Let's say that p is a pointer to memory and the next four bytes in memory (in hex) beginning at p's address are: 89 AB CD EF. After the following code is run on a little-endian computer what are the values of x, y, and the four bytes in memory beginning at p's address. Answer for x and y in four digits of upper-case hexadecimal (eg, 0AC6). Answer for memory as eight digits of upper-case hexadecimal with a single space between each byte (eg, 89 AB CD EF). x: y: mem at p:
Let's say that p is a pointer to memory and the next four bytes in memory (in hex) beginning at p's address are: 89 AB CD EF. After the following code is run on a little-endian computer what are the values of x, y, and the four bytes in memory beginning at p's address. Answer for x and y in four digits of upper-case hexadecimal (eg, 0AC6). Answer for memory as eight digits of upper-case hexadecimal with a single space between each byte (eg, 89 AB CD EF). x: y: mem at p:
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Let's say that p is a pointer to memory and the next four bytes in memory (in hex) beginning at p's address are: 89 AB CD EF. After the following code is run on a little-endian computer what are the values of x, y, and the four bytes in memory beginning at p's address.
Answer for x and y in four digits of upper-case hexadecimal (eg, 0AC6). Answer for memory as eight digits of upper-case hexadecimal with a single space between each byte (eg, 89 AB CD EF).
x:
y:
mem at p:
![uint16_t *q
uint16_t x =
(uint16_t *)p;
q[0];
uint16_t y =
q[1];
q[0] = 0x01;
Øx02;
q[1]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0f84b485-5d6d-484c-881f-5450394bcf53%2F3f0c42c0-75f9-4f2f-bf50-9b0b69a39a98%2Fdfz169q_processed.png&w=3840&q=75)
Transcribed Image Text:uint16_t *q
uint16_t x =
(uint16_t *)p;
q[0];
uint16_t y =
q[1];
q[0] = 0x01;
Øx02;
q[1]
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