Concept explainers
Add the following member function to the ADT class DigitalTime defined in Displays 12.1 and 12.2:
void DigitalTime::intervalSince(const DigitalTime& aPreviousTime,
int& hoursInInterval, int& minutesInInterval) const
This function computes the time interval between two values of type DigitalTime. One of the values of type DigitalTime is the object that calls the member function intervalSince, and the other value of type DigitalTime is given as the first argument. For example, consider the following code:
DigitalTime current(5, 45), previous(2, 30);
int hours, minutes;
current.intervalSince(previous, hours, minutes);
cout << “The time interval between ” << previous
<< “ and ” << current << endl
<< “is ” « hours « “ hours and ”
<< minutes << “ minutes.\n”;
In a
The time interval between 2:30 and 5:45
is 3 hours and 15 minutes.
Allow the time given by the first argument to be later in the day than the time of the calling object. In this case, the time given as the first argument is assumed to be on the previous day. You should also write a program to test this revised ADT class.
Compute interval between two times
Program Plan:
- The interface file “dtime.h” is same as in the Display 12.1 however user needs to add the member function “void intervalSince(const DigitalTime& aPreviousTime, int& hoursInInterval, int& minutesInInterval)const;”.
- In the implementation file “dtime.cpp”, add the function definition for function “intervalSince”.
- In this function, first initializes the variables “hoursInInterval” and “minutesInInterval” to “0”.
- Declare a variable for compute the difference in “DigitalTime”.
- Compute the hour difference using “hour - aPreviousTime.hour”.
- Compute the minute difference using “minute - aPreviousTime.minute”.
- Check the condition of “hour” and “minutes”.
- Finally store the hours and minutes difference in their respective variables.
- In the application file that is “main.cpp”.
- Include the directive “dtime.h”.
- Define main function.
- Initializes the time for current and previous.
- Declare “int” variables for “hours” and “minutes”.
- Call “intervalSince()” function.
- Display the interval between two times.
The below C++ program is used to compute interval between two values of type “DigitalTime”.
Explanation of Solution
Program:
Modified code for “dtime.h”:
//DISPLAY 12.1 Interface File for DigitalTime
//Header file dtime.h: This is the INTERFACE for the class DigitalTime.
//Values of this type are times of day. The values are input and output in
//24-hour notation, as in 9:30 for 9:30 AM and 14:45 for 2:45 PM.
#include <iostream>
using namespace std;
class DigitalTime
{
public:
friend bool operator ==(const DigitalTime& time1, const DigitalTime& time2);
//Returns true if time1 and time2 represent the same time;
//otherwise, returns false.
DigitalTime(int theHour, int theMinute);
//Precondition: 0 <= theHour <= 23 and 0 <= theMinute <= 59.
//Initializes the time value to theHour and theMinute.
DigitalTime( );
//Initializes the time value to 0:00 (which is midnight).
void advance(int minutesAdded);
//Precondition: The object has a time value.
//Postcondition: The time has been changed to minutesAdded minutes later.
void advance(int hoursAdded, int minutesAdded);
//Precondition: The object has a time value.
//Postcondition: The time value has been advanced
//hoursAdded hours plus minutesAdded minutes.
void intervalSince(const DigitalTime& aPreviousTime, int& hoursInInterval, int& minutesInInterval)const;
//Precondition: The object has a time value.
//Precondition: The aPreviousTime object has a time value
//Postcondition: The hoursInInterval represents the number of hours that have passed and the minutesInInterval represents the number of minutes that have passed
friend istream& operator >>(istream& ins, DigitalTime& theObject);
//Overloads the >> operator for input values of type DigitalTime.
//Precondition: If ins is a file input stream, then ins has already been
//connected to a file.
friend ostream& operator <<(ostream& outs, const DigitalTime& theObject);
//Overloads the << operator for output values of type DigitalTime.
//Precondition: If outs is a file output stream, then outs has already been
//connected to a file.
private:
int hour;
int minute;
};
Modified code for “dtime.cpp”:
//DISPLAY 12.2 Implementation File for DigitalTime
//Implementation file dtime.cpp (Your system may require some
//suffix other than .cpp): This is the IMPLEMENTATION of the ADT DigitalTime.
//The interface for the class DigitalTime is in the header file dtime.h.
#include <iostream>
#include <cctype>
#include <cstdlib>
#include "dtime.h"
using namespace std;
//These FUNCTION DECLARATIONS are for use in the definition of
//the overloaded input operator >>:
void readHour(istream& ins, int& theHour);
//Precondition: Next input in the stream ins is a time in 24-hour notation,
//like 9:45 or 14:45.
//Postcondition: theHour has been set to the hour part of the time.
//The colon has been discarded and the next input to be read is the minute.
void readMinute(istream& ins, int& theMinute);
//Reads the minute from the stream ins after readHour has read the hour.
int digitToInt(char c);
//Precondition: c is one of the digits '0' through '9'.
//Returns the integer for the digit; for example, digitToInt('3') returns 3.
bool operator ==(const DigitalTime& time1, const DigitalTime& time2)
{
return (time1.hour == time2.hour && time1.minute == time2.minute);
}
//Uses iostream and cstdlib:
DigitalTime::DigitalTime(int theHour, int theMinute)
{
if (theHour < 0 || theHour > 23 || theMinute < 0 || theMinute > 59)
{
cout << "Illegal argument to DigitalTime constructor.";
exit(1);
}
else
{
hour = theHour;
minute = theMinute;
}
}
DigitalTime::DigitalTime( ) : hour(0), minute(0)
{
//Body intentionally empty.
}
void DigitalTime::advance(int minutesAdded)
{
int grossMinutes = minute + minutesAdded;
minute = grossMinutes%60;
int hourAdjustment = grossMinutes/60;
hour = (hour + hourAdjustment)%24;
}
void DigitalTime::advance(int hoursAdded, int minutesAdded)
{
hour = (hour + hoursAdded)%24;
advance(minutesAdded);
}
//Uses iostream:
ostream& operator <<(ostream& outs, const DigitalTime& theObject)
{
outs << theObject.hour << ':';
if (theObject.minute < 10)
outs << '0';
outs << theObject.minute;
return outs;
}
//Uses iostream:
istream& operator >>(istream& ins, DigitalTime& theObject)
{
readHour(ins, theObject.hour);
readMinute(ins, theObject.minute);
return ins;
}
int digitToInt(char c)
{
return ( static_cast<int>(c) - static_cast<int>('0') );
}
//Uses iostream, cctype, and cstdlib:
void readMinute(istream& ins, int& theMinute)
{
char c1, c2;
ins >> c1 >> c2;
if (!(isdigit(c1) && isdigit(c2)))
{
cout << "Error illegal input to readMinute\n";
exit(1);
}
theMinute = digitToInt(c1)*10 + digitToInt(c2);
if (theMinute < 0 || theMinute > 59)
{
cout << "Error illegal input to readMinute\n";
exit(1);
}
}
//Uses iostream, cctype, and cstdlib:
void readHour(istream& ins, int& theHour)
{
char c1, c2;
ins >> c1 >> c2;
if ( !( isdigit(c1) && (isdigit(c2) || c2 == ':' ) ) )
{
cout << "Error illegal input to readHour\n";
exit(1);
}
if (isdigit(c1) && c2 == ':')
{
theHour = digitToInt(c1);
}
else //(isdigit(c1) && isdigit(c2))
{
theHour = digitToInt(c1)*10 + digitToInt(c2);
ins >> c2;//discard ':'
if (c2 != ':')
{
cout << "Error illegal input to readHour\n";
exit(1);
}
}
if ( theHour < 0 || theHour > 23 )
{
cout << "Error illegal input to readHour\n";
exit(1);
}
}
/*Function definition compute interval between the two values of type DigitalTime */
void DigitalTime::intervalSince(const DigitalTime& aPreviousTime, int& hoursInInterval, int& minutesInInterval) const
{
/* Initializes the value of hours in interval to "0" */
hoursInInterval = 0;
/* Initializes the value of minutes in interval to "0" */
minutesInInterval = 0;
/* Declare the variable for time difference */
DigitalTime diff;
/* Compute hour difference */
diff.hour = hour - aPreviousTime.hour;
/* Compute minutes difference */
diff.minute = minute - aPreviousTime.minute;
/* Check condition */
if (hour < aPreviousTime.hour || hour == aPreviousTime.hour && minute < aPreviousTime.minute)
{
//Display given message
cout << "Preceding time is in the preceding day" << endl;
diff.hour = 24 + (hour - aPreviousTime.hour);
}
/* Check the condition hour */
if (diff.minute < 0)
{
diff.hour--;
diff.minute = diff.minute + 60;
}
/* Store hours and minutes interval in respective variable */
hoursInInterval = diff.hour;
minutesInInterval = diff.minute;
return;
}
Modified “main.cpp”:
//Header file
#include <iostream>
//Header file for "dtime.h"
#include "dtime.h"
//For standard input and output
using namespace std;
//Main function
int main( )
{
//Declare time
DigitalTime current(5, 45), previous(2, 30);
//Declare "int" variables
int hours, minutes;
/* Call intervalSince() function */
current.intervalSince(previous, hours, minutes);
/* Display given time interval */
cout << "The time interval between " << previous << " and " << current << endl << "is " << hours << " hours and " << minutes << " minutes.\n";
return 0;
}
The time interval between 2:30 and 5:45
is 3 hours and 15 minutes.
Want to see more full solutions like this?
Chapter 12 Solutions
Problem Solving with C++ (10th Edition)
Additional Engineering Textbook Solutions
Java How to Program, Early Objects (11th Edition) (Deitel: How to Program)
SURVEY OF OPERATING SYSTEMS
Starting Out With Visual Basic (8th Edition)
Starting Out with C++ from Control Structures to Objects (9th Edition)
Starting Out with Programming Logic and Design (5th Edition) (What's New in Computer Science)
Mechanics of Materials (10th Edition)
- can u solve this questionarrow_forward1. Unsigned Integers If we have an n-digit unsigned numeral dn-1d n-2...do in radix (or base) r, then the value of that numeral is n−1 r² di Σi=0 which is basically saying that instead of a 10's or 100's place we have an r's or r²'s place. For binary, decimal, and hex r equals 2, 10, and 16, respectively. Just a reminder that in order to write down a large number, we typically use the IEC or SI prefixing system: IEC: Ki = 210, Mi = 220, Gi = 230, Ti = 240, Pi = 250, Ei = 260, Zi = 270, Yi = 280; SI: K=103, M = 106, G = 109, T = 10¹², P = 1015, E = 10¹8, Z = 1021, Y = 1024. 1.1 Conversions a. (15 pts) Write the following using IEC prefixes: 213, 223, 251, 272, 226, 244 21323 Ki8 Ki 223 23 Mi 8 Mi b. (15 pts) Write the following using SI prefixes: 107, 10¹7, 10¹¹, 1022, 1026, 1015 107 10¹ M = 10 M = 1017102 P = 100 P c. (10 pts) Write the following with powers of 10: 7 K, 100 E, 21 G 7 K = 7*10³arrow_forwardanswer shoul avoid using AI and should be basic and please explainarrow_forward
- Node A is connected to node B by a 2000km fiber link having a bandwidth of 100Mbps. What is the total latency time (transmit + propagation) required to transmit a 4000 byte file using packets that include 1000 Bytes of data plus 40 Bytes of header.arrow_forwardanswer should avoid using AI and should be basic and explain pleasearrow_forwardasnwer should avoid using AIarrow_forward
- answer should avoid using AI (such as ChatGPT), do not any answer directly copied from AI would and explain codearrow_forwardWrite a c++ program that will count from 1 to 10 by 1. The default output should be: 1, 2, 3, 4, 5, 6 , 7, 8, 9, 10 There should be only a newline after the last number. Each number except the last should be followed by a comma and a space. To make your program more functional, you should parse command line arguments and change behavior based on their values. Argument Parameter Action -f, --first yes, an integer Change place you start counting -l, --last yes, an integer Change place you end counting -s, --skip optional, an integer, 1 if not specified Change the amount you add to the counter each iteration -h, —help none Print a help message including these instructions. -j, --joke none Tell a number based joke. So, if your program is called counter, counter -f 10 --last 4 --skip 2 should produce 10, 8, 6, 4 Please use the last supplied argument. If your code is called counter, counter -f 4 -f 5 -f 6 should count from 6. You should…arrow_forwardshow workarrow_forward
- C++ Programming: From Problem Analysis to Program...Computer ScienceISBN:9781337102087Author:D. S. MalikPublisher:Cengage LearningC++ for Engineers and ScientistsComputer ScienceISBN:9781133187844Author:Bronson, Gary J.Publisher:Course Technology PtrMicrosoft Visual C#Computer ScienceISBN:9781337102100Author:Joyce, Farrell.Publisher:Cengage Learning,
- Programming Logic & Design ComprehensiveComputer ScienceISBN:9781337669405Author:FARRELLPublisher:CengageEBK JAVA PROGRAMMINGComputer ScienceISBN:9781337671385Author:FARRELLPublisher:CENGAGE LEARNING - CONSIGNMENTSystems ArchitectureComputer ScienceISBN:9781305080195Author:Stephen D. BurdPublisher:Cengage Learning