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4) Which of the following is/are valid ways to allocate memory for an integer by dynamic memory allocation in CPP?
a. int *p = new int(100);
b. int *p; p = new int; *p = 100;
c. int *p = NULL; p = new int; *p=100;
d. Only 1,2
e. All of these
5) Choose the correct option?
#include<iostream>
using namespace std;
class Base {};
class Derived: public Base {};
int main()
{
Base *bp = new Derived;
Derived *dp = new Base;
}
a. No Compiler Error
b.Compiler Error in line “Base *bp = new Derived;”
c. Compiler Error in line ” Derived *dp = new Base;”
d. Runtime Error
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- Memory Allocation: Select all of the following statements that are true. The basic idea of contiguous allocation is that each process is contained in a single contiguous section of memory. When using a single-partition allocation scheme, the main memory is usually divided into three partitions. Memory partitioning schemes allow for multiprogramming. When applying the multiple-partition allocation scheme "Fixed Partitions", all partitions always have the same size. When applying the multiple-partition allocation scheme "Variable Partitions", the FirstFit allocation strategy performs best in practice. The Buddy system is prone to internal fragmentation.// add.ll define void @add(i32* %ptr1, i32* %ptr2, i32* %val) {ret void} Fill add.ll function to do the following operation: void add(int *ptr1, int *ptr2, int *val) { *ptr1 += *val; *ptr2 += *val; } This is the full question. It is related to LLVM. If it's going to help there is one more code given which is: #include <stdio.h> void add(int *ptr1, int *ptr2, int *val); int main(int argc, char **argv) {FILE *f = fopen(argv[1], "r");int a, b, c;fscanf(f, "%d %d %d", &a, &b, &c);add(&a, &b, &c);printf("%d %d\n", a, b);fclose(f); return 0;}#include <stdio.h>int main() {long local;int (*mainptr)(void); /* declare a "pointer to function returning int" */mainptr = main; /* address of main */printf("%p,%p\n",&local,mainptr); /* print the two addresses */return 0;}The printf statement outputs the address of the variable “local”, which should be on the stack rightafter the return address, and then the address of the code for “main”.Start by creating a source file aslrtest.c on your VM containing the above C program. Compileit into an executable with the same name:bash$ gcc -o aslrtest aslrtest.c(As always, the “bash$” in the above represents the prompt and is not part of the command youtype.)a. Turn off ASLR with the command:echo 0 | sudo tee /proc/sys/kernel/randomize_va_spaceOnce you have ASLR disabled, run aslrtest 10 times. The following shell script will do that,(assuming your shell is bash, which it is unless you’ve changed it yourself):bash$ for i in {1..10}> do> ./aslrtest> doneWhat range of output…
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- Address of Array//Write the output of each cout statement.//Assume the size of ints to be 4 bytes.//Assume the size of pointers to ints to be 8 bytes.int main() {int a[3][3]={{1,2,3},{4,5,6},{7,8,9}};cout << sizeof(a) << endl: //1: ___________________________cout << sizeof(a+0) << endl: //2: _____________________________cout << sizeof(*(a+0)) << endl: //3: _____________________________cout << sizeof(**a) << endl; //4: ____________________________cout << sizeof(&a) << endl; //5: _____________________________Assume the address of "a" is 0x7ffee2fef9e0cout << a + 1 << endl; //6: ________________________________________cout << *a + 1 << endl; //7: ________________________________________cout << **a + 1 << endl; //8: _______________________________________cout << &a << endl; //9: __________________________________________cout << &a + 1 << endl; //10:…Using C++ Programming language: Given the following definitions: int num1=1, num2=2; int *ptr1=NULL, *ptr2=NULL; Write the following statements: Assign ptr1 to the address of num1. Assign ptr2 to the address of num2. Using pointer notation, write a statement that compares the values that the pointers "point" to to see which is larger. (There is more than one way to do this. You choose.)Code: //POinters #include<iostream> using namespace std; int main() { int var = 9; int *ip; double *dp; float *fp; char *ch; ip=&var; //&= address of (saves address of variable var in ip) cout<<"Variable Value "<<var<<endl; cout<<"Address saved in pointer "<<ip<<endl; cout<<"Pointer pointing to value "<<*ip<<endl; *ip=50; cout<<"\nVariable Value "<<var<<endl; cout<<"Address saved in pointer "<<ip<<endl; cout<<"Pointer pointing to value "<<*ip<<endl; } //Dynamic Memory Allocation #include<iostream> using namespace std; int main() { int size,i; int *ptr; cout<<"Enter size of an array"<<endl; cin>>size; ptr = new int[size]; cout<<"Enter Array Elements to store\n"; for(i=0;i<size;i++) { cin>>ptr[i]; } cout<<"Values of Array are\n"; for(i=0;i<size;i++) { cout<<ptr[i]<<endl; } for(i=0;i<size;i++) {…
- Fill-in-the-Blank The __________ operator is used to dynamically allocate memory.Segmentation: Select all of the following statements that are true. In segmentation, a logical address always has a length of 32 bit. In order to translate logical into physical addresses, the memory management unit uses the segment part of the logical address to determine the start address in the segment table and adds the offset to this to get the physical address. In segmentation, the logical address consists of a segment part and an offset. The segment length is limited by the maximum possible segment number. When applying segmentation, processes are only allowed to access the memory within their segments. Segments can be assigned access rights and privilege levels.C++ Language Write two statements that each use malloc to allocate an int location for each pointer. Sample output for given program:numPtr1 = 44, numPtr2 = 99 #include <stdio.h>#include <stdlib.h> int main(void) { int* numPtr1 = NULL; int* numPtr2 = NULL; /* Your solution goes here */ scanf("%d", numPtr1); scanf("%d", numPtr2); printf("numPtr1 = %d, numPtr2 = %d\n", *numPtr1, *numPtr2); free(numPtr1); free(numPtr2); return 0;}