# Basic and Advance C Question: **Download Questions PDF**

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## How can I find the day of the week given the date?

## Answers:

**Answer #1**

Here are three methods:

1. Use mktime or localtime # . Here is a code fragment which computes the day of the week for February 29, 2000:

#include <stdio.h>

#include <time.h>

char *wday[] = {"Sunday", "Monday", "Tuesday", "Wednesday",

"Thursday", "Friday", "Saturday"};

struct tm tm;

tm.tm_mon = 2 - 1;

tm.tm_mday = 29;

tm.tm_year = 2000 - 1900;

tm.tm_hour = tm.tm_min = tm.tm_sec = 0;

tm.tm_isdst = -1;

if(mktime(&tm) != -1)

printf("%sn", wday[tm.tm_wday]);

When using mktime like this, it's usually important to set tm_isdst to -1, as shown (especially if tm_hour is 0), otherwise a daylight saving time correction could push the time past midnight into another day. # Use Zeller's congruence, which says that if

J is the number of the century [i.e. the year / 100],

K the year within the century [i.e. the year % 100],

m the month,

q the day of the month,

h the day of the week [where 1 is Sunday];

**Answer #2**

and if January and February are taken as months 13 and 14 of the previous year [affecting both J and K]; then h for the Gregorian calendar is the remainder when the sum

q + 26(m + 1) / 10 + K + K/4 + J/4 - 2J

is divided by 7, and where all intermediate remainders are discarded. The translation into C is straightforward:

h = (q + 26 * (m + 1) / 10 + K + K/4 + J/4 + 5*J) % 7;

(where we use +5*J instead of -2*J to make sure that both operands of the modulus operator % are positive; this bias totalling 7*J will obviously not change the final value of h, modulo 7).

# Use this elegant code by Tomohiko Sakamoto:

int dayofweek(int y, int m, int d) /* 0 = Sunday */

{

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

y -= m < 3;

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

}

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