1 00:00:00,000 --> 00:00:02,850 [MUSIC]. Now I'll tell a little bit about, x86-64. 2 00:00:02,850 --> 00:00:05,538 I'm also going to show you how to do the swap example we saw in the previous 3 00:00:05,538 --> 00:00:12,778 video. But now, with, x86-64 insertions as 4 00:00:12,778 --> 00:00:19,400 opposed to the 32 bit. Instructions, okay? 5 00:00:19,400 --> 00:00:28,225 So x86-64 has 16 registers, okay? And just, just like I32, there's some 6 00:00:28,225 --> 00:00:32,320 registers that have special use, but there's a lot more general purpose 7 00:00:32,320 --> 00:00:37,238 registers. Like all of these registers here are only 8 00:00:37,238 --> 00:00:44,488 present in the 64 bit of x86. And also know that while these registers 9 00:00:44,488 --> 00:00:50,240 Are 64 bits wide. You can also access the lower 32 bits of 10 00:00:50,240 --> 00:00:57,176 them as a register, as well. For example, you know, eax, which is also 11 00:00:57,176 --> 00:01:03,153 present in IA-32, is the low order 32 bits of register. 12 00:01:04,160 --> 00:01:07,210 Rax in x86-64. Okay. 13 00:01:07,210 --> 00:01:12,034 So each look when we think about any existing registers to 64 bits plus adding 14 00:01:12,034 --> 00:01:17,976 eight new registers. and now the way we look at operands in 32 15 00:01:17,976 --> 00:01:23,696 versus 64 bit modes the following Remember that we had, in, in I 32, look 16 00:01:23,696 --> 00:01:29,520 at move l. And l referred to moving 4 bytes worth of 17 00:01:29,520 --> 00:01:33,813 data. Now what we have in x86-64 is move q, 18 00:01:33,813 --> 00:01:42,435 it's a new type of data size that it's 8 bytes or 64 bits long. 19 00:01:42,435 --> 00:01:45,370 Okay. Now we also have other instructions, like 20 00:01:45,370 --> 00:01:48,451 addl that's presently in 32. Now we have addq. 21 00:01:48,451 --> 00:01:51,426 And then shift l and we can have shift q, and so on. 22 00:01:51,426 --> 00:01:57,478 So and also note that x86-64 can also use all the 32-bits instructions that 30, 23 00:01:57,478 --> 00:02:03,430 that generate 32-bit results. Okay? 24 00:02:03,430 --> 00:02:06,390 So ends like I showed you in the previous slide. 25 00:02:06,390 --> 00:02:10,050 We can use, we can reference the low order 32 bits of some 64 bit registers 26 00:02:10,050 --> 00:02:13,700 with the same name as we use in 32-bit ISA. 27 00:02:13,700 --> 00:02:21,053 And the, the high order 32 bits are set to zero. 28 00:02:23,060 --> 00:02:25,818 here's the example that we saw in a previous video. 29 00:02:25,818 --> 00:02:32,010 we had this swap function, that received two pointers, xp and xy, as parameters. 30 00:02:32,010 --> 00:02:35,700 And then it swaps the contents of these pointers, okay? 31 00:02:35,700 --> 00:02:39,988 remember that these two instructions here, what they were doing is they were 32 00:02:39,988 --> 00:02:44,724 reading The pointers that were stored in the stack because parameters in the I32, 33 00:02:44,724 --> 00:02:50,980 I, I32ABI passed by a, to stack. 'Kay, that's, that's one thing to note. 34 00:02:50,980 --> 00:03:02,084 And we've used all movl instructions. Because we're moving four bytes worth of 35 00:03:02,084 --> 00:03:03,680 data. Okay? 36 00:03:03,680 --> 00:03:07,328 So now, let's look at how this function, this piece of code will look like in 64 37 00:03:07,328 --> 00:03:12,415 bits. Okay, bam, the first thing to notice, 38 00:03:12,415 --> 00:03:15,910 what we have, movl. Four of them. 39 00:03:15,910 --> 00:03:22,110 And also, note that we are passing the parameters. 40 00:03:22,110 --> 00:03:25,514 My registers. Okay, so directly as opposed to this 41 00:03:25,514 --> 00:03:28,733 text. So XP, the XP parameter is stored in 42 00:03:28,733 --> 00:03:34,222 register RDI. NYP is stored in register RSI, and these 43 00:03:34,222 --> 00:03:43,575 are 64 bit pointers because adversaries in 64, in access 664 are 64 bit. 44 00:03:43,575 --> 00:03:46,610 long. And look how interesting. 45 00:03:46,610 --> 00:03:51,442 Now we have only four instructions. We're no longer reading the parameters 46 00:03:51,442 --> 00:03:54,430 from the stack, which is a big advantage, right. 47 00:03:54,430 --> 00:03:57,954 So it's, it's much faster, don't have to store it to memory and read it back. 48 00:03:57,954 --> 00:04:03,239 And we have much fewer instructions. And now here, what we're doing is this 49 00:04:03,239 --> 00:04:07,517 first instruction gets the contents of the, the of xp, and stores it in a 50 00:04:07,517 --> 00:04:12,410 temporary variable. In this case t0, right? 51 00:04:12,410 --> 00:04:16,620 That happens to be mapped to adx. We get yp, it's contents, right? 52 00:04:16,620 --> 00:04:18,320 That's why we're using the parentheses here. 53 00:04:18,320 --> 00:04:23,794 We're just referencing the, the address. Storing the other temporary variable, t1, 54 00:04:23,794 --> 00:04:29,512 happens to be map, mapped to eax. And staying with just storing the 55 00:04:29,512 --> 00:04:32,860 temporary variable back into the pointers, and then we return. 56 00:04:32,860 --> 00:04:38,090 It's much simpler, right? And because the main reason it's simpler 57 00:04:38,090 --> 00:04:43,030 is because we don't have to, to use the stack at all. 58 00:04:43,030 --> 00:04:48,255 No stack operations required, and note that we still moved. 59 00:04:48,255 --> 00:04:52,472 32 bits worth of data. Why is that? 60 00:04:52,472 --> 00:04:58,930 Well, because this integer's here. Happens to be four bytes long. 61 00:04:58,930 --> 00:05:02,650 Even though the pointer itself is 64 bits, the unit of data that it points to 62 00:05:02,650 --> 00:05:05,305 is a four byte. Integer. 63 00:05:05,305 --> 00:05:14,020 and, and we also use Move L tool. Because we're moving 4 bytes of data. 64 00:05:14,020 --> 00:05:16,722 We're still using the same type of instructions. 65 00:05:16,722 --> 00:05:23,083 Now suppose that we had long ends. Now if you're saying long ends here. 66 00:05:23,083 --> 00:05:26,515 We're saying that the union of data is 8 bytes here. 67 00:05:26,515 --> 00:05:29,543 'Kay? So, even if there is a byte, now the move 68 00:05:29,543 --> 00:05:33,870 operation itself has to be eight bytes long. 69 00:05:33,870 --> 00:05:37,765 Remember that q refers to eight byte words. 70 00:05:37,765 --> 00:05:41,427 Okay? So and now since the long int is eight 71 00:05:41,427 --> 00:05:46,694 bytes as well, we have to use movq instructions. 72 00:05:46,694 --> 00:05:54,542 'Kay? So, alright see you next time.