 85 00:06:23,605 --> 00:06:29,365 constant minus 147 in decimal here, and store to eax in parentheses and as I just 86 00:06:29,365 --> 00:06:34,85 told you, the previous slide This says that now we're going to store it in a 87 00:06:34,85 --> 00:06:43,250 memory location specified by the address stored in register eax. 88 00:06:43,250 --> 00:06:47,400 Now, I'll continue our examples here. We could have register as a source and a 89 00:06:47,400 --> 00:06:50,550 register as a destination, that's the simplest case, that's a simple case, 90 00:06:50,550 --> 00:06:54,303 right. And here we have, we're going to get the 91 00:06:54,303 --> 00:07:00,812 contents of eax and copy to register edx. Now, we can go from register to memory. 92 00:07:00,812 --> 00:07:05,622 So, for example, we can get the contents of eax and store into, remember this is 93 00:07:05,622 --> 00:07:10,506 in parentheses, so, into the memory location specified by the asterisk stored 94 00:07:10,506 --> 00:07:18,340 in register edx. Finally, we could have memory as a source 95 00:07:18,340 --> 00:07:23,160 operand and registered as a destination operand. 96 00:07:23,160 --> 00:07:26,346 This will be a type of load, you're loading memory into, loading memory 97 00:07:26,346 --> 00:07:30,332 dating to a register. Okay, so here we're going to get the 98 00:07:30,332 --> 00:07:35,196 contents of the location specified by register or eax in store it in register 99 00:07:35,196 --> 00:07:39,818 edx. You cannot do memory-memory transfer with 100 00:07:39,818 --> 00:07:45,120 a single instruction. So, how would you do that, then? 101 00:07:45,120 --> 00:07:50,218 Well, you move it to a register. You could get data from memory to a 102 00:07:50,218 --> 00:07:53,708 register. That's step number one, and then you can 103 00:07:53,708 --> 00:07:59,180 get data from a register. So memory, that will be step number two. 104 00:07:59,180 --> 00:08:02,540 That's how you move data from memory to memory. 105 00:08:02,540 --> 00:08:06,516 So now let's see what's the C Analog of all of these operations. 106 00:08:06,516 --> 00:08:10,836 So again, I have for example variable eight that happens to be mapped in a 107 00:08:10,836 --> 00:08:14,634 register. and we store constant hex four and 108 00:08:14,634 --> 00:08:18,885 another example here. For example, I have this, this pointer 109 00:08:18,885 --> 00:08:22,770 p_a. We're going to de-reference the pointer 110 00:08:22,770 --> 00:08:29,230 and say we're going to store the constant minus 147 into the register, into the 111 00:08:29,230 --> 00:08:37,140 location specified by pointer p_a. Okay, so now since memory. 112 00:08:37,140 --> 00:08:40,624 That means that since this is a pointer, it's on the left-hand side, it's going to 113 00:08:40,624 --> 00:08:43,746 be the destination. That's why here in the destination, we 114 00:08:43,746 --> 00:08:48,792 have a memory operand. And if we look at the, the address held 115 00:08:48,792 --> 00:08:56,790 by pointer p_a, that's what's contained in register eax. 116 00:08:56,790 --> 00:09:00,560 So now these other three cases here are also pretty simple. 117 00:09:00,560 --> 00:09:05,450 They've had two register, two variables mapped into register, var d and var_a. 118 00:09:05,450 --> 00:09:09,737 We're just copying the value from var_a to var_d. 119 00:09:09,737 --> 00:09:13,769 And then here I have if I have var_a, which could be stored in a register is 120 00:09:13,769 --> 00:09:18,860 stored in a register. And I copy to a pointer, same thing here. 121 00:09:18,860 --> 00:09:24,908 Pointer p_d syncs memory since restoring pointers point somewhere in memory, the 122 00:09:24,908 --> 00:09:30,620 destination often has to be memory. And finally, here what we're doing is 123 00:09:30,620 --> 00:09:34,160 loading data in this last example here. We're loading data from memory so we have 124 00:09:34,160 --> 00:09:37,931 a pointer here on the right-hand side. Inter-registering, in this case is 125 00:09:37,931 --> 00:09:42,100 [INAUDIBLE] these happen to be matched to a register. 126 00:09:42,100 --> 00:09:47,468 So let's look at the various ways we have to address memory. 127 00:09:47,468 --> 00:09:52,417 So these are the basic addressing modes. There are many others, but the first one 128 00:09:52,417 --> 00:09:57,586 is called indirect access. Indirect just says that now the the 129 00:09:57,586 --> 00:10:02,968 register, contains an address, and then we're going to use that we're going, to 130 00:10:02,968 --> 00:10:09,370 use the contents of the register as an address to memory. 131 00:10:09,370 --> 00:10:12,746 So memory, as, as we said memory was just like an array, okay? 132 00:10:12,746 --> 00:10:16,894 So, that means that I will have an index here, and inside this index, is the 133 00:10:16,894 --> 00:10:23,24 contents of the register specified inside the address expression. 134 00:10:23,24 --> 00:10:28,24 An example here so before, probably I'm repeating this for the third time. 135 00:10:28,24 --> 00:10:33,544 Now this register ecx holds an address using the parentheses here, we say now we 136 00:10:33,544 --> 00:10:40,195 want the contents of memory whose address is stored in ECX. 137 00:10:40,195 --> 00:10:44,334 Now this is a basic one. The other type of addressing, we'll just 138 00:10:44,334 --> 00:10:47,245 call it displacement. And the way we specify is we have the 139 00:10:47,245 --> 00:10:51,80 little you know a constant outside our parenthesis here. 140 00:10:51,80 --> 00:10:55,790 Now all it does is it adds that constant to the base register. 141 00:10:55,790 --> 00:11:00,551 So register R specifies a memory address, but the final memory address is obtained 142 00:11:00,551 --> 00:11:06,350 by adding the con consultant register with with this constant D. 143 00:11:06,350 --> 00:11:11,370 For example, here what we're doing is we're getting these the contents of ebp. 144 00:11:11,370 --> 00:11:19,610 Okay, so the contents of ebp, epb we're adding eight. 145 00:11:19,610 --> 00:11:26,678 Okay, and we are using that as the address to memory, and we're storing that 146 00:11:26,678 --> 00:11:30,808 into edx. Easy, right? 147 00:11:30,808 --> 00:11:37,10 So now let's use this basic addressing modes in example. 148 00:11:37,10 --> 00:11:39,576 So you have this C function called swap that does default. 149 00:11:39,576 --> 00:11:48,221 It takes two pointers as a parameter and it gets the contents of XP, stores and, 150 00:11:48,221 --> 00:11:58,627 and stores into a temporary variable c0. It gets the contents of y, of the pointer 151 00:11:58,627 --> 00:12:03,710 yp and stores into, into temporary variable t1. 152 00:12:03,710 --> 00:12:06,920 And then stores t1 into xp and t0 into yp. 153 00:12:06,920 --> 00:12:11,120 So just swapping the contents pointed, swapping the locations pointed by these 154 00:12:11,120 --> 00:12:15,896 pointers. Now, as you see here, so okay, now this, 155 00:12:15,896 --> 00:12:20,270 this, this is the assembly version of our C code, the C code we just saw. 156 00:12:20,270 --> 00:12:23,900 And so we have three basic pieces here. One's called a set up. 157 00:12:23,900 --> 00:12:28,524 It's getting and rerading the parameters and, and, and setting up the stack and so 158 00:12:28,524 --> 00:12:31,526 on. We have the body, that is the body of the 159 00:12:31,526 --> 00:12:34,58 function. The body of the execution, and then we 160 00:12:34,58 --> 00:12:38,990 have the finish up, where we restore the stack to its original location. 161 00:12:38,990 --> 00:12:43,509 We restore register values and and then return to the caller. 162 00:12:46,350 --> 00:12:52,750 So let's see how, how stack works. Now, we have our the body, let's focus on 163 00:12:52,750 --> 00:12:56,990 the body here, that's the body of of our swap function. 164 00:12:56,990 --> 00:13:03,730 And we have six assembly instructions. So let's go now step by step. 165 00:13:03,730 --> 00:13:08,176 And the registers we're using here, here's the mapping of register to values. 166 00:13:08,176 --> 00:13:13,663 So, yp here happens to be mapped into register ecx, xp here happens to be 167 00:13:13,663 --> 00:13:21,718 mapped into register edx. Now t0 was mapped into register eax. 168 00:13:21,718 --> 00:13:26,614 And t1 was mapped into register ebx. And this mapping is a choice of the 169 00:13:26,614 --> 00:13:29,655 compiler, okay? The compiler chooses how to map the 170 00:13:29,655 --> 00:13:33,811 variables into, into registers. Okay, so now on the, on the right hand 171 00:13:33,811 --> 00:13:38,980 side here we have what each one of these assembly instructions is representing. 172 00:13:38,980 --> 00:13:46,228 And this is the state of our stack in the beginning of our execution. 173 00:13:46,228 --> 00:13:52,298 Okay, so now we are ready to, to, to see how this gets executed step by step. 174 00:13:52,298 --> 00:13:56,514 Okay, so now, we have, here we have our registers that's that, when we begin 175 00:13:56,514 --> 00:14:01,385 executing. And recall that both x, yp, and xp are 176 00:14:01,385 --> 00:14:06,2 addresses, right? So, they're pointers, that means that 177 00:14:06,2 --> 00:14:10,354 the, they can't, they're pointers, so they're variables that hold addresses and 178 00:14:10,354 --> 00:14:15,938 the addresses the address of yp is 120 which is right here. 179 00:14:15,938 --> 00:14:22,100 So and that means that yp points to this location and xp is 124. 180 00:14:22,100 --> 00:14:27,180 That means that xp points to this location here. 181 00:14:27,180 --> 00:14:33,646 And our registers and our ebp or stack pointer, is located in address 104. 182 00:14:33,646 --> 00:14:36,451 So, ebp points to here. Now we're going to see that a lot of the 183 00:14:36,451 --> 00:14:38,911 values are relative, a lot of memory locations are relative to the stack 184 00:14:38,911 --> 00:14:42,990 pointer. Let's execute a versus structure now. 185 00:14:42,990 --> 00:14:45,358 When we execute this move here, what are we doing? 186 00:14:45,358 --> 00:14:50,450 Well, we are storing the, the contents of yp as a pointer into ecx. 187 00:14:50,450 --> 00:14:57,15 So, since yp is stored 12-bytes away from ebp, we're going to use as a source for 188 00:14:57,15 --> 00:15:04,885 print a displacements address mode. So, the way the address is obtained is by 189 00:15:04,885 --> 00:15:08,980 getting ebp adding 12 and we're going to use that memory that adds up and then 190 00:15:08,980 --> 00:15:13,629 store it in this address, that's what happens. 191 00:15:13,629 --> 00:15:24,158 So we've got 120 And store it, store it in ecx. 192 00:15:24,158 --> 00:15:28,903 Great, so what we did there, we just got then the address of yp, and stored in 193 00:15:28,903 --> 00:15:32,596 ecx. The same thing for xp, but now note that 194 00:15:32,596 --> 00:15:36,725 this changed to 8 now, because xp's stored 8 bytes away. 195 00:15:36,725 --> 00:15:40,600 From EBP. Great. 196 00:15:40,600 --> 00:15:45,144 So now I read the X codes at one twenty four which is the address held by pointer 197 00:15:45,144 --> 00:15:49,480 XP. So what, what we want to do now is store 198 00:15:49,480 --> 00:15:55,900 the contents of the pointer YP. Into register eax. 199 00:15:55,900 --> 00:16:01,50 But remember, that we had just put yp into ecx, so now ecx holds an address. 200 00:16:01,50 --> 00:16:04,300 That means if we just use ecx as, as an address, we're actually getting the data 201 00:16:04,300 --> 00:16:08,571 pointed by YP. And since yp was high out /g, add this 202 00:16:08,571 --> 00:16:13,378 one to any, that is pointing right here. Okay? 203 00:16:13,378 --> 00:16:17,228 When we execute this instruction we're getting the contents pointed by yp, that 204 00:16:17,228 --> 00:16:21,10 happens to be stored in ecx, and loading it into eax. 205 00:16:21,10 --> 00:16:26,4 That's what happened here, okay? All right, so now that's the next step, 206 00:16:26,4 --> 00:16:29,998 we're doing the same thing for xp now. Were getting the contents of the address 207 00:16:29,998 --> 00:16:36,142 pointed by xp and loading into ebx. But the point of xp was stored in 208 00:16:36,142 --> 00:16:43,461 register edx so we use a memory. Memory Source over here to get that in 209 00:16:43,461 --> 00:16:48,280 stored in edx and so we get and that happens to be. 210 00:16:48,280 --> 00:16:53,990 We have edx right which is yes. So edx is a destination, edx is 124. 211 00:16:53,990 --> 00:16:56,372 That's an address so we're going to use that. 212 00:16:56,372 --> 00:16:59,728 So, we got this as the address here. We got that contents there, and store it 213 00:16:59,728 --> 00:17:02,884 in edx, because that's the destination register. 214 00:17:02,884 --> 00:17:07,428 So, what are we going to do now? Well, now we're going to get the contents 215 00:17:07,428 --> 00:17:11,396 of eax, which was contents of ip and store in xp because we're swapping it, 216 00:17:11,396 --> 00:17:15,653 right? And we still have the address of xp in 217 00:17:15,653 --> 00:17:18,372 edx. So, now we have this as the destination 218 00:17:18,372 --> 00:17:21,578 operand. So when, when, when we do that, well 219 00:17:21,578 --> 00:17:26,760 what's going to happen, this gets updated with the, with the contents of eax. 220 00:17:26,760 --> 00:17:30,360 And same thing happens when we're storing to yp. 221 00:17:30,360 --> 00:17:34,140 And note that we just left the addresses of the pointers into, into the register, 222 00:17:34,140 --> 00:17:38,633 because we use them to read them, and then later to write to them.