C - Write the routines day_of_year and month_day with pointers

缺乏、安全感 2022-11-14 11:55 169阅读 0赞

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/*
     * Write the routines day_of_year and month_day with pointers.
     *
     * Tail.c - by FreeMan
     */
    
    #include <stdio.h>
    static char daytable[2][13] =
    {
    	{0, 31, 28, 31, 30, 31, 30, 31, 31, 30 ,31, 30, 31},
    	{0, 31, 29, 31, 30, 31, 30, 31, 31, 30 ,31, 30, 31}
    };
    
    int day_of_year(int, int, int);
    void month_day(int, int, int *, int *);
    
    int main()
    {
    	int m, d;
    
    	printf("day_of_year(2021, 1, 1) is %d\n", day_of_year(2021, 1, 1));
    	printf("day_of_year(2021, 4, 3) is %d\n", day_of_year(2021, 4, 3));
    	printf("day_of_year(2021, 12, 31) is %d\n", day_of_year(2021, 12, 31));
    
    	month_day(2021, 1, &m, &d);
    	printf("month_day(2021, 1) is: %d %d\n", m, d);
    	month_day(2021, 100, &m, &d);
    	printf("month_day(2021, 100) is: %d %d\n", m, d);
    	month_day(2021, 365, &m, &d);
    	printf("month_day(2021, 365) is: %d %d\n", m, d);
    
    	return 0;
    }
    
    int day_of_year(int year, int month, int day)
    {
    	int leap;
    	char *p;
    
    	leap = year % 4 == 0 && year % 100 != 0 || year % 400 == 0;
    	p = &daytable[leap][1];
    	while (--month > 0)
    	{
    		day += *p++;
    	}
    
    	return day;
    }
    
    void month_day(int year, int dayofyear, int *pmonth, int *pday)
    {
    	int leap;
    	char *p;
    
    	leap = year % 4 == 0 && year % 100 != 0 || year % 400 == 0;
    	p = &daytable[leap][1];
    	while (dayofyear > *p)
    	{
    		dayofyear -= *p++;
    	}
    	*pmonth = 1 + p - &daytable[leap][1];
    	*pday = dayofyear;
    }
    
    // Output:
    /*
    day_of_year(2021, 1, 1) is 1
    day_of_year(2021, 4, 3) is 93
    day_of_year(2021, 12, 31) is 365
    month_day(2021, 1) is: 1 1
    month_day(2021, 100) is: 4 10
    month_day(2021, 365) is: 12 31
    
    */

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