/* jd.c (conversions between calendar date and Julian day number) */ /*********************************************************************** * This code is part of GLPK (GNU Linear Programming Kit). * * Copyright (C) 2000-2013 Andrew Makhorin, Department for Applied * Informatics, Moscow Aviation Institute, Moscow, Russia. All rights * reserved. E-mail: . * * GLPK is free software: you can redistribute it and/or modify it * under the terms of the GNU General Public License as published by * the Free Software Foundation, either version 3 of the License, or * (at your option) any later version. * * GLPK is distributed in the hope that it will be useful, but WITHOUT * ANY WARRANTY; without even the implied warranty of MERCHANTABILITY * or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public * License for more details. * * You should have received a copy of the GNU General Public License * along with GLPK. If not, see . ***********************************************************************/ #include #include "jd.h" /*********************************************************************** * NAME * * jday - convert calendar date to Julian day number * * SYNOPSIS * * #include "jd.h" * int jday(int d, int m, int y); * * DESCRIPTION * * The routine jday converts a calendar date, Gregorian calendar, to * corresponding Julian day number j. * * From the given day d, month m, and year y, the Julian day number j * is computed without using tables. * * The routine is valid for 1 <= y <= 4000. * * RETURNS * * The routine jday returns the Julian day number, or negative value if * the specified date is incorrect. * * REFERENCES * * R. G. Tantzen, Algorithm 199: conversions between calendar date and * Julian day number, Communications of the ACM, vol. 6, no. 8, p. 444, * Aug. 1963. */ int jday(int d, int m, int y) { int c, ya, j, dd; if (!(1 <= d && d <= 31 && 1 <= m && m <= 12 && 1 <= y && y <= 4000)) return -1; if (m >= 3) m -= 3; else m += 9, y--; c = y / 100; ya = y - 100 * c; j = (146097 * c) / 4 + (1461 * ya) / 4 + (153 * m + 2) / 5 + d + 1721119; jdate(j, &dd, NULL, NULL); if (d != dd) return -1; return j; } /*********************************************************************** * NAME * * jdate - convert Julian day number to calendar date * * SYNOPSIS * * #include "jd.h" * int jdate(int j, int *d, int *m, int *y); * * DESCRIPTION * * The routine jdate converts a Julian day number j to corresponding * calendar date, Gregorian calendar. * * The day d, month m, and year y are computed without using tables and * stored in corresponding locations. * * The routine is valid for 1721426 <= j <= 3182395. * * RETURNS * * If the conversion is successful, the routine returns zero, otherwise * non-zero. * * REFERENCES * * R. G. Tantzen, Algorithm 199: conversions between calendar date and * Julian day number, Communications of the ACM, vol. 6, no. 8, p. 444, * Aug. 1963. */ int jdate(int j, int *d_, int *m_, int *y_) { int d, m, y; if (!(1721426 <= j && j <= 3182395)) return 1; j -= 1721119; y = (4 * j - 1) / 146097; j = (4 * j - 1) % 146097; d = j / 4; j = (4 * d + 3) / 1461; d = (4 * d + 3) % 1461; d = (d + 4) / 4; m = (5 * d - 3) / 153; d = (5 * d - 3) % 153; d = (d + 5) / 5; y = 100 * y + j; if (m <= 9) m += 3; else m -= 9, y++; if (d_ != NULL) *d_ = d; if (m_ != NULL) *m_ = m; if (y_ != NULL) *y_ = y; return 0; } #ifdef GLP_TEST #include #include #include int main(void) { int jbeg, jend, j, d, m, y; jbeg = jday(1, 1, 1); jend = jday(31, 12, 4000); for (j = jbeg; j <= jend; j++) { assert(jdate(j, &d, &m, &y) == 0); assert(jday(d, m, y) == j); } printf("Routines jday and jdate work correctly.\n"); return 0; } #endif /* eof */