Let's say that p is a pointer to memory and the next six bytes in memory (in hex) beginning at p's address are: aa bb cc dd ee ff. What value would be in x if the following code is run on a little- endian computer? uint16_t *q (uint16_t *)p; uint16_t x = q[0]; aa aabb bbaa aabbccdd ddccbbaa

Let's say that p is a pointer to memory and the next six bytes in memory (in hex) beginning at p's address are: aa bb cc dd ee ff. What value would be in x if the following code is run on a little- endian computer? uint16_t q (uint16_t )p; uint16_t x = q[0]; aa aabb bbaa aabbccdd ddccbbaa

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Description: Let's say that p is a pointer to memory and the next six bytes in memory (in hex) beginning at p'saddress are: aa bb cc dd ee ff. What value would be in x if the following code is run on a little-endian computer?uint16_t *quint16_t x =(uint16_t *)p;

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Question

Let's say that p is a pointer to memory and the next six bytes in memory (in hex) beginning at p's
address are: aa bb cc dd ee ff. What value would be in x if the following code is run on a little-
endian computer?
uint16_t *q
uint16_t x =
(uint16_t *)p;
q[0];
aa
aabb
bbaa
aabbccdd
ddccbbaa

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Using C Language In this function, b has the address of a memory buffer that is num_bytes long. The function should repeatedly copy the 16 byte pattern that pattern16 points at into the memory buffer until num_bytes have been written. If num_bytes is not a multple of 16, the final write of the 16 byte pattern should be truncated to finish filling the buffer. void memset16(void *b, int num_bytes, void *pattern16) For example if the 16 bytes that pattern16 points at is 00 11 22 33 44 55 66 77 88 99 aa bb cc dd ee ff, then memset(b, 20, pattern16) should write to the buffer pointed at by p the 20 bytes 00 11 22 33 44 55 66 77 88 99 aa bb cc dd ee ff 00 11 22 33. Use SSE instructions to improve efficiency. Here's pseudocode. x = SSE unaligned load from pattern16while (num_bytes >= 16) SSE unaligned store x to p advance p by 16 bytes decrement num_bytes by 16while (num_bytes > 0) store 1 byte from pattern16 to p advance p by 1 byte advance pattern16 by 1 byte…
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