Assume that the following variables have been properly declared and initialized.

Answer:

Below is the Python code with suitable comments.

Explanation:

#Input file name acquisition

filename=input('Enter the input file name: ')

#Opening the input file

inputFile = open(filename,"r+")

#Dictionary definition.

list={}

#Read and split file content using a loop

for word in inputFile.read().split():

#Check if the word exists in the file.

if word not in list:

list[word] = 1

#Increment count by 1

else:

list[word] += 1

#Closing the file.

inputFile.close();

#Output a blank line

print();

#Sorting words according to their ASCII values.

for i in sorted(list):

#Display unique words along with their

#frequencies in alphabetical order.

print("{0} {1} ".format(i, list[i]));

Answer:

Given Data:

myBytes BYTE 10h, 20h, 30h, 40h

myWords WORD 3 DUP(?), 2000h

myString BYTE "ABCDE"

From the supplied information, we can derive that:

(a).     a. EAX = 1

         b. EAX = 4

         c. EAX = 4

         d. EAX = 2

         e. EAX = 4

         f. EAX = 8

         g. EAX = 5

Response: 1,500,000 bytes.

Clarification:

If we assume the image dimensions are 4000 pixels in width and 3000 pixels in height, the total uncompressed image will consist of 4000*3000= 12,000,000 pixels.

In the case of a binary image, each pixel can have only two values, which necessitates one bit for each pixel.

This indicates that we need to accommodate 12,000,000 bits.

Given that 1 byte equals 8 bits.

So, to store an uncompressed binary image sized 4000 x 3000 pixels, 12,000,000/8 bytes is required ⇒ 1,500,000 bytes.

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