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
Answer:
First I/O instruction format -> 256 ports
Second I/O instruction format -> 65536 ports
Explanation:
- The initial instruction format can address 256 ports, computed from eight bits for port addressing as 2^8 = 256 ports.
- The second instruction format allows for 65536 ports, which can be calculated using sixteen bits for addressing as 2^16 = 65536 ports.
Changing the opcode facilitates the selection between the first and second instruction format, allowing one input or output operation at a given moment.
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.
I have attached my code in the image below. Good luck.
