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ivanzaharov [21]
4 years ago
7

Suppose that you are given the following partial data segment, which starts at address 0x0700 : .data idArray DWORD 1800, 1719,

1638, 1557, 1476, 1395, 1314, 1233, 1152, 1071, 990 u DWORD LENGTHOF idArray v DWORD SIZEOF idArray What value does EAX contain after the following code has executed? (Ignore the .0000 that Canvas sticks on the end) mov esi, OFFSET idArray mov eax, [esi+0*TYPE idArray]
Computers and Technology
1 answer:
Digiron [165]4 years ago
5 0

Answer:

The value EAX contain after the code has been executed is 1233

Explanation:

Solution

Now,

The below instruction describes the MOV operation.

The MOV esi, OFFSET idArray .

It shows the movement, the offset value of the idArray in the esi.

Thus,

Follow the below instruction

MOV eax, [esi+7*TYPE idArray]

It will move the address of the idArray at that position into eax.

Value at the 7th position is moved to eax.

Therefore, the value at eax will be 1233.

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Which of the following is an example of hypertext pattern reading?
Murrr4er [49]
A is the correct answer
5 0
3 years ago
Which is true about POP3 and IMAP for incoming email?
IceJOKER [234]
<h2>Answer:</h2>

Option A: Both POP3 and IMAP  keep email on an email server by default.

is the correct answer.

<h2>Explanation:</h2>

Following points will make the idea of POP3 and IMAP clear:

<h3>POP3:</h3>
  • POP3  is the acronym for Post Office Protocol 3.
  • POP3 is the method of receiving emails in which the emails received on the app can be downloaded on to the computer by having an internet connection.
  • These downloaded emails can be viewed offline whenever needed and managed as well.
  • POP3 makes it possible that the storage space of the default server do not run short as the downloaded emails are deleted from the server.
<h3>IMAP:</h3>
  • IMAP stands for Internet Message Access Protocol.
  • It is the method in which the mails are viewed and managed directly on the internet server instead of downloading them on the computer.
  • IMAP makes sure to manage the mails so carefully and timely so that unimportant mails are deleted to make sure that the storage space does not run short.
<h3>Conclusion:</h3>

So from these points we can make sure that both ways keep emails on email server but POP3 have option to download mails from server while IMAP dont have.

<h2>I hope it will help you!</h2>
8 0
3 years ago
What is inputted into a computer system?
Anna [14]

Answer:

Hey mate.....

Explanation:

This is ur answer.....

<em>Input refers to any information, or data, that is sent to a computer for processing. Input is often sent to the computer from a device such as a keyboard, mouse, or other input device. Putting it simple, input is the act of entering data into a computer.</em>

<em />

Hope it helps!

Mark me brainliest....

FOLLOW ME!!!! :)

4 0
3 years ago
ossless compression tools generally use either Huffman coding or Lempel-Ziv-Welch (LZW) coding. Discuss the advantages and disad
anygoal [31]

Answer:

It we were asked to develop a new data compression tool, it is recommended to use Huffman coding since it is easy to implement and it is widely used.

Explanation:

The pros and the cons of Huffman coding

Huffman coding is one of the most simple compressing encoding schemes and can be implemented easily and efficiently. It also has the advantage of not being patented like other methods (e.g. arithmetic codingfor example) which however are superior to Huffman coding in terms of resulting code length.

One thing not mentioned so far shall not be kept secret however: to decode our 96 bit of “brief wit” the potential receiver of the bit sequence does need the codes for all letters! In fact he doesn’t even know which letters are encoded at all! Adding this information, which is also called the “Huffman table” might use up more space than the original uncompressed sentence!

However: for longer texts the savings outweigh the added Huffman table length. One can also agree on a Huffman table to use that isn’t optimized for the exact text to be transmitted but is good in general. In the English language for example the letters “e” and “t” occur most often while “q” and “z” make up the least part of an average text and one can agree on one Huffman table to use that on average produces a good (=short) result. Once agreed upon it doesn’t have to be transmitted with every encoded text again.

One last thing to remember is that Huffman coding is not restricted to letters and text: it can be used for just any symbols, numbers or “abstract things” that can be assigned a bit sequence to. As such Huffman coding plays an important role in other compression algorithms like JPG compression for photos and MP3 for audio files.

The pros and the cons of Lempel-Ziv-Welch

The size of files usually increases to a great extent when it includes lots of repetitive data or monochrome images. LZW compression is the best technique for reducing the size of files containing more repetitive data. LZW compression is fast and simple to apply. Since this is a lossless compression technique, none of the contents in the file are lost during or after compression. The decompression algorithm always follows the compression algorithm. LZW algorithm is efficient because it does not need to pass the string table to the decompression code. The table can be recreated as it was during compression, using the input stream as data. This avoids insertion of large string translation table with the compression data.

3 0
4 years ago
CHALLENGE ACTIVITY |
fredd [130]

Answer:

maxSum = FindMax(numA, numB) + FindMax(numY, numZ);  

The maxSum is a double type variable which is assigned the maximum of the two variables numA numB PLUS the maximum of the two variables numY numZ using FindMax function. The FindMax() method is called twice in this statement one time to find the maximum of numA and numB and one time to find the maximum of numY numZ. When the FindMax() method is called by passing numA and numB as parameters to this method, then method finds if the value of numA is greater than that of numB or vice versa. When the FindMax() method is called by passing numY and numZ as parameters to this method, then method finds if the value of numY is greater than that of numZ or vice versa. The PLUS sign between the two method calls means that the resultant values returned by the FindMax() for both the calls are added and the result of addition is assigned to maxSum.

Explanation:

This is where the statement will fit in the program.

#include <iostream>

using namespace std;

double FindMax(double num1, double num2) {

  double maxVal = 0.0;    

  if (num1 > num2) { // if num1 is greater than num2,

     maxVal = num1;  // then num1 is the maxVal.    }

  else {          

     maxVal = num2;  // num2 is the maxVal.   }

  return maxVal; }  

int main() {

  double numA;

  double numB;

  double numY;

  double numZ;

  double maxSum = 0.0;  

maxSum = FindMax(numA, numB) + FindMax(numY, numZ);  

  cout << "maxSum is: " << maxSum << endl;   }

Lets take an example to explain this. Lets assign values to the variables.

numA = 6.0

numB = 3.0

numY = 4.0

numZ = 9.0

maxSum =0.0

maxSum = FindMax(numA, numB) + FindMax(numY, numZ);  

FindMax(numA,numB) method call checks if numA>numB or numB>numA and returns the maximum of the two. Here numA>numB because 6.0 is greater than 3.0. So the method returns numA i.e. 6.0

FindMax(numY, numZ) method call checks if numY>numZ or numZ>numY and returns the maximum of the two. Here numZ>numY because 9.0 is greater than 4.0. So the method returns numZ i.e. 9.0

FindMax(numA, numB) + FindMax(numY, numZ) this adds the two values returned by the method FindMax for each call.

FindMax returned maxVal from numA and numB values which is numA i.e. 6.0.

FindMax returned maxVal from numY and numZ values which is numZ i.e. 9.0.

Adding these two values 9.0 + 6.0 = 15

So the complete statement  FindMax(numA, numB) + FindMax(numY, numZ) gives 15.0  as a result.

This result is assigned to maxSum. Hence

maxSum = 15

To see the output on the screen you can use the print statement as:

cout<<maxSum;

This will display 15

6 0
3 years ago
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