C++ Programming: From Problem Analysis to Program Design
C++ Programming: From Problem Analysis to Program Design
8th Edition
ISBN: 9781337102087
Author: D. S. Malik
Publisher: Cengage Learning
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Chapter 12, Problem 18SA

Explanation of Solution

The program has been explained in the in-lined comments:

#include <iostream> 

using namespace std; 

int main()  

{  

    //declare int pointers

    int *intArrayPtr;

    int *temp;

    //creates an array of five components of type int and

    //stores the base address of the array 

    //into the pointer intArrayPtr

    intArrayPtr = new int[5];

    //contents of the base address of the array is 

    //assigned a value of 7 i.e. intArrayPtr[0] is 7

    *intArrayPtr = 7;

    //temp is assigned the value of intArrayPtr

    //so it also points to array base address

    //i.e. intArrayPtr[0]

    temp = intArrayPtr;

    //modify the contents of the array inside for loop

    for (int i = 1; i < 5; i++)

    {

        //increment pointer to the next address

        //i.e. next element in the array

        intArrayPtr++;

        //assign the value of the current element

        //pointed to by intArrayPtr with the value

        //on RHS ...

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int j; for (int i = 0; i < 5; i++)   {j = 2*j*i;} What is j?
int sum = 0; for (int i 0; i < 5; i++){ sum += i; } cout << sum;
#include using namespace std; const int y = 1; int main () { int static y = 2; int i = 3, j = 4, m = 5, n = 6; %3D int a = [] (int x, int i = 1) { return x * i; } (y, 3) ; %3D int b = [=] (int x) { return [=] (int b) { return b + j; } (x) % 7; } (a); int c = [=] (int x) mutable ->int { m = 6; return [6] (int j) mutable { y = a * b; return y / j; } (x) -m; } (b) ; cout << a << endl; cout << b << endl; cout << c << endl; cout << m << endl; cout << y < endl; return 0; This program outputs 5 lines. What are they? Please explain your work and your answer.
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