Chapter 7, Problem 44E

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 '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: sub mov $8, %rsp %rsp, %rdi call gets 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 each 8-byte (fill in section) Address Value (8 bytes) 0x7fffffffdfc0 0x7fffffffdfb8 0x7fffffffdfb0 0x7fffffffdfa8 0x7fffffffdfa0 0x7fffffffdf98 0x7fffffffdf90 0x7fffffffdf88 Gadgets Address Gadget Ox4006a7 pop %rdi; ret Ox4006a9…

Problem 1 [15 points] The code below is buggy. Assume the code compiles. Briefly: 1). Identify the problem with the code (e.g., can access memory out of bounds) and 2). Suggest a solution (e.g., check the length). Question 1 1 #define BLENGTH 5 2 int b[BLENGTH]; 3 void copy_from_global_int_array_b (int n, int* dest) { 4 5 } *dest = b[n]; ==

Which statement regarding SGA_MAX_SIZE is true? SGA_MAX_SIZE is modifiable after an instance is started, only when Automatic Memory Management is used. SGA_MAX_SIZE is not dyamically modifiable. SGA_MAX_SIZE is ignored when MEMORY_TARGET > 0. SGA-MAX_SIZE must be specified when SGA_TARGET > 0

Explian this C program  #include <stdio.h> unsigned int rotateRight(unsigned int num, unsigned int bits) { unsignedint bit_count =sizeof(unsignedint) *8; bits = bits % bit_count; // Handle cases where bits >= bit_count return (num >> bits) | (num << (bit_count - bits)); } int main() { unsignedint num, bits; printf("Enter a number: "); scanf("%u", &num); printf("Enter the number of bits to shift: "); scanf("%u", &bits); printf("After rotation: %u\n", rotateRight(num, bits)); return0; }

Explian thiS C program #include<stdio.h> int countSetBits(int n) {    int count = 0;    while (n) {        count += n & 1;        n >>= 1;    }    return count;} int main() {    int num;    printf("Enter a number: ");    scanf("%d", &num);    printf("Output: %d units\n", countSetBits(num));    return 0;}

Please provide the Mathematica code

Explian this C program code. #include <stdio.h> void binary(unsigned int n) { if (n /2!=0) { binary(n /2); } printf("%d", n %2); } int main() { unsignedint number =33777; unsignedchar character ='X';   printf("Number: %u\n", number); printf("Binary: "); binary(number); printf("\nDecimal: %u\nHexadecimal: 0x%X\n\n", number, number);   printf("Character: %c\n", character); printf("ASCII Binary: "); binary(character); printf("\nASCII Decimal: %u\nASCII Hexadecimal: 0x%X\n", character, character);   return0; }

Design a dynamic programming algorithm for the Longest Alternating Subsequence problem described below: Input: A sequence of n integers Output: The length of the longest subsequence where the numbers alternate between being larger and smaller than their predecessor The algorithm must take O(n²) time. You must also write and explain the recurrence. Example 1: Input: [3, 5, 4, 1, 3, 6, 5, 7, 3, 4] Output: 8 ([3, 5, 4, 6, 5, 7, 3, 4]) Example 2: Input: [4,7,2,5,8, 3, 8, 0, 4, 7, 8] Output: 8 ([4, 7, 2, 5, 3, 8, 0,4]) (Take your time with this for the subproblem for this one)

Design a dynamic programming algorithm for the Coin-change problem described below: Input: An amount of money C and a set of n possible coin values with an unlimited supply of each kind of coin. Output: The smallest number of coins that add up to C exactly, or output that no such set exists. The algorithm must take O(n C) time. You must also write and explain the recurrence. Example 1: Input: C24, Coin values = = [1, 5, 10, 25, 50] Output: 6 (since 24 = 10+ 10+1+1 +1 + 1) Example 2: Input: C = 86, Coin values = [1, 5, 6, 23, 35, 46, 50] Output: 2 (since 86 = 46+35+5)