Robert Lafore 4th Edition Solution Manual Chapter 4 Structures

CODE:


1. A phone number, such as (212) 767-8900, can be thought of as having three parts: the area
  code (212), the exchange (767), and the number (8900). Write a program that uses a structure to
  store these three parts of a phone number separately. Call the structure phone. Create two structure
  variables of type phone. Initialize one, and have the user input a number for the other one. Then
  display both numbers. The interchange might look like this: 
  Enter your area code, exchange, and number: 415 555 1212
  My number is (212) 767-8900
  Your number is (415) 555-1212


#include<iostream.h>
#include<conio.h>
using namespace std;
struct phone{int area_code, exchange, number;};
int main()
{
 
 phone input, mine={212, 767, 8900};
 cout<<"Enter your area code, exchange, and number: "; 
 cin >>input.area_code>>input.exchange>>input.number;
 cout<<"My number is ("<<mine.area_code<<") "<<mine.exchange<<"-"<<mine.number<<endl;
 cout<<"Your number is ("<<input.area_code<<") "<<input.exchange<<"-"<<input.number;
 
}
  code (212), the exchange (767), and the number (8900). Write a program that uses a structure to
  store these three parts of a phone number separately. Call the structure phone. Create two structure
  variables of type phone. Initialize one, and have the user input a number for the other one. Then
  display both numbers. The interchange might look like this: 
  Enter your area code, exchange, and number: 415 555 1212
  My number is (212) 767-8900
  Your number is (415) 555-1212


#include<iostream.h>
#include<conio.h>
using namespace std;
struct phone{int area_code, exchange, number;};
int main()
{
 
 phone input, mine={212, 767, 8900};
 cout<<"Enter your area code, exchange, and number: "; 
 cin >>input.area_code>>input.exchange>>input.number;
 cout<<"My number is ("<<mine.area_code<<") "<<mine.exchange<<"-"<<mine.number<<endl;
 cout<<"Your number is ("<<input.area_code<<") "<<input.exchange<<"-"<<input.number;
 
}

CODE:


2. A point on the two-dimensional plane can be represented by two numbers: an x coordinate and
  a y coordinate. For example, (4,5) represents a point 4 units to the right of the vertical axis, and 5
  units up the horizontal axis. The sum of two points can be defined as a new point whose x
  coordinate is the sum of the x coordinates of the two points, and whose y coordinate is the sum of
  the y coordinates. 
  Write a program that uses a structure called point to model a point. Define three points, and have
  the user input values to two of them. Then set the third point equal to the sum of the other two, and
  display the value of the new point. Interaction with the program might look like this:

  Enter coordinates for p1: 3 4
  Enter coordinates for p2: 5 7
  Coordinates of p1+p2 are: 8, 11


#include<iostream.h>
#include<conio.h>
using namespace std;
struct point{int x, y;};
int main()
{
 point p1, p2, sum;
 cout<<"Enter coordinates for p1: "; 
        cin >>p1.x>>p1.y;
 cout<<"Enter coordinates for p2: ";  
        cin >>p2.x>>p2.y;
 sum.x=p1.x+p2.x; sum.y=p1.y+p2.y;
 cout<<"Coordinates of p1+p2 are: "<<sum.x<<", "<<sum.y;
 
 
}

CODE:


3.Create a structure called Volume that uses three variables of type Distance (from the ENGLSTRC
  example) to model the volume of a room. Initialize a variable of type Volume to specific dimensions,
  then calculate the volume it represents, and print out the result. To calculate the volume, convert
  each dimension from a Distance variable to a variable of type float representing feet and fractions of a
  foot, and then multiply the resulting three numbers.


#include<iostream.h>
#include<conio.h>
using namespace std;
struct Distance{int feet; float inches;};
struct Volume{Distance x, y, z;};
int main()
{
 
 Volume dimension; 
        char c;
 cout<<"Enter x, y & z ...\n(EX: 3.6=3 feet and 6 inches, default: 'x.0 y.0 z.0' x, y & z are in feet) : \n"; 
 cin >>dimension.x.feet>>c>>dimension.x.inches
  >>dimension.y.feet>>c>>dimension.y.inches
  >>dimension.z.feet>>c>>dimension.z.inches;
 /*float result=(dimension.x.feet+dimension.x.inches/12)*
     (dimension.y.feet+dimension.y.inches/12)*
     (dimension.z.feet+dimension.z.inches/12);*/
 cout<<"the volume is : "
     <<(dimension.x.feet+dimension.x.inches/12)*
    (dimension.y.feet+dimension.y.inches/12)*
    (dimension.z.feet+dimension.z.inches/12);
}

CODE:


4. Create a structure called employee that contains two members: an employee number (type int)
  and the employee’s compensation (in dollars; type float). Ask the user to fill in this data for three
  employees, store it in three variables of type struct employee, and then display the information for
  each employee.



#include<iostream.h>
#include<conio.h>
#include<iomanip.h>
using namespace std;
struct employee{int number; float compensation;};
int main()
{
 employee  e[3]; int i;

 
 for(i=0; i<3; i++) 
        {
 cout<<"Enter the number of employee number       "<<i+1<<" : "; cin>>e[i].number;
 cout<<"Enter the compensation of employee number "<<i+1<<" : "; cin>>e[i].compensation;}
 cout<<"Employee number"<<"   Employee compensation\n";
 for(i=0; i<3; i++) cout<<setw(15)<<e[i].number<<setw(24)<<e[i].compensation<<endl;
 
}

CODE:


5.Create a structure of type date that contains three members: the month, the day of the month, and
  the year, all of type int. (Or use day-month-year order if you prefer.) Have the user enter a date in
  the format 12/31/2001, store it in a variable of type struct date, then retrieve the values from the
  variable and print them out in the same format.


#include<iostream.h>
#include<conio.h>
#include<iomanip.h>
using namespace std;
struct date{int day; int month; int year;};
int main()
{
 
 date  x; 
        char c;
        cout<<"Enter the date : "; 
        cin >>x.day>>c>>x.month>>c>>x.year;
 cout<<"The date is    : "; 
        cout<<x.day<<c<<x.month<<c<<x.year;
 
}

CODE:


6. We said earlier that C++ I/O statements don’t automatically understand the data types of
  enumerations. Instead, the (>>) and (<<) operators think of such variables simply as integers. You
  can overcome this limitation by using switch statements to translate between the user’s way of
  expressing an enumerated variable and the actual values of the enumerated variable. For example,
  imagine an enumerated type with values that indicate an employee type within an organization:
  
  enum etype { laborer, secretary, manager, accountant, executive,
  researcher };


  Write a program that first allows the user to specify a type by entering its first letter (‘l’, ‘s’, ‘m’, and
  so on), then stores the type chosen as a value of a variable of type enum etype, and finally displays
  the complete word for this type.
  
  Enter employee type (first letter only)
     laborer, secretary, manager,
     accountant, executive, researcher): a
  Employee type is accountant.
  You’ll probably need two switch statements: one for input and one for output.*/



#include<iostream>
using namespace std;
#include<conio.h>
enum etype{laborer, secretary, manager, accountant, executive, researcher};
int main()
{
 etype x; char *ret;
 cout<<"Enter employee type (first letter only)"<<endl
  <<"(laborer, secretary, manager, accountant, executive, researcher): ";
 switch(getche())
{
 case 'l': x=laborer   ; 
        break;
 case 's': x=secretary ; 
        break;
 case 'm': x=manager   ; 
        break;
 case 'a': x=accountant; 
        break;
 case 'e': x=executive ; 
        break;
 case 'r': x=researcher;
       }
 switch(x)
{
 case 0: ret = "laborer"   ; 
        break;
 case 1: ret = "secretary" ; 
        break;
 case 2: ret = "manager"   ; 
        break;
 case 3: ret = "accountant"; 
        break;
 case 4: ret = "executive" ; 
        break;
 case 5: ret = "researcher";
     }
 cout<<"\nEmployee type is "<<ret<<".";
 
}

CODE:




7.  Add a variable of type enum etype (see Exercise 5), and another variable of type struct date (see
  Exercise 3) to the employee class of Exercise 4. Organize the resulting program so that the user
  enters four items of information for each of three employees: an employee number, the employee’s
  compensation, the employee type, and the date of first employment. The program should store this
  information in three variables of type employee, and then display their contents.*/


#include<iostream>
#include<conio.h>
#include<iomanip>
using namespace std;
enum etype{laborer, secretary, manager, accountant, executive, researcher};
struct date{int day; int month; int year;};
struct employee{int number; float compensation; date d; char *ret;};
int main()
{

 employee e[3]; 
        int i; 
        char c; 
        etype x;
 
 for(i=0; i<3; i++) 
     {
 cout<<"\nEnter the number of employee number "<<i+1<<" : "; 
        cin>>e[i].number;
 cout<<"Enter the compensation of employee number "<<i+1<<" : "; 
        cin>>e[i].compensation;
 cout<<"Enter employee type (first letter only) of employee number "<<i+1<<" : "<<endl
  <<"(laborer, secretary, manager, accountant, executive, researcher): ";
 switch(getche())
        {
 case 'l': x=laborer   ;  
        break;
 case 's': x=secretary ; 
        break;
 case 'm': x=manager   ; 
        break;
 case 'a': x=accountant; 
        break;
 case 'e': x=executive ; 
        break;
 case 'r': x=researcher;
        }
 switch(x)
        {
 case 0: e[i].ret = "laborer"   ; 
        break;
 case 1: e[i].ret = "secretary" ; 
        break;
 case 2: e[i].ret = "manager"   ; 
        break;
 case 3: e[i].ret = "accountant"; 
        break;
 case 4: e[i].ret = "executive" ; 
        break;
 case 5: e[i].ret = "researcher"; 
        break;
 default: e[i].ret = "Unknow";}
 cout<<"\nEnter the date of employee number "<<i+1<<" : "; cin >>e[i].d.day>>c>>e[i].d.month>>c>>e[i].d.year;} cout<<"\nEmployee number"<<"   compensation"<<"           type"<<"   date of first employment"<<endl;
 for(i=0; i<3; i++) cout<<setw(15)<<e[i].number
         <<setw(15)<<e[i].compensation
         <<setw(15)<<e[i].ret
         <<setw(21)<<e[i].d.day<<c<<e[i].d.month<<c<<e[i].d.year<<endl;
 
}

CODE:


8: Start with the fraction—adding program of Exercise 9 in Chapter 2, “C++ Programming Basics.”
  This program stores the numerator and denominator of two fractions before adding them, and may
  also store the answer, which is also a fraction. Modify the program so that all fractions are stored in
  variables of type struct fraction, whose two members are the fraction’s numerator and denominator
  (both type int). All fraction-related data should be stored in structures of this type.


#include<iostream>
#include<conio.h>
using namespace std;
struct fraction{int numerator; int denominator;};
int main()
{

 fraction equ[2];
 char operation;
 
 
 cout<<"Enter first  fraction: ";
 cin >>equ[0].numerator>>operation>>equ[0].denominator; //if (operation != '/') 
        {raise error event}
 cout<<"Enter second fraction: ";
 cin >>equ[1].numerator>>operation>>equ[1].denominator; //if (operation != '/') {raise error event}
 cout<<"Sum = "<<(equ[0].numerator*equ[1].denominator+equ[0].denominator*equ[1].numerator)
  <<operation<<(equ[0].denominator*equ[1].denominator)<<endl;
 
 
}
  This program stores the numerator and denominator of two fractions before adding them, and may
  also store the answer, which is also a fraction. Modify the program so that all fractions are stored in
  variables of type struct fraction, whose two members are the fraction’s numerator and denominator
  (both type int). All fraction-related data should be stored in structures of this type.


#include<iostream>
#include<conio.h>
using namespace std;
struct fraction{int numerator; int denominator;};
int main()
{

 fraction equ[2];
 char operation;
 
 
 cout<<"Enter first  fraction: ";
 cin >>equ[0].numerator>>operation>>equ[0].denominator; //if (operation != '/') 
        {raise error event}
 cout<<"Enter second fraction: ";
 cin >>equ[1].numerator>>operation>>equ[1].denominator; //if (operation != '/') {raise error event}
 cout<<"Sum = "<<(equ[0].numerator*equ[1].denominator+equ[0].denominator*equ[1].numerator)
  <<operation<<(equ[0].denominator*equ[1].denominator)<<endl;
 
 
}

CODE:

9. Create a structure called time. Its three members, all type int, should be called hours, minutes, and
  seconds. Write a program that prompts the user to enter a time value in hours, minutes, and seconds.
  This can be in 12:59:59 format, or each number can be entered at a separate prompt (“Enter
  hours:”, and so forth). The program should then store the time in a variable of type struct time, and
  finally print out the total number of seconds represented by this time value: 

  long totalsecs = t1.hours*3600 + t1.minutes*60 + t1.seconds



#include<iostream>
#include<conio.h>
using namespace std;
struct time{int hours; int minutes; int seconds;};
int main()
{

 time t1; char c;
 
 cout<<"Enter a time value in hours, minutes, and seconds [hh:mm:ss] format: ";
 cin >>t1.hours>>c>>t1.minutes>>c>>t1.seconds;
 cout<<"The total number of seconds is: "<<t1.hours*3600 + t1.minutes*60 + t1.seconds;

}

CODE:


10.  Create a structure called sterling that stores money amounts in the old-style British system
  discussed in Exercises 8 and 11 in Chapter 3, “Loops and Decisions.” The members could be
  called pounds, shillings, and pence, all of type int. The program should request the user to enter a
  money amount in new-style decimal pounds (type double), convert it to the old-style system, store it
  in a variable of type struct sterling, and then display this amount in pounds-shillings-pence format.


#include<iostream>
#include<conio.h>
using namespace std;
struct sterling{int pounds; int shillings; int pence;};
int main()
{
 
 float decpounds; float decfrac; //From old programme.
 sterling s1;
 cout<<"Enter a money amount in new-style decimal pounds: ";
 cin >>decpounds;
 s1.pounds = static_cast<int>(decpounds);
    decfrac = 240*(decpounds-s1.pounds);
 s1.shillings = (static_cast<int>(decfrac))%12;                  //Ignore fracions in pence.
 decfrac = static_cast<int>((decfrac-s1.shillings)/12);          //Ignore fracions in pence.
 cout<<"Equivalent in old notation = \x9c"<<s1.pounds<<"."<<decfrac<<"."<<s1.shillings<<endl;
 
}

CODE:

11.Use the time structure from Exercise 9, and write a program that obtains two time values from
  the user in 12:59:59 format, stores them in struct time variables, converts each one to seconds (type
  int), adds these quantities, converts the result back to hours-minutes-seconds, stores the result in a
  time structure, and finally displays the result in 12:59:59 format.


#include<iostream>
#include<conio.h>
using namespace std;
struct time{int hours; int minutes; int seconds;};
int main()
{
 time t1, t2, t3; char c; long tmp; //t3 coz he want to store the result in variable.
 cout<<"In [hh:mm:ss] format;\n";
 cout<<"Enter first time value : "; 
        cin >>t1.hours>>c>>t1.minutes>>c>>t1.seconds;
 cout<<"Enter second time value: "; 
        cin >>t2.hours>>c>>t2.minutes>>c>>t2.seconds;
 tmp=t1.hours*3600+t1.minutes*60+t1.seconds+t2.hours*3600+t2.minutes*60+t2.seconds;
 t3.seconds=tmp%60; t3.minutes=((tmp-t3.seconds)%3600)/60; t3.hours=tmp/3600;
 //those lines for true input at first then true output.
 if(t3.seconds>59) 
        {t3.seconds-=59; t3.minutes++;} //Check seconds over.
 if(t3.minutes>59) 
        {t3.minutes-=59; t3.hours++;}   //Check minutes over.
 //hours check not needed .. at 25 hours I haven't a format for days.
 cout<<"The result is: "<<t3.hours<<":"<<t3.minutes<<":"<<t3.seconds;
}

CODE:

12.Revise the four-function fraction calculator program of Exercise 12 in Chapter 3 so that each
  fraction is stored internally as a variable of type struct fraction, as discussed in Exercise 8 in this
  chapter.


#include<iostream>
#include<conio.h>
using namespace std;
struct fraction{int numerator; int denominator;};
int main()
{
 fraction f[2]; char c, op;
 cout<<"Enter your task : ";
 cin >>f[0].numerator>>c>>f[0].denominator>>op>>f[1].numerator>>c>>f[1].denominator;
 if(!f[0].denominator || !f[1].denominator) {cout<<"Illeagle fraction !"<<endl; op=false;}
 switch(op) 
{
 case '+':
 cout<<"Answer = "<<(f[0].numerator*f[1].denominator+f[0].denominator*f[1].numerator)<<c
  <<(f[0].denominator*f[1].denominator); 
        break;
 case '-':
 cout<<"Answer = "<<(f[0].numerator*f[1].denominator-f[0].denominator*f[1].numerator)<<c
  <<(f[0].denominator*f[1].denominator); 
        break;
 case '*':
 cout<<"Answer = "<<f[0].numerator*f[1].numerator<<c<<f[0].denominator*f[1].denominator; 
        break;
 case '/':
  if(f[0].numerator != 0) cout<<"Answer = "<<f[0].numerator*f[1].denominator<<c
             <<f[1].numerator*f[0].denominator;
  else              cout<<"Math error !"<<endl; 
break;
 default:
  cout<<"Unknow operator please try again !"<<endl;}
 
}