MODERN DATABASE MGMT.(PEARSON CHANNEL)
13th Edition
ISBN: 9780136709299
Author: Hoffer
Publisher: PEARSON
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Chapter 8, Problem 8.51PAE
Program Plan Intro
The types of indexes recommended for a given
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Chapter 8 Solutions
MODERN DATABASE MGMT.(PEARSON CHANNEL)
Ch. 8 - Prob. 8.1RQCh. 8 - Prob. 8.2RQCh. 8 - Prob. 8.3RQCh. 8 - Prob. 8.4RQCh. 8 - Prob. 8.5RQCh. 8 - Prob. 8.6RQCh. 8 - Prob. 8.7RQCh. 8 - Prob. 8.8RQCh. 8 - Explain why you sometimes have to reserve much...Ch. 8 - Why are field values sometimes coded?
Ch. 8 - Prob. 8.11RQCh. 8 - Prob. 8.12RQCh. 8 - Explain why normalized relations may not comprise...Ch. 8 - Prob. 8.14RQCh. 8 - List three common situations that suggest that...Ch. 8 - Explain the reasons why some experts are against...Ch. 8 - Prob. 8.17RQCh. 8 - Prob. 8.18RQCh. 8 - Prob. 8.19RQCh. 8 - Prob. 8.20RQCh. 8 - Prob. 8.21RQCh. 8 - Prob. 8.22RQCh. 8 - One of the strongest recommendations regarding...Ch. 8 - Explain why an index is useful only if there is...Ch. 8 - Indexing can clearly be very beneficial. Why...Ch. 8 - Prob. 8.26RQCh. 8 - Prob. 8.27RQCh. 8 - Describe the role of data dictionary in the...Ch. 8 - Prob. 8.29RQCh. 8 - Prob. 8.30RQCh. 8 - Explain how creating a view may increase data...Ch. 8 - Prob. 8.32RQCh. 8 - Prob. 8.33RQCh. 8 - Prob. 8.34RQCh. 8 - Prob. 8.35RQCh. 8 - Prob. 8.36RQCh. 8 - Prob. 8.37RQCh. 8 - Prob. 8.38RQCh. 8 - Prob. 8.39RQCh. 8 - How can views be used as part of data security?...Ch. 8 - Prob. 8.41RQCh. 8 - Prob. 8.42RQCh. 8 - Consider the following two relations for...Ch. 8 - Prob. 8.44PAECh. 8 - Prob. 8.45PAECh. 8 - Prob. 8.46PAECh. 8 - Prob. 8.47PAECh. 8 - Suppose you are designing a default value for the...Ch. 8 - When a student has not chosen a major at a...Ch. 8 - Prob. 8.50PAECh. 8 - Prob. 8.51PAECh. 8 - Consider the relations in Problem and Exercise...Ch. 8 - Prob. 8.53PAECh. 8 - Prob. 8.54PAECh. 8 - Prob. 8.55PAECh. 8 - Prob. 8.56PAECh. 8 - Prob. 8.57PAECh. 8 - Consider the relations specified in Problem and...Ch. 8 - Prob. 8.59PAECh. 8 - Prob. 8.60PAECh. 8 - Prob. 8.61PAECh. 8 - Prob. 8.62PAECh. 8 - Prob. 8.63PAECh. 8 - Prob. 8.64PAECh. 8 - Problems and Exercises 8-65 through 8-68 refer to...Ch. 8 - Prob. 8.66PAECh. 8 - Problems and Exercises 8-65i5 through 8-68 refer...Ch. 8 - Refer to Figure 4-5 0. For each of the following...Ch. 8 - Prob. 8.69PAECh. 8 - Prob. 8.70PAECh. 8 - Prob. 8.71PAECh. 8 - Prob. 8.72PAECh. 8 - Prob. 8.73PAECh. 8 - Prob. 8.74PAECh. 8 - Prob. 8.75PAECh. 8 - Prob. 8.76PAECh. 8 - Prob. 8.77PAE
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- The next problem concerns the following C code: /copy input string x to buf */ void foo (char *x) { char buf [8]; strcpy((char *) buf, x); } void callfoo() { } foo("ZYXWVUTSRQPONMLKJIHGFEDCBA"); Here is the corresponding machine code on a Linux/x86 machine: 0000000000400530 : 400530: 48 83 ec 18 sub $0x18,%rsp 400534: 48 89 fe mov %rdi, %rsi 400537: 48 89 e7 mov %rsp,%rdi 40053a: e8 d1 fe ff ff 40053f: 48 83 c4 18 add callq 400410 $0x18,%rsp 400543: c3 retq 0000000000400544 : 400544: 48 83 ec 08 sub $0x8,%rsp 400548: bf 00 06 40 00 mov $0x400600,%edi 40054d: e8 de ff ff ff callq 400530 400552: 48 83 c4 08 add $0x8,%rsp 400556: c3 This problem tests your understanding of the program stack. Here are some notes to help you work the problem: • strcpy(char *dst, char *src) copies the string at address src (including the terminating '\0' character) to address dst. It does not check the size of the destination buffer. You will need to know the hex values of the following characters:arrow_forwardA ROP (Return-Oriented Programming) attack can be used to execute arbitrary instructions by chaining together small pieces of code called "gadgets." Your goal is to create a stack layout for a ROP attack that calls a function located at '0x4018bd3'. Below is the assembly code for the function 'getbuf', which allocates 8 bytes of stack space for a 'char' array. This array is then passed to the 'gets' function. Additionally, you are provided with five useful gadgets and their addresses. Use these gadgets to construct the stack layout. Assembly for getbuf 1 getbuf: 2 sub $8, %rsp 3 mov %rsp, %rdi 4 call gets 56 add $8, %rsp ret #Allocate 8 bytes for buffer #Load buffer address into %rdi #Call gets with buffer #Restore the stack pointer #Return to caller. Stack Layout (fill in Gadgets each 8-byte section) Address Gadget Address Value (8 bytes) 0x4006a7 pop %rdi; ret 0x7fffffffdfc0 Ox4006a9 pop %rsi; ret 0x7fffffffdfb8 0x4006ab pop %rax; ret 0x7fffffffdfb0 0x7fffffffdfa8 Ox4006ad mov %rax,…arrow_forwardIn each of the following C code snippets, there are issues that can prevent the compilerfrom applying certain optimizations. For each snippet:• Circle the line number that contains compiler optimization blocker.• Select the best modification to improve optimization.1. Which line prevents compiler optimization? Circle one: 2 3 4 5 6Suggested solution:• Remove printf or move it outside the loop.• Remove the loop.• Replace arr[i] with a constant value.1 int sum( int ∗ ar r , int n) {2 int s = 0 ;3 for ( int i = 0 ; i < n ; i++) {4 s += a r r [ i ] ;5 p r i n t f ( ”%d\n” , s ) ;6 }7 return s ;8 }2. Which line prevents compiler optimization? Circle one: 2 3 4 5 6Suggested solution:• Move or eliminate do extra work() if it’s not necessary inside the loop.• Remove the loop (but what about scaling?).• Replace arr[i] *= factor; with arr[i] = 0; (why would that help?).1 void s c a l e ( int ∗ ar r , int n , int f a c t o r ) {2 for ( int i = 0 ; i < n ; i++) {3 a r r [ i ] ∗= f a c t o r…arrow_forward
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