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Lena [83]
2 years ago
9

Which pdu is processed when a host computer is de-encapsulating a message at the transport layer of the tcp/ip model?.

Computers and Technology
1 answer:
yulyashka [42]2 years ago
5 0

The PDU that is processed when a host computer is de-encapsulating a message at the transport layer of the tcp/ip model is segment.

<h3>What is this PDU about?</h3>

Note that in the transport layer, a host computer is said to often de-encapsulate  what we call a segment so that they can be able to put back data to an acceptable or given format through the use of the application layer protocol that belongs to the TCP/IP model.

Therefore, The PDU that is processed when a host computer is de-encapsulating a message at the transport layer of the tcp/ip model is segment as it is the right thing to do.

Learn more about host computer from

brainly.com/question/553980

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suppose a malloc implementation returns 8-byte aligned addresses and uses an explicit free list where the next and previous poin
Naya [18.7K]

4 bytes because even if you call malloc and free a little more frequently, that might still occur. malloc frequently uses the extra space for managing (a linked list of all memory blocks and their sizes) when it takes a few bytes more than what is requested.

There is a good chance that you will tamper with the internal management structures and cause the subsequent malloc of free will to crash if you write some bytes either before or after your allocated block.

4004 because malloc internally always allocates at least four bytes. Your program would be 4000 if you added four bytes, making it 4004. may crash only when you access byte n+1. Additionally, the operating system typically only protects pages of memory.

Your process may be able to read-write the remainder of the page and will only crash when accessing the next memory page if your malloc-ed byte is in the middle of a page with a size of 512 bytes. But keep in mind: Even if this works, the behavior is not clear.

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Full Question = Suppose a malloc implementation returns 8-byte aligned addresses and uses an explicit free list where the next and previous pointers are each 32-bits. Blocks have a 32-bit header and 32-bit footer, where the low-order bit of the header and footer are used to indicate whether the block is allocated (1) or free (0). Furthermore, the block size (which includes the header, payload, footer, and any necessary padding) is rounded up to the nearest multiple of 8, and this size (in bytes) is stored in the header and footer. Assume any padding must be between the payload and the footer.

a) If we call malloc(1), what block size will be allocated, in bytes?

b) With the same conditions, and assuming we've already called malloc(1), if the heap used by malloc starts at address 0x4000 (16384 in decimal), what address would be returned if we then called malloc(32)?

8 0
1 year ago
Phil wants to make a dark themed superhero movie. What could be his target demographic
maria [59]

Answer:

d

Explanation:

i think, because it makes the most sense to me.

5 0
3 years ago
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(1) Prompt the user to enter two words and a number, storing each into separate variables. Then, output those three values on a
Sergeu [11.5K]

Answer:

import java.util.Scanner;

public class num3 {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       System.out.println("Enter Favorite color");

       String word1 = in.next();

       System.out.println("Enter Pet's name");

       String word2 = in.next();

       System.out.println("Enter a number");

       int num = in.nextInt();

       //Point One

       System.out.println("You entered: "+word1+" "+word2+

               " "+num);

       //Point Two

       String passwordOne = word1+"_"+word2;

       String passwordTwo = num+word1+num;

       System.out.println("First password: "+passwordOne);

       System.out.println("Second password: "+passwordTwo);

       //Point Three

       int len_passwrdOne = passwordOne.length();

       int len_passwrdTwo = passwordTwo.length();

       System.out.println("Number of characters in "+passwordOne+" :" +

               " "+len_passwrdOne);

       System.out.println("Number of characters in "+passwordTwo+" :" +

               " "+len_passwrdTwo);

   }

}

Explanation:

  • This question is solved using java programming language
  • The scanner class is used to receive the three variables (Two string and one integer)
  • Once the values have been received and stored in the respective variables (word1, word2 and num), Use different string concatenation to get the desired output as required by the question.
  • String concatenation in Java is acheived with the plus (+) operator.
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What is the full form of RPM in computer ​
alex41 [277]

Full from of RPM in computer is Revolutions Per Minute.

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Answer:

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Explanation:

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Data validation is a tool in Access used to put a constraint on how data is entered in the table. The input mask is a data validation type that forces users to enter data in a specified format for a given field or column.

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