12. Write the inverse of Exercise 10, so that the user enters an amount in Great Britain’s new decimal-pounds notation (pounds and pence), and the program converts it to the old pounds-shillings-pence notation.

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Chapter 2: 

C++ Programming Basics 

Programming Exercise

Problem # 12:

Write the inverse of Exercise 10, so that the user enters an amount in Great Britain’s new decimal-pounds notation (pounds and pence), and the program converts it to the old pounds-shillings-pence notation. An example of interaction with the program might be
Enter decimal pounds: 3.51
Equivalent in old notation = £3.10.2.
Make use of the fact that if you assign a floating-point value (say 12.34) to an integer variable, the decimal fraction (0.34) is lost; the integer value is simply 12. Use a cast to avoid a compiler warning. You can use statements like
float decpounds;    // input from user (new-style pounds)
int pounds;            // old-style (integer) pounds
float decfrac;        // decimal fraction (smaller than 1.0)
pounds = static_cast<int>(decpounds); // remove decimal fraction
decfrac = decpounds - pounds;  // regain decimal fraction
You can then multiply decfrac by 20 to find shillings. A similar operation obtains pence.

Solution:

#include<iostream>
#include<conio.h>
using namespace std;
int main()
{
    int pounds,shillings,pence ;
    float decpounds,decfrac,shillingsfrac,pencefrac;
    cout<<"Enter decimal pounds: ";
    cin>>decpounds;
    pounds = (int)decpounds;
    decfrac = decpounds - pounds;
    shillingsfrac = decfrac*20;
    shillings = (int)shillingsfrac;
    shillingsfrac=shillingsfrac-shillings;
    pencefrac=shillingsfrac*12;
      
    cout<<"Equivalent in old notation = \x9c"<<pounds<<"."<<shillings<<"."<<pence;
  
    return 0;
}

Let me know in the comment section if you have any question.

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