Robert Lafore 4th Edition Solution Manual Chapter 5 - Functions

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1.  Refer to the CIRCAREA program in Chapter 2, “C++ Programming Basics.” Write a function
  called circarea() that finds the area of a circle in a similar way. It should take an argument of type float
  and return an argument of the same type. Write a main() function that gets a radius value from the
  user, calls circarea(), and displays the result.




#include<iostream>
#include<conio.h>
using namespace std;

float circarea(float radius);
int main()
{
 float rad;
 cout<<"Enter radius of circle: "; 
cin>> rad;
 cout <<"Area is "<<circarea(rad);
 
}

float circarea(float radius) 
{ return 3.14159F*radius*radius;}

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2.  Raising a number n to a power p is the same as multiplying n by itself p times. Write a function
  called power() that takes a double value for n and an int value for p, and returns the result as a double
  value. Use a default argument of 2 for p, so that if this argument is omitted, the number n will be
  squared. Write a main() function that gets values from the user to test this function.



#include<iostream>
#include<conio.h>
using namespace std;
double power(double n, int p=2);

int main(void)
{
 
 double n; int p=2;
 cout<<"Enter n: "; cin>> n;
 cout<<"Enter p: "; cin>> p;
 cout <<"The power is "<<power(n, p);
}

double power(double n, int p)
{ for(int ret=1; p>0; p--) 
ret*=n; 
return ret;}
  called power() that takes a double value for n and an int value for p, and returns the result as a double
  value. Use a default argument of 2 for p, so that if this argument is omitted, the number n will be
  squared. Write a main() function that gets values from the user to test this function.



#include<iostream>
#include<conio.h>
using namespace std;
double power(double n, int p=2);

int main(void)
{
 
 double n; int p=2;
 cout<<"Enter n: "; cin>> n;
 cout<<"Enter p: "; cin>> p;
 cout <<"The power is "<<power(n, p);
}

double power(double n, int p)
{ for(int ret=1; p>0; p--) 
ret*=n; 
return ret;}

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3.  Write a function called zeroSmaller() that is passed two int arguments by reference and then sets
  the smaller of the two numbers to 0. Write a main() program to exercise this function.



#include<iostream>
#include<conio.h>
using namespace std;
int main()
{
 int n1, n2;
 cout<<"Enter n1: "; 
        cin>> n1;
 cout<<"Enter n2: "; 
 cin>> n2;
 cout <<"The number assigned to zero is ";
 if(zeroSmaller(n1, n2)) cout<<"n1"; 
 else cout<<"n2";
}

bool zeroSmaller(int& n1, int& n2)
{ 
 if(n1<n2) {n1=0; return true;}
 else {n2=0;return false;}
}
  the smaller of the two numbers to 0. Write a main() program to exercise this function.



#include<iostream>
#include<conio.h>
using namespace std;
int main()
{
 int n1, n2;
 cout<<"Enter n1: "; 
        cin>> n1;
 cout<<"Enter n2: "; 
 cin>> n2;
 cout <<"The number assigned to zero is ";
 if(zeroSmaller(n1, n2)) cout<<"n1"; 
 else cout<<"n2";
}

bool zeroSmaller(int& n1, int& n2)
{ 
 if(n1<n2) {n1=0; return true;}
 else {n2=0;return false;}
}

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4.Write a function that takes two Distance values as arguments and returns the larger one. Include
  a main() program that accepts two Distance values from the user, compares them, and displays the
  larger.



#include<iostream>
#include<conio.h>
using namespace std;
int distWatch(int d1, int d2);
int main()
{
 int d1, d2;
 cout<<"Enter d1: "; 
 cin>> d1;
 cout<<"Enter d2: "; 
 cin>> d2;
 cout <<"The largest distance is "<<distWatch(d1, d2);
}

int distWatch(int d1, int d2)
{ if(d1<d2) return d2; 
else return d1;}

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5.Write a function called hms_to_secs() that takes three int values—for hours, minutes, and
  seconds—as arguments, and returns the equivalent time in seconds (type long). Create a program
  that exercises this function by repeatedly obtaining a time value in hours, minutes, and seconds from
  the user (format 12:59:59), calling the function, and displaying the value of seconds it returns.



#include<iostream>
#include<conio.h>
using namespace std;
long hms_to_secs(int hours, int minutes, int seconds);
int main()
{
 int h, m, s; 
 char sep;
 cout<<"Enter the time in format hh:mm:ss : "; 
 cin>>h>>sep>>m>>sep>>s;
 cout <<"The equivalent time in seconds is : "<<hms_to_secs(h, m, s);
}

long hms_to_secs(int hours, int minutes, int seconds)
{ return seconds+minutes*60+hours*3600;}

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6.Start with the program from Exercise 11, Chapter 4, “Structures,” which adds two struct time
  values. Keep the same functionality, but modify the program so that it uses two functions. The first,
  time_to_secs(), takes as its only argument a structure of type time, and returns the equivalent in
  seconds (type long). The second function, secs_to_time(), takes as its only argument a time in
  seconds (type long), and returns a structure of type time.



#include<iostream>
#include<conio.h>
using namespace std;
struct time{int hours; int minutes; int seconds;};
long time_to_secs(time t);
time secs_to_time(long s);
int main()
{
 time t1, t2, t3; 
 char c;
 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;
 t3=secs_to_time(time_to_secs(t1)+time_to_secs(t2));
 cout<<"The result is: "<<t3.hours<<":"<<t3.minutes<<":"<<t3.seconds;
}

long time_to_secs(time t){ return t.hours*3600+t.minutes*60+t.seconds;}
time secs_to_time(long s){
 time t;
 t.seconds=s%60; t.minutes=((s-t.seconds)%3600)/60; t.hours=s/3600;
 if(t.seconds>59) {t.seconds-=59; t.minutes++;} //Check seconds over.
 if(t.minutes>59) {t.minutes-=59; t.hours++;}   //Check minutes over.
 return t;
}

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7.Start with the power () function of Exercise 2, which works only with type double. Create a series
  of overloaded functions with the same name that, in addition to double, also work with types char,
  int, long, and float. Write a main() program that exercises these overloaded functions with all
  argument types.


#include<iostream>
#include<conio.h>
using namespace std;
double power(double n, int p=2);
char   power(char   n, int p=2);
int    power(int    n, int p=2);
long   power(long   n, int p=2);
float  power(float  n, int p=2);

int main()
{
 char sep; int p=2; double d_n; char c_n; int i_n; long l_n; float f_n;
 cout<<"In [n^p] format;\n";
 cout<<"Enter a double type n : "; 
 cin >>d_n>>sep>>p;
 cout<<"The power is "<<power(d_n, p)<<endl;
 cout<<"Enter a char type n   : "; 
 cin >>c_n>>sep>>p;
 cout<<"The power is "<<power(c_n, p)<<endl;
 cout<<"Enter a int type n    : "; 
 cin >>i_n>>sep>>p;
 cout<<"The power is "<<power(i_n, p)<<endl;
 cout<<"Enter a long type n   : "; 
 cin >>l_n>>sep>>p;
 cout<<"The power is "<<power(l_n, p)<<endl;
 cout<<"Enter a float type n  : "; 
 cin >>f_n>>sep>>p;
 cout<<"The power is "<<power(f_n, p)<<endl;
}

double power(double n, int p)
{ for(int ret=1; p>0; p--) ret*=n; return ret;}
char   power(char   n, int p)
{ for(int ret=1; p>0; p--) ret*=n; return ret;}
int    power(int    n, int p)
{ for(int ret=1; p>0; p--) ret*=n; return ret;}
long   power(long   n, int p)
{ for(int ret=1; p>0; p--) ret*=n; return ret;}
float  power(float  n, int p)
{ for(int ret=1; p>0; p--) ret*=n; return ret;}

  of overloaded functions with the same name that, in addition to double, also work with types char,
  int, long, and float. Write a main() program that exercises these overloaded functions with all
  argument types.


#include<iostream>
#include<conio.h>
using namespace std;
double power(double n, int p=2);
char   power(char   n, int p=2);
int    power(int    n, int p=2);
long   power(long   n, int p=2);
float  power(float  n, int p=2);

int main()
{
 char sep; int p=2; double d_n; char c_n; int i_n; long l_n; float f_n;
 cout<<"In [n^p] format;\n";
 cout<<"Enter a double type n : "; 
 cin >>d_n>>sep>>p;
 cout<<"The power is "<<power(d_n, p)<<endl;
 cout<<"Enter a char type n   : "; 
 cin >>c_n>>sep>>p;
 cout<<"The power is "<<power(c_n, p)<<endl;
 cout<<"Enter a int type n    : "; 
 cin >>i_n>>sep>>p;
 cout<<"The power is "<<power(i_n, p)<<endl;
 cout<<"Enter a long type n   : "; 
 cin >>l_n>>sep>>p;
 cout<<"The power is "<<power(l_n, p)<<endl;
 cout<<"Enter a float type n  : "; 
 cin >>f_n>>sep>>p;
 cout<<"The power is "<<power(f_n, p)<<endl;
}

double power(double n, int p)
{ for(int ret=1; p>0; p--) ret*=n; return ret;}
char   power(char   n, int p)
{ for(int ret=1; p>0; p--) ret*=n; return ret;}
int    power(int    n, int p)
{ for(int ret=1; p>0; p--) ret*=n; return ret;}
long   power(long   n, int p)
{ for(int ret=1; p>0; p--) ret*=n; return ret;}
float  power(float  n, int p)
{ for(int ret=1; p>0; p--) ret*=n; return ret;}

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8.  Write a function called swap() that interchanges two int values passed to it by the calling program.
  (Note that this function swaps the values of the variables in the calling program, not those in the
  function.) You’ll need to decide how to pass the arguments. Create a main() program to exercise the
  function.



#include<iostream>
#include<conio.h>
using namespace std;
void swap(int& a, int& b);
int main()
{
 int a, b;
 cout<<"Enter a : "; 
 cin >>a;
 cout<<"Enter b : "; 
 cin >>b; swap(a, b);
 cout<<"Now a value is : "<<a<<" and b value is : "<<b;
}

void swap(int& a, int& b){int c=a; a=b; b=c;}

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9.  This exercise is similar to Exercise 8, except that instead of two int variables, have the swap()
  function interchange two struct time values (see Exercise 6).


#include<iostream>
#include<conio.h>
using namespace std;
struct time{int hours; int minutes; int seconds;};
void swap(time& t1, time& t2);

int main()
{
 time t1, t2; char c;
 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;
 swap(t1, t2);
 cout<<"Now first time is : "  <<t1.hours<<c<<t1.minutes<<c<<t1.seconds
  <<" and second time is : "<<t2.hours<<c<<t2.minutes<<c<<t2.seconds;
}

void swap(time& a, time& b){time c=a; a=b; b=c;}

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10.  Write a function that, when you call it, displays a message telling how many times it has been
  called: “I have been called 3 times”, or whatever. Write a main() program that calls this function at
  least 10 times. Try implementing this function in two different ways. First, use an external variable to
  store the count. Second, use a local static variable. Which is more appropriate? Why can’t you use
  an automatic variable?


#include<iostream>
#include<conio.h>
using namespace std;
void caller_counter(void);
int main()
{
 int outer_counter=0;
 outer_counter++; caller_counter();
 cout<<"\nThe main programme counter value is: "<<outer_counter;
}

void caller_counter(void)
{
 static int inner_counter=0; inner_counter++;
 cout<<"I have been called "<<inner_counter<<" times";
}

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11.  Write a program, based on the sterling structure of Exercise 10 in Chapter 4, “Structures,” that
  obtains from the user two money amounts in old-style British format (£9:19:11), adds them, and
  displays the result, again in old-style format. Use three functions. The first should obtain a pounds-
  shillings-pence value from the user and return the value as a structure of type sterling. The second
  should take two arguments of type sterling and return a value of the same type, which is the sum of
  the arguments. The third should take a sterling structure as its argument and display its value.*/


#include<iostream>
#include<conio.h>
using namespace std;
struct sterling{int pounds; int shillings; int pence;};
sterling psp_to_sterling(int pounds, int shillings, int pence);
sterling sterling_add(sterling s1, sterling s2);
void sterling_disp(sterling s);
char c;
int main()
{
 int x, y, z; 
 sterling x1, x2;
 cout<<"In [9:19:11] format;\n";
 cout<<"Enter first money amount in old-style British  : \x9c";
 cin>>x>>c>>y>>c>>z; 
 x1=psp_to_sterling(x, y, z);
 cout<<"Enter second money amount in old-style British : \x9c";
 cin>>x>>c>>y>>c>>z; x2=psp_to_sterling(x, y, z);
 sterling_disp(sterling_add(x1, x2));
}

sterling psp_to_sterling(int pounds, int shillings, int pence)
{
 sterling x; 
 x.pounds=pounds; x.shillings=shillings; 
 x.pence=pence; return x;
}

sterling sterling_add(sterling s1, sterling s2)
{
 s1.pounds += s2.pounds; 
 s1.shillings += s2.shillings; 
 s1.pence += s2.pence;
 if(s1.pence>11){s1.shillings += static_cast<int>(s1.pence/12); s1.pence %= 12;}
 if(s1.shillings>19){s1.pounds += static_cast<int>(s1.shillings/20); s1.shillings %= 20;}
 return s1;}
 
void sterling_disp(sterling s){
 cout<<"Total is                                       : \x9c"
  <<s.pounds<<c<<s.shillings<<c<<s.pence;}
  obtains from the user two money amounts in old-style British format (£9:19:11), adds them, and
  displays the result, again in old-style format. Use three functions. The first should obtain a pounds-
  shillings-pence value from the user and return the value as a structure of type sterling. The second
  should take two arguments of type sterling and return a value of the same type, which is the sum of
  the arguments. The third should take a sterling structure as its argument and display its value.*/


#include<iostream>
#include<conio.h>
using namespace std;
struct sterling{int pounds; int shillings; int pence;};
sterling psp_to_sterling(int pounds, int shillings, int pence);
sterling sterling_add(sterling s1, sterling s2);
void sterling_disp(sterling s);
char c;
int main()
{
 int x, y, z; 
 sterling x1, x2;
 cout<<"In [9:19:11] format;\n";
 cout<<"Enter first money amount in old-style British  : \x9c";
 cin>>x>>c>>y>>c>>z; 
 x1=psp_to_sterling(x, y, z);
 cout<<"Enter second money amount in old-style British : \x9c";
 cin>>x>>c>>y>>c>>z; x2=psp_to_sterling(x, y, z);
 sterling_disp(sterling_add(x1, x2));
}

sterling psp_to_sterling(int pounds, int shillings, int pence)
{
 sterling x; 
 x.pounds=pounds; x.shillings=shillings; 
 x.pence=pence; return x;
}

sterling sterling_add(sterling s1, sterling s2)
{
 s1.pounds += s2.pounds; 
 s1.shillings += s2.shillings; 
 s1.pence += s2.pence;
 if(s1.pence>11){s1.shillings += static_cast<int>(s1.pence/12); s1.pence %= 12;}
 if(s1.shillings>19){s1.pounds += static_cast<int>(s1.shillings/20); s1.shillings %= 20;}
 return s1;}
 
void sterling_disp(sterling s){
 cout<<"Total is                                       : \x9c"
  <<s.pounds<<c<<s.shillings<<c<<s.pence;}

CODE:

12.  Revise the four-function fraction calculator from Exercise 12, Chapter 4, so that it uses
  functions for each of the four arithmetic operations. They can be called fadd(), fsub(), fmul(), and fdiv
  (). Each of these functions should take two arguments of type struct fraction, and return an argument
  of the same type.



#include<iostream>
#include<conio.h>
using namespace std;
struct fraction{int numerator; int denominator;};
fraction fadd(fraction a, fraction b);
fraction fsub(fraction a, fraction b);
fraction fmul(fraction a, fraction b);
fraction fdiv(fraction a, fraction b);
int main()
{
 fraction f[3]; 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 '+': 
 f[2]=fadd(f[0], f[1]); 
 break;
 case '-': 
 f[2]=fsub(f[0], f[1]); 
 break;
 case '*': 
 f[2]=fmul(f[0], f[1]); 
 break;
 case '/': 
 f[2]=fdiv(f[0], f[1]); 
 break;
 default: 
 cout<<"Unknow operator please try again !"<<endl;}
 cout<<"Answer = "<<f[2].numerator<<c<<f[2].denominator;
}

fraction fadd(fraction a, fraction b)
{
 fraction f;
 f.numerator  =a.numerator*b.denominator+a.denominator*b.numerator;
 f.denominator=a.denominator*b.denominator; return f;
}

fraction fsub(fraction a, fraction b)
{
 fraction f;
 f.numerator  =a.numerator*b.denominator-a.denominator*b.numerator;
 f.denominator=a.denominator*b.denominator; return f;
}

fraction fmul(fraction a, fraction b)
{
 fraction f;
 f.numerator  =a.numerator*b.numerator;
 f.denominator=a.denominator*b.denominator; return f;
}

fraction fdiv(fraction a, fraction b)
{
 fraction f;
 if(b.numerator != 0)
{
 f.numerator  =a.numerator*b.denominator;
 f.denominator=b.numerator*a.denominator;
}
 else              cout<<"Math error !"<<endl; return f;
}