Pointer in C++:
A pointer is a variable whose value will be another variable’s address. Generally, a pointer variable is declared as follows:
type *var-name;
Here, type is the pointer’s base type and var-name is the pointer variable name. The asterisk is used to designate a variable as a pointer.
Given Program:
//Include libraries
#include <iostream>
//Use namespace
using namespace std;
//Define main method
int main()
{
// Declare pointer variables
int *p1, *p2;
//Assign value
p1 = new int;
//Assign value
p2 = new int;
//Assign value
*p1 = 10;
//Assign value
*p2 = 20;
//Display value
cout << *p1 << " " << *p2 << endl;
//Assign value
*p1 = *p2;
//Display value
cout << *p1 << " " << *p2 << endl;
//Assign value
*p1 = 30;
//Display value
cout << *p1 << " " << *p2 << endl;
//Pause console window
system("pause");
//Return 0 value
return 0;
}
Want to see the full answer?
Check out a sample textbook solutionChapter 9 Solutions
Problem Solving with C++ (9th Edition)
- int mystery (int u, int v) { int a; a = u - V; u = a; V = u; return u + v; choose the output of the follos cout << mystery (9,7); Select one: a. 7 b. 4 C. 9arrow_forwardVoid Do1 (int: &, a. int &b) { a = 5; a = a + b; b = a + 2; } Int main() { Int x = 10; Do1 (x,x); Cout << x << endl; } The output of this program isarrow_forwardNonearrow_forward
- What is the output of the following code? int *myPtr = new int;int *yourPtr;*myPtr = 15;yourPtr = new int;delete myPtr;myPtr = yourPtr;*yourPtr = 28;myPtr = new int;*myPtr = 49;*yourPtr = 34;delete yourPtr;yourPtr = myPtr;myPtr = new int;*myPtr = 54;cout << *myPtr << " " << *yourPtr << endl;arrow_forwardIn Java:arrow_forward#include using namespace std; void some_action_1(int); int main() { cout<arrow_forward#include<stdio.h> #include<stdarg.h> void fun1(int num, ...); void fun2(int num, ...); int main() { fun1(1, "Apple", "Boys", "Cats", "Dogs"); fun2(2, 12, 13, 14); return 0; } void fun1(int num, ...) { char *str; va_list ptr; va_start(ptr, num); str = va_arg(ptr, char *); printf("%s ", str); } void fun2(int num, ...) { va_list ptr; va_start(ptr, num); num = va_arg(ptr, int); printf("%d", num); }.arrow_forwardJava: Abstraction using Interfacearrow_forwardQ1. The circle has two data members, a Point representing the center of the circle and a float value representing the radius as shown below. typedef struct{ Point center; float radius;} Circle; typedef struct{ int x; int y;}Point; Implement the following functions:a. float distance (Line l1);- computes and returns the distance between the two points of the line.distance = sqrt((x2-x1)^2+(y2-y1)^2) b. float diameter(Circle circ);- computes the diameter of a circle.. .arrow_forwardQ5: Correct the following code fragment and what will be the final results of the variable a and b class A {protected int x1,y1,z; public: A(a, b,c):x1(a+2),y1 (b-1),z(c+2) { for(i=0; i<5;i++) x1++;y1++;z++;}}; class B {protected: int x,y; public: B(a,b):x(a+1),y(b+2) { for(i=0; i<5;i++) x+=2; y+=1;}}; class D:public B, virtual public A { private: int a,b; public: D(k,m,n): a(k+n), B(k,m),b(n+2),A(k,m,n) { a=a+1;b=b+1;}); int main() {Dob(4,2,5);)arrow_forwardThis is what I currently have : void displaySentence(char * sentence); char * sentence;arrow_forward2- Trace the following code and write the output: class Test1 { Test1(int x) { System.out.println("Test Calls " + x); class Test2 { Testi t1 = new Test1(10); Test2(int i) { t1 = new Test1(i); } public static void main(String[] args) { Test2 t2 = new Test2(5);arrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
- EBK JAVA PROGRAMMINGComputer ScienceISBN:9781337671385Author:FARRELLPublisher:CENGAGE LEARNING - CONSIGNMENT