Modern Database Management
13th Edition
ISBN: 9780134773650
Author: Hoffer
Publisher: PEARSON
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Chapter 8, Problem 8.17RQ
Program Plan Intro
Advantages and Disadvantages of horizontal and vertical partitioning.
Expert Solution & Answer
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1 typedef struct node* {
2
struct node* next;
3
char* key;
4
char* val;
5} node_t;
6
7 char* find_node (node_t* node, char* key_to_find) {
while(strcmp (node->key, key_to_find ) != 0 ) {
node = node->next;
8
9
10
}
11
return node->val;
12 }
Match each of the assembler routines on the left with the equivalent C function on
the right. Write the name of the label (e.g., foo) to the right of the corresponding
function. Note: shrq is the logical right shift instruction, and sarq is the arithmetic
right shift instruction.
foo1:
leaq
0(,%rdi, 8), %rax
long choice1 (long x)
{
ret
return x -
8 >8;
foo3:
}
movq
sarq
%rdi, %rax
$8, %rax
long choice4 (long x)
ret
{
return x*256;
}
foo4:
long choice5 (long x)
leaq
-8 (%rdi), %rax
{
ret
return x-8;
}
long choice6 (long x)
foo5:
{
leaq
-8 (%rdi), %rax
return x+8;
shrq
$63, %rax
}
ret
Given the variables and code in the text below, identify where in memory they will
live once the code is compiled.
1 char
big_array [1L<<24]; /* 16 MB */
2 GB *
:/
2 char huge_array [1L<<31]; /*
3
4 int global = 0;
5
6 int useless () { return 0; }
7
8 int main()
9 {
10
void *p1, p2, *p3, *p4;
int local =
0;
malloc (1L << 28); /* 256 MB *,
11
12
p1
13
p2
=
malloc (1L << 8);
/* 256 B *
14
p3
15
p4
=
malloc (1L << 32);
malloc (1L << 8);
/* 4
GB *
*/
/* 256
B */
16 }
Note: *pN is the thing at which pN points.
1. big_array
2. huge_array
3. global
4. useless
5. void* p1
6. *p1
7. void* p2
8. *p2
9. void* p3
10. *p3
11. void* p4
12. *p4
Chapter 8 Solutions
Modern Database Management
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 di fe ff ff callq 400410 40053f: 48 83 c4 18 add $0x18,%rsp 400543: c3 retq 400544: 0000000000400544 : 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_forwardConsider the following assembly code for a C for loop: movl $0, %eax jmp .L2 .L3: addq $1, %rdi addq %rsi, %rax subq $1, %rsi .L2: cmpq %rsi, %rdi jl .L3 addq ret %rdi, %rax Based on the assembly code above, fill in the blanks below in its corresponding C source code. Recall that registers %rdi and %rsi contain the first and second, respectively, argument of a function. (Note: you may only use the symbolic variables x, y, and result in your expressions below do not use register names.) long loop (long x, long y) { long result; } for ( } return result; __; y--) {arrow_forwardIn each of the following C code snippets, there are issues that can prevent the compiler from 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 6 Suggested solution: ⚫ Remove printf or move it outside the loop. Remove the loop. • Replace arr[i] with a constant value. 1 int sum (int *arr, int n) { 2 int s = 0; 3 for (int i = 0; i < n; i++) { 4 5 6 } 7 8 } s = arr[i]; printf("%d\n", s); return s; 234206 2. Which line prevents compiler optimization? Circle one: 2 3 4 5 6 Suggested 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 scale (int *arr, int n, int factor) { 5 6 } for (int i = 0; i < n; i++) { rr[i] = factor; do_extra_work ();arrow_forward
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