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 15SA

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

#include <iostream> 

using namespace std; 

int main()  

{  

    //pointer variables declared and memory allocated

    double *firstPtr = new double; //Line 1

    double *nextPtr = new double; //Line 2

    //value assigned to pointer variable

    *firstPtr = 62; //Line 3

    //nextPtr is assigned the pointer firstPtr

    //so nextPtr points to a memory location

    //having a stored value of 62

    nextPtr = firstPtr; //Line 4

    //memory pointed to by both the pointers are

    //deallocated

    delete firstPtr; //Line 5

    delete nextPtr; //Line 6

    //fresh memory i...

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C++ A robot is initially located at position (0; 0) in a grid [?5; 5]  [?5; 5]. The robot can move randomly in any of the directions: up, down, left, right. The robot can only move one step at a time. For each move, print the direction of the move and the current position of the robot. If the robot makes a circle, which means it moves back to the original place, print "Back to the origin!" to the console and stop the program. If it reaches the boundary of the grid, print \Hit the boundary!" to the console and stop the program. A successful run of your code may look like:Down (0,-1)Down (0,-2)Up (0,-1)Left (-1,-1)Left (-2,-1)Up (-2,0)Left (-3,0)Left (-4,0)Left (-5,0)Hit the boundary! or Left (-1,0)Down (-1,-1)Right (0,-1)Up (0,0)Back to the origin! About: This program is to give you practice using the control ow, the random number generator, and output formatting. You may use <iomanip> to format your output.ˆ ˆ You may NOT use #include "stdafx.h".
if L1={ab,b} L2={abb,bb} then L1L2= * O {ababb,abbb,abbab,bbab} O {ab,b,abb,bb} O {ababb,abbb,babb,bbb} O {abbab,abbb,bbab,bbb}
double tab1[5] = {2,3,4,5,6};      double tab2[5] = {6,5,4,3,2};      for(int i = 0;i<5;i++)       {         tab1[i] = tab1[i]*tab2[i];         cout<<tab1[i]<<" ";       }     12 15 16 15 16   15 15 16 15 12   12 15 16 15 12   12 15 12 15 12
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