자료구조0318

Mm-hmm.

ご視聴ありがとうございました。

Thank you.

네, 여기가 이거를.

- Oh.

and for a long time.

Thank you.

One second. Last week, I explained how you can evaluate the big O-type complexity. To measure the big O-type complexity, you have to count the number of iterations for every 4 loop, or Y loop, or 2 loop, or some strategies. I gave some examples. conveys your legal competency

The last example is the Pibonacci sequence program. Even if the function, the program, which computes, the Pibonacci frequency is quite simple. There are only two statements, but the complexity is very expensive. The constant time, it computes, the Pibonacci sequence is very, very expensive. Even if the program is quite simple, there are only two statements. Why, professor? Why can you say that the computing time of the Pibonacci sequence is too expensive?

I will explain just right now. If the bigger time complexity is equal to or larger than the exponential time complexity, then we say that the computing time is very very expensive. We call those programs whose bigger complexity is exponential, then we say that the problem is an intractable problem. What do I mean that the problem is an intractable problem, intractable problem, intractable algorithm?

This means that even if there is a solution, there is an algorithm, the constant time is too expensive. So, in reality, it is very difficult or almost impossible to get the solution. The constant time is too expensive. I will explain the reason why. For example, look here the Kimona sequence program.

And here, there are only two statements. Now, let's compute the bigger panchamber here. Now, let's here be the loading time, complete time for the pibou and If n is either 0 or 1, then it creates just constant time. The constant time 1, 2, 3, doesn't matter. Right? So, if n is either 0 or 1, this means that or

If M is 0 or 1, then T0 equals T1 is just constant time. Here, 1, 2, 3, 4, 1, 2, 3, the constant is very dynamic. So we can set T0 equals T1 equals 1. Why? If M is either 0 or 1, then this function P2M would never execute the state number 2. If M is either 0 or 1, then the computer is just constant time.

We will not institute the secondary two. If N is two or three or four or so on, no problem, no problem. If Seng-kyung should be skipped, the problem, the state number, so we face some constant time, constant time, just let's say that one more time. If N is two, three, four, five, so on, no. Each function should be skipped the state number one, so the country pi to exclude the state number one is just one.

Now, when you execute the statement number 2, what is the complete time to execute the statement number 2? The complete time to execute the statement number 2 is the complete time to execute the P4 and minus 1. What is the complete time to execute the P4 and minus 1? That is P and minus 1. Right? Now, what is the complete time to finish, to execute, to complete the statement number 2?

you make P4 and -2 that is P and -2. So when you execute the state number 2 then the committee will need P and -1, plus you should add this addition. So the total committee will be carried who else could save the number for years

E n minus 1 plus E n minus 2 plus 1. So, the total-portal-company-pine to achieve improvement step number 1 and step number 2 is E n minus 1 plus E n minus 2 plus 1 plus 1. So, total-company-pine is T n equal T n minus 1 plus E n minus 2 plus Thank you.

Can you understand this number two? Super. You have to evaluate, you have to find out the recursive solution for this sequence. I am quite quite sure that when you were high school student, then you learned the sequence. 여러분, 이런 수요를 배웠죠?

You know how to evaluate, how to solve this recursive equation when you were a high school student. Right? Can you see it? You can solve this recursive equation. But this recursive equation cannot be solved with the value when you look at the high school student. So, look here. Pbom equals Pbom minus 1 plus Pbom minus 2 by the definition of Pbom equals so.

T_n is larger than or equal to T_n star. Let's do this. Let's offset this inequality equation. Don't add this. Just offset this inequality. So, aha! Q_n is taking this number and this number. Looking to the reference, what is the point of type of list? but on the hand side also expresses a time-compress. So, conclusion.

Now, this quantity tie is between this exponential tie complexity and this exponential tie complexity. So, conclusion. Ah, the quantity tie to compute the P-volarchie-equalcy should be the exponential tie complexity. We would not concern, we would not mention about some co-action by the definition of pure So we will ignore.

the coefficients. Right? So, conclusion, this P-bo-M is between this exponential time complexity and this exponential time complexity. So, conclusion is P-bo-M is exponential time complexity. What do I say? If the bigger time complexity is equal or larger than the exponential time and computer system. What? What do I see?

This is the interactive program. In fact, the interactive program is so even if there is an algorithm, there is a solution, is a program, but in reality, it is very very difficult or even impossible to get as wide the constant times to expand. In fact, for these Kibonat sequence functions, the components that you will compute

people and thousand people. You cannot come to these people one thousand. Why? Maybe, you will get the air message saying the step overflow.

So, I will explain the long time span, maybe at the first or second week of the next month, April, that I will explain a little more about this anyway. This program is an example of the tractor problem. Even if it is quite simple program, there are only two statements, but the constant time takes the period of time complexity. The admin step, e-brief, version of solution, version algorithm, version program,

In reality, if L is a sufficient line in our integer number, then it is almost impossible to get the answer. This is the definition of omega-type complexity. Last week, I explained the omega-type complexity. The definition of B-O-type complexity and omega-type complexity is exactly the same. It's same, except the dimension of inequality. Look here. - Okay. - Okay.

The definition of bigger time complexity and omega time complexity are high, high, exactly the same. Next set, the direction of inequality. Let me define the bigger time complexity definition. We say that Fn = Big O G and E. There exists two constant C and M there. The search station is Fn = C multiplied by Gn whenever M is structured by Gn.

What's the difference between bigger time complexity and omega time complexity? Bigger time complexity is worst time complexity. This means the computing time will not exceed. The computing time will take much bigger time. Omega time complexity is the best time complexity. The computing time will take a release.

omega time complex. This is the law of our body. I have explained about this unless I specify as I say, the time computing is the bigger time complex. This is a good example to measure the bigger time complex here. What is the bigger time complex for this problem? It's quite simple. What do I say? Wherever you compute

the bigger panclexy, then you have to close the number of iterations for every fore loop, or wide loop, or two loop, or so on. Now, if that is fore loop, force, outer fore loop, and inner fore loop, this is left in the corner, what can I say? I have tension on the inner loop force, but how do I get the outer loop?

Here, just look at the inner loop first. How many iterations for this inner loop? Inner loop, exact end times. J becomes 1, 2, 3 and so on. Right? If you think arrives, then we will exit from here. Then, how many iterations for this outer loop, exact end times? How many iterations for this inner loop? J , I , one more time.

The communication for this inner loop is I times. J becomes 1, 2, 3, 4, so on. However, J becomes I then will activate from this inner loop. The communication for this outer loop is L times. I = 0, 1, 2, 3, and so on. If I arrives at the number L, then you will activate from this inner loop.

아무걸로.

I will take my case here. Right? What did I say? When you look into the bigger time complexity, you will only look at the first term whose exponent is largest. You will ignore. You will skip. You will not consider the sixth or sixth form of terms. You will not think the whole stone whose complex is largest.

Now, look at the quite interesting example, whose complexity is row-time complexity. Now, look at this example. Oops, there is no full loop, but there is a wide loop. To measure the income-time complexity for this example, what did I say? to help count the number of iterations for every four loop

or why lose, or do lose, or so on. Now, first, uniquely, J is said to be some reason of the one. Suppose that you are given some could give natural number three, and should be one, or two, or three, or four, or so on. Right? And should be given whose value is equal to or larger than one. Right? Then, what happens?

So we think J is 1, 1 is less than or equal to L. Then A means that subroutine called some function A is a name of method, a name of function, a name of subroutine. Call some subroutine A, call some function A. And you will increase the value of J, J is B multiplied by J. So, now let's up.

compute the number of iterations for this y loop. Initially, j=1. Let's say that you are given some option number say n=50. Now, when you enter the y loop, 1 is less than 15. So, you will execute some function program A, called some function program, some subroutine A and Q.

program A, right? And then J becomes B multiplied by B. So J becomes B. Right? And then, go to the end of the day. Assume that B is less than or equal to L. You will execute

This is the fourth loop. You will execute the second loop. Folder-learned-tone. Folder-learned-tone. And J-requence, B multiplied by B, which is B-square. And go back to this Y-loop. Control B-square and N. Assume that B-square is less than or equal to N. You will execute the third operations. Stop!

the whole methodology, execute the third group in A. And then J becomes B square multiplied by B, which is B cubic. And go back to the Y loop and source. Eventually, eventually, eventually, if B power K, if this guy is "Water than" , if,

This Q is larger than L, then it will exit from this Y loop. So, A = Rook B and E. Let's solve this equation. A = Rook B. Okay, okay.

Okay, so if, let's solve this in equality equation. Let's solve this. So, Google can find out the small info, image number which says find this in equality. One more time. Eventually, if our play is larger than n, and will actually produce

So conclusion, let's solve this inequality. So, find out the smallest natural number, find out the smallest positive integer number which satisfy this inequality. So, find out, when you find out this, then find out the smallest integer number which satisfy this inequality. Can you see?

And I have failed networks, C, A, K, and I have a similar thing. So when you solve this inequality, D equals the opposite state function, log, B, and plus one. What did I say? You can ignore this sub-sequent term. You can ignore some co-efficient part. So conclusion. The bigger pie complexity for this program is log BF. Can you see?

Now, I gave some examples to measure the bigger type of flexi or the given floor. The single foreloop, nested foreloop, or the external type of flexi, or rope type of flexi, or so on.

I don't know whether I have to explain this PowerPoint 1.5 because this material, PowerPoint material 1.5, plays about some basic background about C++ language. So I don't know whether I have to explain this PowerPoint 1.5 or not because when you were freshman students in the last year, maybe last year, you learned the basic fundamentals about C++

So I can omit this. But just briefly talk about, just briefly explain about this. All a collection of a set of homogeneous data types. Just one more time. All a aggregation of a collection of a set. All a collection of homogeneous data types. I'm here.

This is the element of the array. This is the second element of the array. This is the third element array. And this is the last element of the array. The number of every element, the delta type of every member, the delta type of every node should be same. One more time. The second element, member, same terminology. So I will interchangeably use these terms.

element, member, or node, same. Anyway, the type of this guy, the data type of this guy, the data type of this guy, the data type of this guy should be same. So, did you explain, you showed the one-dimensional array? Yes, I do. This is an example of one-dimensional array.

So, consider the mid-dimensional array like this. This is the general set, two-dimensional array. Here, there are two elements, two components, long and two. Right? This is the first element. This is the last element. Practition. How can you access, how can you locate the element of the array?

You can access, you can locate the number of elements using the index for the subscript, for the position, subscript, index, position, same. So this one of the first array is a0. This is the first element, a0. This is the second element. If the Aracuta one members

M elements, M nodes in the one-dimensional array. This is the first element, first member, and this is the last element. The general element is KI. I is the index, or the position, or the subscript. You can rotate, you can find, you can manipulate every member of the one-dimensional one.

can manipulate, you can rotate every member, every element of the array using the position, using the subcript, using the index like this. This is one dimensional array, this is two dimensional array. Can you see? Okay, okay. You will not know about this, okay. The different types of these cards should be safe. All the eggs are cleturo.

members, or flagged member elements, or flagged element nodes with same data types, with homogeneous data types that can be located by the position, by the index, by the subscripts. You have to differentiate the array and the VAC. VAC is the best of example, VAC of integers. VAC of integers is a companion.

which holds the collection of interest back. Before interest, there may be sudden duplicates. Anyway, this is the example of the class that -- oops! Class is a reserved word. You are in the C++ translated. The aim is the name of class. This is the usually defined word, so you can change the name of the class as you want. the class as you want.

Now I define a class, beginning of the class, end of the class. What did I say? Class of the C++ language consists of two sections, public section and private section. Right? In the public section, when you define a method, a function, a member asset function, operations being in the public section, then the every item defined inside the public section will be visible, will be transported.

will be exported outside of the place. Every item defined inside the private section will be invisible, will be transparent outside the class. The only way to access, to manipulate the items defined inside the private section should be done using some operations using some methods, using some functions defined inside the public section.

I have already explained in the last week or two weeks ago, even if I can't remember the exact name. Anyway, now you define, look at this statement in the public statement. I define a new data file in general. CYPEF statement, file definition statement. What do I say? C++ language means, improve, tone, height,

What do I mean to only type programmer need? Identity parameters, the variables, the data type of every variable, the data type of every identifier should be defined before you use those variables. Moreover, the programmer can define new data type as you want. Moreover, the programmer can define sub-aggregated data types,

using the statement stiucit-soft-word class. I will explain the differences between stroke and class. Even if stroke and class are the example of aggregated data types, however, an item defined in the stroke will be in the outside class. In the class, the class and stroke with high symptoms.

However, there are some subtle differences. Class can manipulate, class can embody the notion of OOK. In class, you can separate the data object, data item from the operations, possible operations on this data item. Anyway, now, in the public section, I am inquiring a new data artist whose name is "Vegenton"

"Venue type" is user-defined word. So, you can change the name of "Venue type" you want. If my name is Kim, then I can define public high definition "Venue type" So, this is a name of new data type. Now, I define a new data type whose name is "Venue type". Here, the data type is required as "Venue type".

Using this high definition statement of now, I can regard, I can consider the very data as the same as the mixture data. Bag of clothes, bag of insurbs, a container which holds a collection of single position, the longer floating for now. Bag of catars, same way.

Now, I am very happy to explain the dynamic variable and static value.

Now let's talk about the static variable and dynamic. It's quite quite important. I think that you learn, so you can differentiate, discriminate the static variable and the dynamic variable. Today is March 18th, right? When you're in the middle of the 18th, you can see the same as the whole thing.

이 중목, 허리천, 이 중목, 막 흔들어 봤어요. 자, 20difference, the set variable and dynamic variable 동적인 변수는 뭐죠? 동적인 변수는 뭐예요? 동적인 변수는 막 움직이는 변수야 뭐예요? 여기 중몇에, 이 방안으로 배웠어요. 이 방안이 되었죠? 여기가 안 되었어요? 거기까지, 대유이뎅으로, 잘 되는 거예요. 동적인 변수는 뭐, 동적인 변수는 뭐.

Excuse me, can you explain the static or the dynamic variable? Can you understand the current and the current and the current and the current and current and current and current?

You are from the other country or the German. Right? You don't know the German language. Anyways. One of the dynamic variables. Dynamic variable means that, of course, talk about step variable. Step variable means that the computer system will resolve, will end up.

The minimum space when the compiler executes and the compiler compiles. When the compiler translates the high-level language into the machine language, then the compiler gives the memory allocation information to the load. And then, before the Comsta system executes the program, the project program, the target program, the machine language program, but the load and gauge

the memory space for this static variable. One more time. When the compiler translates the source program into the target program, into the object program, then the compiler gives the memory information the load. Then the load will allocate, will reserve the memory space for this static variable. and then...

This memory space will be allocated before the project program terminates. Even if in Unity, the program does not use the memory space. Once it is allocated, then the memory space will be kept, will be resolved until the computer system terminates.

the user program. In comparison, the dynamic area of the V-step, the computer system will resolve, will allocate the memory space only when the program uses the memory space. How can the computer system know when the user program uses the memory space for this dynamic

That is very good question. Use the pointer variable and use the new statement and delete the statement.

You learn the new statement and the statement. Okay, what's your new statement? You can't do that. When the computer system executes the new statement, one more time, when CPU, when the computer executes the new statement, one more time, the computer system will relocate the new space for this dynamic area. Well, I don't know.

when the computer system executes the new statement. When the computer system executes the data statement, then the computer system will return the allocated memory space, which was allocated, which was reserved for this dynamic variable to the available sheet space so that the other program may be used. Number one: "When the computer system executes the new statement of this,

The memory space for this dynamic variable will be allocated, will be resolved, will be canned. And if the programmer thinks that now I don't need the allocated memory space no longer, I don't need the memory space, I will not use the memory space anymore. Use the letter statement then, computer system will turn the memory space to abandon ship space so that the other program may use.

Can you understand? 방금 내가 얘기한 거, 선생님들, 알아듣겠어요? 못 알아듣겠어요? 우리 여학생, 방금 내가 얘기한 거 알아듣겠어요? 무슨 얘기인지 알아듣겠어요? 우리 학생, 알아듣어요. 알아듣는다, 이건 일인데, 그럼 모르면 우리 학생 들어보세요. 그러면 진짜 말아야, 진짜 잘 지어뒀지 말고, 우리만으로 노약을 해주고 있다.

요약은 안 되겠다는 얘기 탁대요. 그러면 다음 주로는 내가 물어보기 위해 스페인걸로 다 세게. 오케이. 이렇게. 그만 있지 말고 내가 당대는 게 아니라는 거예요. 제가 강수에요. 동생이에요. 내가 바로 부러울지 않고 우리많으로는 요약이 필요하냐는.

Okay, that's me who you claim to give the Korean language summary decision. The Korean language cheese. So let's talk about Korean language cheese. Please allow me to give the Korean language summary decision for a while.

"아니, please allow me to train for a while because many students are requesting." 자, 갑니다. 우리가 변수에는, Identifier에는, 정적 변수와 정적 변수 두 가지가 나와요. 우리가 특별한 언급이 없으면 변수는 정적 변수입니다. 이건 뭐냐 하면 이 변수에 대한 연관이, 기업 공간 스페이스가 언제 이루어져요?

소스 프로그램을 목적 프로그램, 사계 프로그램, 오브젝트 프로그램으로 변환할 때, 번역을 할 때, 변환할 때 로더에게 정보를 줘요. 로더 이리와. 이 변수에 대해서 공간이 이만큼 필요하니까, 내가 컴파일 다 끝난 다음에 네가 이 프로그램을 메모리에 답지할 때 이 변수에 대한 공간을 같이 네가 같이 마련해 줘야 돼. 공간 안 마련해 주면 큰일 나. 선생님한테 맴매야.

하고 그때 결정이 나요. 그럼 로더가 프로그램 시작하기 전에, Q&A CPU가 이 프로그램을 수행하기 전에 로더가 이 공간을 다시 실제로 만들어준다 이런 거. 결정이 문제나. 인터넷한테 이뤄진다. 그걸 로더에게 준단 말이에요. 그런데 도미를 통수로는 공간이 안 만들어줄까. 언제 만들어줄까. 그런 거에요. 유라는 명령을 사용할 때

라는 명령을 사용을 할 때 그때 비로소 공간이 마련해진다. 프로그램은 나이는 더 이상 인생이 변수해서 공간이 필요 없어. 안슬래. 필요 없어. 말도 하지 말고 뭐를 해요. 이 n 이 문장을 쓴다. 그러면 이 동적변수에 할당된 공간이 이용 가능한 히트 공간으로 분한이 돼요. 이 아침부터는 다른 프로그램을 한 이 만이 부족한 내 시기 때문이다. 내가 이 공간을 쓰야 한다.

누가 더 능률적이에요. 얘가 능률이에요. 얘가 능률적이에요. 얘가 능률적이에요. 마치 여러분이 도서관의 책을 빌릴 때 책을 한 번 빌리고 나서는 도로 갈 때까지 빌리면 도로 갈 때까지 못 나가면 제가 볼까 못 볼까. 하지만 내가 도서관의 책을 빌린 다음에 나한테 다 봤어. 피 없어. 그럼 리컬을 해주려는 거예요.

다른 사람이 이용 가능합니다. 오케이 오케이. 그런 면에서, in this sense, dynamic variable is quite quite additional. Right? 그래야 안 그래요. 그러면, 동적 변수를 어떻게 정의를 한다. pointer variable로 정의한다. 그래서 여러분이 정의하는 거예요. There is one night I say at the beginning of all states morning right now that,

I don't know whether I have to explain this PowerPoint material 1.5. Because if you learn the C++ language in the last year, sufficiently there. Ah, professor, let's skip this. I know everything about this. Why do you explain right now? I know everything about this. That is the reason why I say at the beginning of class today this morning,

I don't know whether I have to explain this or not. Because if you know about this static variable, what the direct variable, what the new instruction, what the daily instruction, what the pointer variable, so on, then I may speak. Okay? So the size of the bag is

to remind it at the time of compilation time. The size of dynamic batch is defined when the CONTRA system is executing the new instruction. The size of array will be defined when the compiler translates the source program into target line. The size of dynamic array will can we define

when the computer system extudes the new instruction. OK. OK. 자, dynamic array, stack array, 얘기했어요.

Here is a white pointer. Pointer is not overname of the mongong. Dengdenging. Dengdenging is what is mongong? Dengdenging is not overname. Pointer is memory location. Pointer is memory location. memory location. Let's take 10 minutes break.

그리고 저도 important notice. 여기 통계 듣는 학생도 확률이 듣는 학생이. 자, 오케이. 됐어요. 오늘 오후에 LC 수업은 휴강입니다. 왜? 내가 오늘 히사 인터넷진흥원에서 회의가 있습니다. 그래서, 나는 히사 인터넷을 통계에 대한 공개의 공개가 있습니다.

스테이크 클래스 스테이크 클래스. 스테이크 클래스. Can you see? 자, 오늘 확률 동계 수업은 내가 키다. 인터넷이 원해도 회의가 입수 때문에 키. 스테이크 클래스.

이라면 Some people they were able to build up or design or like it

Point is memory location, memory address for some given variable and poor to have a particular point. Put in the whole words, point is memory location address of the variable or the high-risk. Address, political address for the physical.

Then you define this in general vertical variable. Integer AB. This is vertical variable. Now, I define two variables whose name is A and B. What's the data type of this guy? Integer data. The programmer does not have to know in detail how the integer value A, the integer value B is

specified is defined on some specific computer system. No, the program does not have to know detail. This is the notion of data extraction, information hiding, and calculation. The program, the user, does not have to know detail how the static variable, the vertical variable A use the wads complement or twos complement or the signed magnitude. No, the program does not have to know.

This is an example of musical data. In fact, when the compiler translates, the source program in... Just wait for a while. I won't have it.

and the rest of the day.

Thank you.

When the compiler translates, convert the source program into the target program then, and then when the loader loads place the target program on the main memory of the computer system,

the actual memory location will be determined to one more time. After compiler reports, transmits the source program into the target program and load or place the target program in the main memory. At that time, the exact memory location, the exact memory address this root criteria over A is gamma-1.

So every logical variable has specific memory locations such as 0, 1, 2, and m-1. This is the location, this is the address of the main memory. The main memory, look at this, this is the main memory. Name memory. Okay. Who can you name memory?

White addressable memory and world addressable memory. White addressable memory means that every body has its own address such as 0123. White addressable memory means that every word, not everybody, every word is

Not every byte has its own address memory located such as 0, 1, 2, 3 and so on. So the size of memory is 0, 1, 2, 3 and M-. In general, memory is the byte addressable memory, but that depends on some specific hardware computer system. You journey.

Memory means the byte addressable memory. That means in general, every byte has its own location. Zero address, address one, address two, address one thousand, address one million, address one billion and so. But that depends on specific hardware system. Okay? That's a good idea. So,

Look at it. They are all together very different. So, the size of the main memory is 1. Address 0 and the last address is M-1. So, how many total bytes in this main memory? Except M bytes.

example of one-dimensional array, I can omit this. Right? Proof is double Y10. This is the fourth member is the first element in Y general, the last element in Y9. They are both of them. Y10. Y10. Y10. Y10. Y10. Y10.

this one dimension or a y? Excuse me. How many bytes will be allocated this one dimension or a y? Shira gave more than 0.1 seconds. Excuse me. How many will be allocated this one dimension or a y? It means 4 bytes, I think. It could be 400 bytes. 400 bytes.

How about, he says that 400 miles will be resolved for this one demand or a mile. Because I am very democracy professor, and I am very generous and very valuable professor, so in general, I would follow the majority of period slugs, but in some rare cases,

I should not follow the measuring group. Look here. I was already surprised. Look here. This is an example one-dimensional. What did I say? The root element is Y-general. And the last element is Y-general. There are 10 animals. And every beta time, beta time over, every member should be able to

same data type. Look here, double float y. This means that, for example, what do you mean the double float y? This means that when I explain the single species on floating point and double species on floating point hy? This is a presentation of single species on floating point. For example, float

Look here, this is variable X, this is logical. In fact, when the compiler converts, translates, the source point in the target program, and when the loader plays, the target program in the main memory, that at that time, the exact memory location for this single-plicitial floating point X will be determined. Assume that the memory location is 1000.

And then, this is 4 bytes. Float, X, Y, and here. It is 4 bytes. So, the left-mode bit is result for the side bit. Right? Side bit is followed by 7 bits of exponent power. And this guy will be followed by remaining beans.

This remaining is called Reservable for the Menteja part or the Characteristic part. So, many parts will be reserved for the Menteja part, for the characteristic part of single precision floating point X. Simul fine, good. Just see it. D'olvo. Cinta e pilla. Avez, bevi.

얼굴이 혹시 뭐라고 그러지요? 한여름이 사실 강릉으로. 한여름이 있었어요. 한여름이. I'm a song. I'm a surprise. 이런 것. 여기 보세요. 전체가 32bit씩 5분만. OK OK. 자 그러면 싸이 비트 한 비트. 스프레스 파트 일곱 비트. 그러면 32bit에서 한 비트 빼고 일곱 비트 빼고 몇 비트 하나.

24비트가 남지 24비트를 4로 나누면 얼마가 돼요? 24비트를 8비트로 하면 얼마가 돼요? 3마이트가 아니죠. 내가 뭐라 그랬어요? 싱글 프리션 프로틴 포인트는 인펙티브 유효수자 개수가 몇 개라고 했어요. 몇 개라고 그랬어요. 생각 안 나는데 금방 나와요. 여섯 개예요 왜?

16증법으로 나타날 때, 16증 숫자나 나타낸 데서 몇 비트가 필요해요? 4비트는 4지요. 그냥 4요. OK OK. In order to represent the traditional digit, we need just 4 bits. 5,000,000.

Okay, okay. MQI, H, B, will represent two effective digits. Why? Could represent, for hexa-one hexadecimal digits, we need four digits. So, MQI, H, B, will represent six effective digits.

I said in the last week, when you define a single-tricial problem like this, there would be total only the six effective digits. So, to extend the effective digits, we attach the new work to the total total number of the two.

메디페이자 팟, will be increased from penny to penny to penny for class 32. 그냥 그래요. 이게 뭐예요? 싱글 프리시 존에 메디페이자 팟트의 길이죠. 이게 뭐예요? 더블 프리시 존에 메디페이자 팟트의 길이죠. 그냥 그래요. 오케이 오케이. 그냥 그래요.

수자 개수를 몇 개가 되죠. 24를 4로 나니까 6개에서 이건 얼마에요? 50이죠. 56으로 4로 넣으면 얼마에요? 14가 되죠. 아하 유효수자가 6개에서 14로 판결을 하는다. 이런 얘기에요. 언제? 더블 프로 지정으로 나타내죠. 그리고 그는 5개에서 10로 판결을 하는다. 그는 5개에서 10로 판결이 되는다.

So, to keep, how many bytes will be returned for this Y0? 8 bytes. I'm sure. Now, let's say, single precision on the whole time is 4 bytes. When you define single precision floating point, then the contact system will allocate 4 bytes, not a little bit, but again, that depends on some specific analysis and impact. Usually, in general,

Most of the task will allocate 4 by 32 bits from the single friction floating point. Okay, okay. Okay, okay. Then, when you define the double friction floating point like this set, all together attach 32 plus 32 equals 64. This is the first time we have to do this.

The assistant will allocate, will return, will keep all parallel and it divides for the double friction floating point x. Look here, double float xy. Then, at the time, double friction floating point x will keep, will keep at the time. And then,

So, the first one, the double twist of floating point will take, will kill 8 bytes. Ok, ok. So, then, look here, double float 1. How many bytes will be dissolved for this one-dimensional array? Look here. Y0 will take 8 bytes. Y1 paste 8 bytes and Y9 paste Y bytes.

how will the total bytes will be allocated to this one dimensional double-tri-stone protein-vide? 8+8+8. 이건 몇 번 더해요? 10번 더해요. 이거 10번 더하는 것이 역할. 수구단에서 이렇게 하면 금방 나오죠. 수구단 배워서 안되었어요. -아, 안쪽으로. -아래, 패시리.

Ah, how about 8 bytes will be allocated to this one-dimensional array? Right. Okay? Okay. Okay? Okay. Okay. Okay. I don't know. I don't know. This is an example of multi-dimensional array. Look here. example, each job has 5M.

There are also 5 rows. The row is row is row. The row is row is row. The row is row is row. There are also 10 columns. First column, second column, and Alaska. 0, 1, 2, and 9. Then, how many elements in the 2-dimensional array? There are only 50 elements, 50 members. Then, how many bytes will be allocated to this

2D는 x10. There are 15 members, and every member takes 4 bytes. 그래 안 그래? 그러면 얼마가 돼요? 얼마 금요? 4+50이니까 200 bytes. 2D는. OK, OK. 4, 안 돼. 2D는.

This is initialization of one-dimensional array or two-dimensional. 이거는 2차원 배어를 투기화하는 건데 이건 여러분이 1학년 배웠으니까 승리합시다. This is a way to initialize some one-dimensional array or two-dimensional. This is a way, this is a method, how you can initialize some one-dimensional array.

array. Okay, okay. When you represent multi-dimensional array, there are two ways. Low-eyed representation and colored-eyed depiction. Look here. The problem is that multi-dimensional array looks like this. Conserve is two-dimensional. Flow.

X, 10, 10. How many elements here? There are 200 members. Right? The fourth element at the last member. But this is the main memory. But main memory is one dimensional. So you should retwee this two dimensional array into this one dimensional memory.

So there are two ways, long-wide representation and calamite representation. Long-wide representation is the first. We'll represent this first row at this main memory, and then you will represent the next row, and then you will represent the next row and so on. This is the long-wide representation. "I'm a good person, I'm a good person, I'm a good person, I'm a good person,

the fourth column here. We present the fourth column in the main memory. And then we present the next column and so on. Okay, okay. Okay, okay. Okay, then you can compute the exact memory location for the multi-dimensional array. For example, for example, a second dimensional array double

x100, x1010. Then, assume this. The fourth element, assume that the memory address of x100, x100, assume this. Then, what is the effect of memory location of x1 is address x915.

What is this? You should be able to compute the exact memory location of this element. When you represent the low-wise representation or the color-wise representation, you should be able to compute this exact memory location. This is the way. This is the standard operation of the C-Cup-Cup-Lesson-Aging.

This is stand-up statement. This is the example of the story of the story. It's clear to me that it is true. Okay, they are all together, one dimension already. One hundred members, one hundred elements in this one-member library. The name of, this is the name of struct ui cs_tacmonchip.

The next element would be UICS department 1. The last element is UICS department 99. Every element has three fields. The fourth field, second field, and the last field. Okay, okay. What is the data type of the first field? Stood field. What is the data type of the second field? Stood number. What is the data type of last thing?

gpa 가 먼저 알죠. 점은 gpa 이라면 같은 높게

Thank you.

I will give 5 seconds and please tell me whether I have to explain this page in the name of the page of Rana. I will give 5 seconds and please say yes or no whether I have to explain page of the Rana. I will give 5 seconds and please tell me if I have to explain this page.

and yeah that's one look the last so yeah yeah yeah yeah yeah yeah

Thank you.

Some students say, "Professor, I don't need to explain this, but they don't want to catch me, ask me to explain this paper."

여러분이 사실은 인간에 이 품품 언어를 배웠으면 누워서 똥 먹으면 좋아. 안 돼요. 여러분 누워서 똥 먹으면 쉽다라는 얘기인데 누워서 똥 먹으면 좋아. 좋아. 누워서 똥 먹으면 좋겠어요. 그리고 앞으로도 안 먹으면 좋겠어요. 그리고 앞으로도 안 먹으면 좋겠어요.

누웠을 때는 먹지 마세요. 누워서 떡 먹으면 좋은 게 아니에요. 건강에 나빠요. 그런데 우리가 누워서 떡 먹기라고 하면 쉽다라는 것 같아요. 그렇지? 누워서 떡 먹으면 안 좋아요. 누워서 떡 먹지 마세요. 왜? 누워서 떡 먹으면 이런 병이야. 고기!

Thank you.

입에서 마우스에요. 길다란 관이 있습니다. 처음 음식물이 들어가면 이 저자구에서 애교가 먼다면 스토메크 위에요. 이 마우스에서 스토메크까지 롱뚤크가 롱뚤크가 롱뚤크가 롱뚤크가 롱뚤크가

short, bearded based on the tooth from Afterみ 손develop Estrada What length is about 30 seconds. When you eat some food, you will have a lot of food.

when eating stuff, then stomach emits the juice. The juice emitted from the stomach will fall the stomach acid. Stomach acid is very strong acid. The pH is about 1 or 2. What is the pH? -0 m, soio-nong-to. It's a very important thing.

pH수가 7이면 중성립이고요. pH가 6이 아니고 굉장히 간산이에요. pH가 7분하다면 알칸이에요. -5이고 10에 소기온 5이고 여러분 화학시간에 비어있어요. 이게 위산이 굉장히 간산이에요. 그래서 여러분 음식물을 넣으면 후계의 균, 마케리아 같은 게 헌물 같은 것과 위산이 굉장히 많아요. 복구로 많이, 태도로 많이 사는 것 같아요. 몸집에서 한국에서,

인간을 살만 미치고 움직여야 하는 게 있겠죠. 그 금이 이 금산에 있었다. 위산에는 어떤 생물도 못 사는 줄 알았는데 이 위산에 붙어 사는 금이 있어요. 그걸 뭐라고 하죠? 흰색이 왔다. 흰색이 왔다. 흰색이 왔다. 이 금이 위암을 발생한다는 희향들이 있습니다.

강산에서 위산에서 위생물을 보수할 줄 알았는데 이 강한 위산에서 투자하는 아주 대국한 세균인 증거를 발견한 사람이 호주의 아델레이드 대학에 있는 어떤 문과대학이 있었다. 1912년인가 1912년인가 다 됐다니 한국과정 조판에 이것을 발표를 했다. 그 사람이 공간의 의약사. 그저 이 강산에는 위산에는 또한 위생물을 보수한다. 됐습니다.

1902쯤에서 초반에 호주가 아덴덴이 되어야 할 위과 대학 조수가 이 위산에 있는 증폭한 도미이다. 그 귤을 하고 푸는 것이다. 뭐 넣어먹고 먹으면 위산이 거꾸로 올라가고 갈리아픔이 있어서 못 올라가고 있습니다. 이걸 도라미는 위산이 집도 다다고 고구로는 거니. 이 아이들아. 겨우스 공.

에스포지얼 리플럭스 루스라고 해서 이 위산이 영육성 식도량입니다. 누워서 떡 먹지 마세요. 누워서 떡 먹으면 이게 아닙니다. 에스포에스포지얼 리플럭스는 영육성 식도량입니다. 여러분이 다류 구조에서 앱티비어블 포인트는 몰라도 누워서 떡 먹으면 영육성 식도량입니다. 여러분이 다른데요.

알겠어요? 모두 떵목을 안해. 모두 떵목을 안해. 오케이. 자. 몇 일정. 이 캐슨이 도입래. 이 캐슨이 도입래. 이 캐슨이 도입래. 포인터는 지역 메모리 로케이션. address 0, address 1, address 2, address 10, address 100, and so on. 포인트 매력이 100종

for a pointer of some variables. You can define both the variables using the x-ray just at the left-hand side of the variable. One more time. Look here. Look here means integer x. Unless I specify x-ray, every variable in your step variable. This is a general step in the end.

Right? Now, consider this flow as a A. This is a pointer variable. How can you say that variable A is a set variable? Look here. There is an entry just at the left side of the variable A. One more. As a in place or position of a left side of the Ruch Pagelioblebe.

OK. 자 보세요. 친구 이름 A에 바로 왼쪽에 as a, 이 있죠. 그러니까 이건 뭐예요? 친구 이거 비례를 다 읽고 있어요. A라는 변수는 우리가 일반적으로 쓰는 싱글 포지션 실수를 표현하는 변수가 아니라 뭘 나타내요? 메모리 로케이션. 그 번지수를 가르친다 이런 이름. 다시 한 번. This is an example of pointer very broken.

variable A is defined as the pointer variable, cost of the asterisk. Look here. Asterisk is placed position at the right hand side of the vertical variable name A. So, the vertical variable A is indicated, integrates the memory location of the four-byte, where the four-byte will contain No single pressure floating point.

We present it like this. Look here. The left most of the leaf is the German form of side leaf. Side leaf is followed by seven bit of the story part. Seven bits of the story part is followed by the main and four bits of the character's part. So, how is the video? It is for a...

This memory location. Look here. It's wrong. No, it's in memory. A is should be an exact memory location. Can you see? A라는 변수는 뭐해요? 실수, 100.0을 나타내는 게 아니냐. 이건 뭐해요? 이 어떤 4 파이드를 표현된 실수, 싱글탈전 실수를

which is the "co-safara" and "dain" when "op-tap" to S-M. You have a quick variant, "dainemic variant". So, point variant is "high-tent-time" of variant. This is a huge number of location or sun variant. Point variant is represented by an ethnic just before the left side of the point of the variant. For example,

자, 같이 한번 내가 읽어보겠습니다. I will give 5 seconds. Should I explain this page 14 or page 15? I will give 5 seconds. 이거 중간고사에 나올 가능성이 높아요. There is very high chance to give similar problems 와, 이스크림 도베리오.

"Beat an exam was finally she." "There is a high chance, but I can guarantee just right now." 내가 지금은 고등학은 못하지만, 시험에서 나올 가능성이, 네가 비슷한 문제가 나올 가능성이. How did you five seconds? Thim, fear, cry, survive, eyes. Tell me you come out. You're ready. Can I stream this painting?

Page 14 and 15, what? May I skip? No. No. You're going to need it. Because you are very smart, so you say no. So, what's going on? Should I explain page 14 or page 15, what may I skip? Hmm. It's not going to be fine. It's not going to be fine. But, you can't do it. Now, I define, who value of the answer value of the answer. So,

variable a will take some integer value. Variable b will contain some integer value. We don't know the exact value right now. Now, integer as the P1 as the P1 is not. I define two point of view of the P1. How can you say that?

an exact memory location, for example, 1 million, or 1,000, or 100 and so on. This T1 will indicate an exact memory location of 4 bytes, and 4 bytes of the main memory will contain some integer value, but we don't know the exact integer value right now. So, you find two variables, 1,000, like 2 keys, A = 10, A = 3.

So, mince value A will compare some mince value L and the next value will be comparing mince exact mince value L. Can you see? Now, look here. E1 equals 10% A. This means that. We go. But I'm going to see.

to look at it. The right hand side first. The right hand side, ampersand A. This means that the exact memory location, the exact memory address of the logical variable A will be moved will be moved to the point of the algorithm. After you ask you, the statement of the

The pointer variable p1 now indicates the exact memory location of this 4 byte. And this 4 byte will contain the integer value 10. Here is one more point. When you execute this state 1, you want to look at the right-hand side first. You get the right-hand side first. What do you get right-hand side? 10 percent. It's a sense. We get memory order. We get the right-hand side.

right memory location, like you said, right memory address of the variable B will be moved, will be copied to the point of variable P. So, answer you ask you, these stakeholders, point of variable P1 indicates, 2x, contains the exact memory location, the exact right of the variable which contains the

Okay, okay. Now, look at last statement. S-CQ1 = S-CQ1. What happens? Look at your brain. What did I say? Don't look at the left hand side force. Look at the right hand side force. Now, S-CQ1. What's the S-CQ1? As in now, S-CQ1 conveys 2Q. Okay.

SAP 1 contains the content of the provide which is indicated by the point variable keyword. What is SAP 1? SAP 1 is the content of the venue which is pointed by the point variable keyword. What is the content of the provide which is pointed by which is delivered by?

the variable one is the inter-value path. I have done the left hand side. So now, you get, I have done at the right hand side. And then, look at the left hand side. So, what is the content of the right hand side? Inter-value. So, inter-value path will be moved to the right hand side. What is the right hand side? The right hand side is the

The content, S-E-P-T means the content of the four by, content of the one word, which is point by the point of variable equal. So, the statement of the seven will be moved to the four by, which is point by the point of variable equal. So, the content of the four by, the content of the one word,

Today's object interval to the which is point of value, point of area of view. Now, I will give five similar views to the meaning of the theme or the final theme of the evening. I can't guarantee right now to check everything. What do you think?

Look here, what do I say? We can get the left side force. Whenever you look at this kind of side statement, look at the right side force. This means that what is the actual problem?

SAP1 is the content of the new location, which is point by the point of variable P1. One more time. What is the right-hand side, SAP1? SAP1 is the content of the new location, which is point by the point of variable P1. Can you see? This content will be moved

On-directed side, SAP. SAP is content for the memory, pitch is point by pointer value to... 오케이 오케이. 여러분, 유할드 디퍼런셜이, 크리스틱 번호. 이거 설명을 잘해야 돼요.

Bye.

Thank you.

여러분 이 문장하고 이 문장하고 그걸 잘해야 돼요. You would give a shape by from this guy. What is the result from this guy? 어떻게 돼요? 이걸 수행하면 어떻게 돼요? 여기 하나 빼먹는다. 뭐가 돼요? 여기 하나 빼먹는다. 여기 하나 빼먹는다. 여기 하나 빼먹는다.

You have to differentiate state number A from state number 2. When you execute state number 2, what happens? The content of the right hand side, what is the content of the right hand side? Inject integer value minus. When you execute state number A, integer value minus 10 will be moved to here.

So, the point of variable now P1 desegregates the eject integer value minus 10 when you execute the state number. A, for Q, okay? Okay? Now let's use G1 equal P2. So, that's what we do. Now, what could I say? Look at the right hand side force. What's a pituit? Pituit is a jackpot.

memory location contains the integer value -10. Now, P2 is the memory location, object memory address, the byte address, byte number, which contains the integer value -6. Remember whether by the scanning, it changed by one's complement to one's complement, one's complement, one's side-back, not with my nose.

That's the notion of the elevation hiding, or the location. That's it. Anyway, anyway, what is the reason to do this? If we eat, if we eat, what is the location of this world? In case, I will get the picture value of minus 10. Okay, so, look at this, take the number. Take number P, P1 = P2.

Now the contents of this memory, which is pointed by pointer variable P2, will be moved to this memory location. No. When you expect number two, let's say that the memory location of the P2 is 5,000. 5,000 will be moved to the left hand side. Appreciate it.

So, when you execute the state number B, then what happens? Now, cointervaluable P1 indicates, contains 5,000. What's the 5,000? 5,000 begins, but in the future, the cointervaluable P1 indicates, contains, the cointervaluable minus set, okay, okay. So, when you execute the state number B, Thank you.

point of variable t1 and t2, or two point of variable t2 states with the same memory location. Okay, okay. Okay. You have already read it. I strongly believe that you have already known about this point of variable in the last year when you watch it. Okay.