Oracle 12c: SQL
3rd Edition
ISBN: 9781305251038
Author: Joan Casteel
Publisher: Cengage Learning
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Expert Solution & Answer
Chapter 9, Problem 3HOA
Explanation of Solution
SQL Query:
Refer to the “books”, “orders”, “orderitems” and “customers” tables in the JustLee Books
The SQL statement to perform the stated task with traditional approach is as follows:
SELECT DISTINCT c.lastname, c.customer#
FROM books b, orders o, orderitems i, customers c
WHERE c.customer# = o.customer#
AND o.order# = i.order#
AND i.isbn = b.isbn
AND c.state = 'FL'
AND b.category = 'COMPUTER';
The SQL statement to perform the stated task using the JOIN keyword is as follows:
SELECT DISTINCT c...
Expert Solution & Answer
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Students have asked these similar questions
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
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:
Consider 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--) {
Knowledge Booster
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