Chapter 9
_________
Working with
Structures
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Prog 19.1: Illustrating a structure
#include <stdio.h>
int main (void) {
struct date {
int month;
int day;
int year;
}
// type declaration
struct date today;
today.month = 12;
today.day
= 17;
today.year = 2008;
printf (“Today’s date is %i/%i/%.2i.\n”,
today.day, today.month, today.year);
today.month = 1;
today.day
= 1;
if (today.month == 5 && today.day == 17)
printf(“Happy May 17th!\n”);
return 0;
}
___________________________________________
Today’s date is 17/12/2008.
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Happy New Year!
Prog 19.2 (A): Determining
tomorrow’s date given today’s.
#include <stdio.h>
int main (void) {
struct date {
int month;
int day;
int year;
}
// type declaration
const int daysPerMonth [13] =
{0, 31, 28, 31, 30, 31, 30, 31,
31, 30, 31, 30, 31};
struct date today, tomorrow;
printf (“Today’s date (dd mm yyyy): “);
scanf (“%i%i%i”, &today.day,
&today.month, &today.year);
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Prog 9.2 (B): Determining
tomorrow’s date given today’s.
if (today.day !=
daysPerMonth [today.month]) {
tomorrow.day
= today.day + 1;
tomorrow.month = today.month;
tomorrow.year = today.year;
}
else if (today.month == 12) {
tomorrow.day
= 1;
tomorrow.month = 1;
tomorrow.year = today.year + 1;
}
else {
tomorrow.day
= 1;
tomorrow.month = today.month + 1;
tomorrow.year = today.year;
}
printf (“Tomorrow’s date is ”
“%i/%i/%.2i.\n”, tomorrow.day,
tomorrow.month, tomorrow.year);
return 0;
}
___________________________________________
Today’s date (dd mm yyyy): 20 12 2008
Tomorrow’s date is 20/12/2008.
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Prog 19.3 (A): Revising: Find tomorrow’s
date given today’s.
#include <stdio.h>
struct date {int month;
int day;
int year;};
int main (void) {
struct date today, tomorrow;
int numOfDays (struct date d) ;
printf (“Today’s date (dd mm yyyy): “);
scanf (“%i%i%i”, &today.day,
&today.month, &today.year);
if ( today.day != numOfDaysInMonth (today) )
{
tomorrow.day
= today.day + 1;
tomorrow.month = today.month;
tomorrow.year = today.year;
}
else if (today.month == 12) {
tomorrow.day
= 1;
tomorrow.month = 1;
tomorrow.year = today.year + 1;
}
else {
tomorrow.day
= 1;
tomorrow.month = today.month + 1;
tomorrow.year = today.year;
}
printf (“Tomorrow’s date is %i/%i/%i.\n”,
tomorrow.day,tomorrow.month, tomorrow.year);
return 0;
}
1-5
Prog 19.3 (B): Revising: Find tomorrow’s
date given today’s.
int numOfDaysInMonth (struct date d) {
int days;
bool isLeapYear (struct date d);
const int daysPerMonth [13] =
{0, 31, 28, 31, 30, 31, 30, 31,
31, 30, 31, 30, 31};
if ( isLeapYear(d) && d.month == 2)
days = 29;
else
days = daysPerMonth [d.month];
return days;
}
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Prog 19.3 (C): Revising: Find tomorrow’s
date given today’s.
bool isLeapYear (struct date d);
bool aLeapYear = false;
if ( (d.year % 4
== 0 &&
d.year % 100 != 0) ) ||
(d.year % 400 == 0 ))
aLeapYear = true;
return aLeapYear;
}
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