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Rudiy27
3 years ago
14

Pendant Publishing edits multi-volume manuscripts for many authors. For each volume, they want a label that contains the author'

s name, the title of the work, and a volume number in the form Volume 9 of 9. For example, a set of three volumes requires three labels: Volume 1 of 3, Volume 2 of 3, and Volume 3 of 3. Design an application that reads records that contain an author's name, the title of the work, and the number of volumes. The application must read the records until eof is encountered and produce enough labels for each work.

Computers and Technology
1 answer:
lbvjy [14]3 years ago
3 0

Answer:

Check the explanation

Explanation:

Pseudocode For Reading File:

start

Declarations

InputFIle records.txt

OutputFile result.txt

string authorName

string title

int numOfVol

open InputFile

open OutputFile

input authorName, title, numOfVol from records.txt

while not eof

output authorName

output title

output numOfVol

end while

close records.txt

close results.txt

END

The flowchart for reading the file can be seen below.

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The act of engaging in crime through the use of a computer or similar type of device is called:
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When you take a multiple-choice test, you are relying on ________, a means of retrieving information out of your long-term memor
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Answer:b)Recognition

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3 years ago
The file containing the definitions of the member functions of class DateType is called the ______ file.
Free_Kalibri [48]

Answer:

Implementation file

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6 0
3 years ago
What are the first and the last physical memory addressesaccessible using
Misha Larkins [42]

Answer

For First physical memory address,we add 00000 in segment values.

For Last physical memory address,we add 0FFFF in segment values.

<u>NOTE</u>-For addition of hexadecimal numbers ,you first have to convert it into binary then add them,after this convert back it in hexadecimal.

a)1000

For First physical memory address, we add 00000 in segment value

We add 0 at the least significant bit while calculating.

          1000<u>0</u> +00000 = 1000<u>0</u> (from note)

For Last physical memory address,We add 0 at the least significant bit while calculating.

   we add 0FFFF in segment value

           1000<u>0</u> +0FFFF =1FFFF    (from note)

b)0FFF

For First physical memory address,we add 00000 in segment value

We add 0 at the least significant bit while calculating.

     0FFF<u>0</u> +00000=0FFF0 (from note)

For Last physical memory address,We add 0 at the least significant bit while calculating.

   we add 0FFFF in segment value

           0FFF<u>0</u> +0FFFF =1FFEF (from note)

c)0001

For First physical memory address,we add 00000 in segment value

We add 0 at the least significant bit while calculating.

     0001<u>0</u> +00000=00010 (from note)

For Last physical memory address,We add 0 at the least significant bit while calculating.

   we add 0FFFF in segment value

           0001<u>0</u> +0FFFF =1000F (from note)

d) E000

For First physical memory address,we add 00000 in segment value

 We add 0 at the least significant bit while calculating.

    E000<u>0</u> +00000=E0000 (from note)

For Last physical memory address,We add 0 at the least significant bit while calculating.

   we add 0FFFF in segment value

           E000<u>0</u> +0FFFF =EFFFF (from note)

e) 1002

For First physical memory address,we add 00000 in segment value

 We add 0 at the least significant bit while calculating.

    1002<u>0</u> +00000=10020  (from note)

For Last physical memory address,We add 0 at the least significant bit while calculating.

   we add 0FFFF in segment value

           1002<u>0</u> +0FFFF =2001F (from note)

6 0
3 years ago
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