Database System Concepts
7th Edition
ISBN: 9781260084504
Author: Abraham Silberschatz, S. Sudarshan, Abraham, Henry Korth, Silberschatz Abraham, Silberschatz Professor
Publisher: McGraw-Hill Education
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Chapter 4, Problem 24E
Explanation of Solution
Reason for causing a cycle in the authorization graph
“Yes”. It causes a cycle in the authorization graph.
Reason
- The grant to public results in an edge from A to public.
- The grant to the public operator provides authorization to everyone, B is now authorized...
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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:
A 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,…
In 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…
Chapter 4 Solutions
Database System Concepts
Ch. 4 - Prob. 1PECh. 4 - Prob. 2PECh. 4 - Prob. 3PECh. 4 - Prob. 4PECh. 4 - Prob. 5PECh. 4 - Prob. 6PECh. 4 - Prob. 7PECh. 4 - Prob. 8PECh. 4 - SQL allows a foreign-key dependency to refer to...Ch. 4 -
Given the relations a(name, address, title) and...
Ch. 4 - Prob. 11PECh. 4 - Prob. 12PECh. 4 - Prob. 13PECh. 4 - Prob. 14ECh. 4 - Prob. 15ECh. 4 - Prob. 16ECh. 4 - Prob. 17ECh. 4 - Prob. 18ECh. 4 - Prob. 19ECh. 4 - Show how to define a view tot_credits (year,...Ch. 4 - Prob. 21ECh. 4 - Prob. 22ECh. 4 - Prob. 23ECh. 4 - Prob. 24ECh. 4 - Prob. 25ECh. 4 - Prob. 26E
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- 123456 A 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 'Ox4018bd3'. 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: sub mov $8, %rsp %rsp, %rdi call gets add $8, %rsp 6 ret #Allocate 8 bytes for buffer #Load buffer address into %rdi #Call gets with buffer #Restore the stack pointer #Return to caller Stack each Layout (fill in Gadgets 8-byte section) Address Gadget Address Value (8 bytes) 0x7fffffffdfc0 0x7fffffffdfb8 0x7fffffffdfb0 0x7fffffffdfa8 0x7fffffffdfa0 0x7fffffffdf98 0x7fffffffdf90 0x7fffffffdf88 Original 0x4006a7 pop %rdi;…arrow_forwardCharacter Hex value || Character Hex value | Character Hex value 'A' 0x41 יני Ox4a 'S' 0x53 0x42 'K' 0x4b 'T" 0x54 0x43 'L' Ox4c 0x55 0x44 'M' Ox4d 0x56 0x45 'N' Ox4e 'W' 0x57 0x46 Ox4f 'X' 0x58 0x47 'P' 0x50 'Y' 0x59 'H' 0x48 'Q' 0x51 'Z' Охба 'T' 0x49 'R' 0x52 '\0' 0x00 Now consider what happens on a Linux/x86 machine when callfoo calls foo with the input string "ZYXWVUTSRQPONMLKJIHGFEDCBA". A. On the left draw the state of the stack just before the execution of the instruction at address Ox40053a; make sure to show the frames for callfoo and foo and the exact return address, in Hex at the bottom of the callfoo frame. Then, on the right, draw the state of the stack just after the instruction got executed; make sure to show where the string "ZYXWVUTSRQPONMLKJIHGFEDCBA" is placed and what part, if any, of the above return address has been overwritten. B. Immediately after the ret instruction at address 0x400543 executes, what is the value of the program counter register %rip? (That is…arrow_forwardDraw out the way each of these structs looks in memory, including padding! Number the offsets in memory. 1 struct okay Name 2 { short a; 3 4 long number; 5 int also_a_number; 6 7 }; char* text; 1 struct badName 2 { 3 4 5 }; short s; struct okay Name n;arrow_forward
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