week_of_year.c

/*
 * (C)2012 Michael Duane Rice All rights reserved.
 *
 * Redistribution and use in source and binary forms, with or without
 * modification, are permitted provided that the following conditions are
 * met:
 *
 * Redistributions of source code must retain the above copyright notice, this
 * list of conditions and the following disclaimer. Redistributions in binary
 * form must reproduce the above copyright notice, this list of conditions
 * and the following disclaimer in the documentation and/or other materials
 * provided with the distribution. Neither the name of the copyright holders
 * nor the names of contributors may be used to endorse or promote products
 * derived from this software without specific prior written permission.
 *
 * THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS"
 * AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
 * ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT OWNER OR CONTRIBUTORS BE
 * LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR
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 * SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS
 * INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN
 * CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE)
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 */

/* $Id$ */

/*
    Return the week of year, where 'base' represents the first day of the week.
    In the USA, the week is generally considered to start on Sunday (base = 0),
    while in Europe it is generally considered to be Monday (base = 1).

    Return value ranges from 0 to 52.
*/

#include <time.h>

uint8_t
week_of_year(const struct tm * timestruct, uint8_t base)
{
    int             first, n;

    /* find the first base day of the year (start of week 1) */
    first = 7 + timestruct->tm_yday - timestruct->tm_wday + base;
    first %= 7;

    /* find days since that first base day*/
    n = timestruct->tm_yday - first;

    /* if negative, we are in week 0 */
    if (n < 0)
        return 0;

    return n / 7 + 1;
}