Answered step by step
Verified Expert Solution
Link Copied!

Question

1 Approved Answer

#include Date.h Date::Date() { //constructor to assign todays date to date char data[9]; //holder for the date _strdate_s(data); //gets the current date mm/dd/yy ( THIS

#include "Date.h"

Date::Date() {

//constructor to assign todays date to date

char data[9]; //holder for the date

_strdate_s(data); //gets the current date mm/dd/yy (THIS IS ONE OF THE MISTAKES)

string date=data; //copy to a string for parsing

month=stoi(date.substr(0,2)); //gets characters 0 and 1 of date and converts to int

day=stoi(date.substr(3,2)); //gets characters 3 and 4 of date and converts to int

year=stoi(date.substr(6,2))+2000; //gets characters 6 and 7 of date and converts to int

if(leapYear()) days2[2]=29; else days2[2]=28;

for(int m=0;m<13;m++) days[m]=days2[m];

}//constructor for today

static bool validDate(int m, int d, int y) { (THIS IS THE OTHER MISTAKE)

bool valid=true; //assume it is valid until found to be invalid

if(y<1000) valid=false;

if(m<1 || m>12) valid=false;

if(leapYear(y)) days2[2]=29; else days2[2]=28;

if(d<1 || d>days2[m]) valid=false;

return valid;

}//validDate

Date::Date(int m, int d, int y) {

//constructor to assign values to month day and year

if(validDate(m,d,y)) {

month=m;

day=d;

year=y;

}

else {

month=day=1;

year=1970; //Unix time starting point

} //not valid: set to default valid date

for(int m=0;m<13;m++) days[m]=days2[m];

} //constructor with assigned values

Date::Date(int julian) {

//Fliegel-Van Flandern algorithm to convert Julian date to Gregorian number month, day, and year

gregorian(julian,month,day,year);

if(leapYear()) days2[2]=29; else days2[2]=28;

for(int m=0;m<13;m++) days[m]=days2[m];

}//Date Julian

Date::Date (string str) { //constructor for todays date as "mm/dd/year

//Parse str by adding one char at a s time to the token until a "/" is encounter.

//When "/" is encountered start the next token

//int p=0;

int count=0;

int num[3];

string token[3];

int len=str.length();

for(int p=0; p

if(str.substr(p,1)=="/") count++;

else token[count]+=str.substr(p,1);

}//parse str to create array of tokens

bool error=false;

for(int p=0;p<3;p++) {

try {

num[p]=stoi(token[p]);

}//try to convert to int

catch(invalid_argument&) {

num[p]=-1;

error=true;

} //catch

}//each of the 3 tokens

if(!error && validDate(num[0],num[1],num[2])) {

month=num[0];

day=num[1];

year=num[2];

} //no error

else {

month=day=1;

year=1970; //Unix time starting point

} //not valid: set to default valid date

for(int m=0;m<13;m++) days[m]=days2[m];

}//constructor with string such as "10/31/2016"

Date Date::operator=(const Date& otherDate) {

//assigns another instance of the date class to this.

month=otherDate.month;

day=otherDate.day;

year=otherDate.year;

return *this; //allows date1=date=date3;

}//overloaded operator =

Date Date::operator+(int numDays) {

//Adds the number of days to the Julian date.

Date other(month,day,year); //make copy of the date

int jd=other.julian(); //find the Julian date

jd+=numDays; //add the number of days to the Julian date

gregorian(jd,other.month,other.day,other.year); //Convert the Julian date back to Gregorian

if(other.leapYear()) days2[2]=29; else days2[2]=28;

for(int m=0;m<13;m++) other.days[m]=days2[m];

return other;

} //operator +

int Date::dayofYear() const {

//returns the day of the year, ie 2/1/???? is the 32 day of the year

int total=day;

for(int m=1;m

return total;

}//dayofYear

void gregorian(int julian, int &mth, int &d, int &y) {

//Fliegel-Van Flandern algorithm to convert Julian date to Gregorian month, day, and year

int p,q,r,s,t,u,v;

p = julian + 68569;

q = 4*p/146097;

r = p - (146097*q + 3)/4;

s = 4000*(r+1)/1461001;

t = r - 1461*s/4 + 31;

u = 80*t/2447;

v = u/11;

y = 100*(q-49)+s+v;

mth = u + 2 - 12*v;

d = t - 2447*u/80;

} //Gregorian

int Date::julian() const {

int jd= day-32075+1461*(year+4800+(month-14)/12)/4+

367*(month-2-(month-14)/12*12) /12-3*((year+4900+(month-14)/12)/100)/4;

return jd;

}

bool Date::leapYear() const {

bool leap=false;

if(year%4==0) leap=true;

if(year%100==0 && year%400!=0) leap=false;

return leap;

}//leapYear

SO I HAVE 2 SEMANTIC ERRORS THAT I CANNOT SEEM TO FIX OR UNDERSTAND, PLEASE LET ME KNOW HOW

bool leapYear(int yr) {

bool leap=false;

if(yr%4==0) leap=true;

if(yr%100==0 && yr%400!=0) leap=false;

return leap;

}//leapYear

int Date::weekday() const {

//returns 0 for Sunday, 1 for Monday, etc.

static int t[] = { 0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4 };

int y =year;

y-= month < 3;

return ( y + y/4 - y/100 + y/400 + t[month-1] + day) % 7;

}//weekday

int Date::getMonth() const {

//private variables cannot be accessed directly but require "getter" functions

return month;

}//getMonth

int Date::getDay() const {

return day;

}//

int Date::getYear() const {

return year;

}//getYear

string Date::toString() const {

stringstream oss; //a stream to append the values

oss<

return oss.str();

}

bool Date::operator==(const Date& otherDate) {

return (month==otherDate.month && day==otherDate.day && year==otherDate.year);

}//operator ==

bool Date::operator<(const Date& otherDate) {

//A date is less than another date if is earlier

bool result=false; //assume false until proven true

if(year

else if(year==otherDate.year && month

else if(year==otherDate.year && month==otherDate.month && day

return result;

}//operator

bool Date::operator>(const Date& otherDate) {

//Convert both dates to Julian and compare the Julian dates

int jd1=julian();

int jd2=otherDate.julian();

return jd1>jd2;

}//operator

ostream& operator << (ostream &output, const Date &d) {

output << d.toString();

return output;

} // operator <<

istream& operator >> (istream &input, Date &d) {

string s;

input >> s;

Date other(s); //create a new Date

d=other; //assign the new Date to d

return input;

} // operator >>

Step by Step Solution

There are 3 Steps involved in it

Step: 1

blur-text-image

Get Instant Access to Expert-Tailored Solutions

See step-by-step solutions with expert insights and AI powered tools for academic success

Step: 2

blur-text-image

Step: 3

blur-text-image

Ace Your Homework with AI

Get the answers you need in no time with our AI-driven, step-by-step assistance

Get Started

Recommended Textbook for

More Books

Students also viewed these Databases questions

Question

What are the advantages of licensing as described in the case? L01

Answered: 1 week ago

Question

Perform an Internet search. Discuss a company that uses EPLI.

Answered: 1 week ago

Question

How do you feel about employment-at-will policies? Are they fair?

Answered: 1 week ago