WEBVTT 00:05.870 --> 00:12.020 So guys, now let us try to understand the concept of transfer of data and transfer of computation and 00:12.020 --> 00:13.730 what is the difference between them. 00:14.000 --> 00:20.060 So in case of transfer of data, let us suppose that we have two entities the entity one and the entity 00:20.060 --> 00:20.630 two. 00:20.660 --> 00:23.900 The entity one and entity could be the two threads. 00:23.900 --> 00:25.700 There could be the two process. 00:25.700 --> 00:29.540 There could be the two process running on the same machine or different machine. 00:29.540 --> 00:30.320 Right. 00:30.410 --> 00:36.740 Entity one and entity two are just two bodies which wants to communicate with each other. 00:36.740 --> 00:37.400 Right? 00:38.570 --> 00:45.560 Now let us suppose that entity one owns two integers, let us say an integer A and integer B, whereas 00:45.590 --> 00:48.080 Entity two owns the computation part. 00:48.110 --> 00:51.950 That is entity two knows how to multiply two numbers. 00:51.980 --> 00:52.640 Right. 00:53.930 --> 00:59.870 Let us assume that entity one does not know how to multiply two numbers and it relies on the entity 00:59.870 --> 01:01.640 two to compute two numbers. 01:01.640 --> 01:02.300 Right. 01:02.540 --> 01:09.590 Also, let us assume that entity two has a local storage in which it can store the result which it obtains 01:09.590 --> 01:11.410 after multiplying the two numbers. 01:11.420 --> 01:11.990 Right. 01:12.000 --> 01:15.950 So C is the memory location, which is a local storage of the entity. 01:15.950 --> 01:22.430 Two So on whichever machine the entity two is running C is a local storage of that machine. 01:22.610 --> 01:28.130 Now let us suppose that we need to compute the multiplication of two numbers A and B, right? 01:28.130 --> 01:34.160 So it is quite obvious that entity one has to send these two numbers to somebody who can multiply, 01:34.160 --> 01:34.850 right? 01:34.850 --> 01:40.760 So it means that there should be a transfer of data from entity one to entity two. 01:40.790 --> 01:47.450 Once the entity two receives the values of A and B, it can do the computation on A and B and whatever 01:47.450 --> 01:53.280 final results it obtains, it can store this result to its local storage right. 01:53.850 --> 02:00.930 So you can see there is a transfer of data from entity one to entity two, and the computation is stationary, 02:00.930 --> 02:03.900 which is present on entity two side only. 02:03.900 --> 02:04.650 Right? 02:05.580 --> 02:08.960 Now let us understand what is transfer of computation. 02:08.970 --> 02:13.440 In fact, the transfer of computation is exactly opposite to the transfer of data. 02:13.440 --> 02:20.370 So in this case also we have two entities and one entity owns the data and the other entity owns the 02:20.370 --> 02:21.330 computation. 02:21.910 --> 02:24.220 Now the end goal is same. 02:24.220 --> 02:30.370 That is, the multiplication of these two numbers need to be stored in the local storage of the entity 02:30.370 --> 02:31.350 Two's memory. 02:31.360 --> 02:32.020 Right. 02:32.590 --> 02:40.270 But in this case, instead of sending the values of A and B from an entity, one to entity to entity 02:40.270 --> 02:45.820 two itself chooses to send the computation to the entity one. 02:45.820 --> 02:46.500 Right? 02:46.510 --> 02:51.370 That is, it is the function which is being sent from entity two to entity one. 02:51.370 --> 02:55.000 And once the entity one receives this computation logic. 02:56.360 --> 03:03.170 It can feed its data members as an argument to this computation logic and whatever result is produced, 03:03.170 --> 03:08.150 this result can be stored in the local storage, which is owned by the entity too. 03:08.330 --> 03:12.710 Now you must be saying that there is a transfer of data, right? 03:12.740 --> 03:17.270 The result Si is being sent from entity one to entity two. 03:17.270 --> 03:25.100 So no, actually there is no transfer of data because here we have a constraint that entity one and 03:25.100 --> 03:30.410 entity two must reside in the same virtual address space, right? 03:30.440 --> 03:37.310 That is entity one must have an access to the local storage of the entity two and entity two must have 03:37.310 --> 03:39.830 an access to the local storage of the entity. 03:39.830 --> 03:41.240 One Right. 03:41.600 --> 03:47.690 So transfer of computation is feasible only when entity one and two are in the same virtual address 03:47.690 --> 03:48.320 space. 03:48.320 --> 03:54.530 And now we already know what are the two entities who lives in the same virtual address space that is 03:54.530 --> 03:56.130 threads right? 03:57.000 --> 04:03.300 So it simply means that transfer of computation is possible only when entity one and entity two are 04:03.300 --> 04:05.370 the two threads of the same process. 04:05.610 --> 04:06.320 Right? 04:06.330 --> 04:10.290 They live in the same virtual address space of a process, right. 04:10.620 --> 04:16.710 So it simply means that the result Si is being stored by entity one into the memory location. 04:16.740 --> 04:21.510 Si which is owned by entity two without any transfer of data. 04:21.540 --> 04:27.120 The address of this Si memory location is accessible to the entity one also right? 04:27.120 --> 04:31.050 So it can directly write the result into this memory location. 04:31.740 --> 04:37.230 So in a nutshell, if you were to summarize what is transfer of computation, Transfer of computation 04:37.230 --> 04:41.880 is nothing, but it is just a function call through a pointer, right? 04:42.030 --> 04:47.550 So many of you must already be knowing that what is a function pointer and how does this work? 04:47.580 --> 04:50.040 I have just given you a broader perspective. 04:50.070 --> 04:54.660 That function pointer is actually an example of transfer of computation. 04:55.230 --> 05:00.540 Transfer of computation can also be implemented in different ways, not necessarily through function 05:00.540 --> 05:04.470 pointers, but that is the topic of some other course I guess. 05:05.080 --> 05:11.080 So now in the next lecture video, we will go through the example which demonstrates the transfer of 05:11.080 --> 05:12.010 computation.