method

<=>

v1_9_3_392 - Show latest stable - 1 note - Class: Date

<=>(p1) public

Compares the two dates and returns -1, zero, 1 or nil. The other should be a date object or a numeric value as an astronomical Julian day number.

For example:

Date.new(2001,2,3) <=> Date.new(2001,2,4) #=> -1
Date.new(2001,2,3) <=> Date.new(2001,2,3) #=> 0
Date.new(2001,2,3) <=> Date.new(2001,2,2) #=> 1
Date.new(2001,2,3) <=> Object.new         #=> nil
Date.new(2001,2,3) <=> Rational(4903887,2)#=> 0

See also Comparable.

Show source

static VALUE
d_lite_cmp(VALUE self, VALUE other)
{
    if (!k_date_p(other))
        return cmp_gen(self, other);

    {
        get_d2(self, other);

        if (!(simple_dat_p(adat) && simple_dat_p(bdat) &&
              m_gregorian_p(adat) == m_gregorian_p(bdat)))
            return cmp_dd(self, other);

        if (have_jd_p(adat) &&
            have_jd_p(bdat)) {
            VALUE a_nth, b_nth;
            int a_jd, b_jd;

            a_nth = m_nth(adat);
            b_nth = m_nth(bdat);
            if (f_eqeq_p(a_nth, b_nth)) {
                a_jd = m_jd(adat);
                b_jd = m_jd(bdat);
                if (a_jd == b_jd) {
                    return INT2FIX(0);
                }
                else if (a_jd < b_jd) {
                    return INT2FIX(-1);
                }
                else {
                    return INT2FIX(1);
                }
            }
            else if (a_nth < b_nth) {
                return INT2FIX(-1);
            }
            else {
                return INT2FIX(1);
            }
        }
        else {
#ifndef USE_PACK
            VALUE a_nth, b_nth;
            int a_year, b_year,
                a_mon, b_mon,
                a_mday, b_mday;
#else
            VALUE a_nth, b_nth;
            int a_year, b_year,
                a_pd, b_pd;
#endif

            a_nth = m_nth(adat);
            b_nth = m_nth(bdat);
            if (f_eqeq_p(a_nth, b_nth)) {
                a_year = m_year(adat);
                b_year = m_year(bdat);
                if (a_year == b_year) {
#ifndef USE_PACK
                    a_mon = m_mon(adat);
                    b_mon = m_mon(bdat);
                    if (a_mon == b_mon) {
                        a_mday = m_mday(adat);
                        b_mday = m_mday(bdat);
                        if (a_mday == b_mday) {
                            return INT2FIX(0);
                        }
                        else if (a_mday < b_mday) {
                            return INT2FIX(-1);
                        }
                        else {
                            return INT2FIX(1);
                        }
                    }
                    else if (a_mon < b_mon) {
                        return INT2FIX(-1);
                    }
                    else {
                        return INT2FIX(1);
                    }
#else
                    a_pd = m_pc(adat);
                    b_pd = m_pc(bdat);
                    if (a_pd == b_pd) {
                        return INT2FIX(0);
                    }
                    else if (a_pd < b_pd) {
                        return INT2FIX(-1);
                    }
                    else {
                        return INT2FIX(1);
                    }
#endif
                }
                else if (a_year < b_year) {
                    return INT2FIX(-1);
                }
                else {
                    return INT2FIX(1);
                }
            }
            else if (f_lt_p(a_nth, b_nth)) {
                return INT2FIX(-1);
            }
            else {
                return INT2FIX(1);
            }
        }
    }
}

March 12, 2009

1 thank

Comparing Date with Numeric in mixed sort

While:

Date#<=>(other)

can accept a Numeric object as other, the reverse is not true:

Numeric#<=>(other)

cannot accept a Date object as other.

So if you are sorting a list containing a mix of dates and numbers, you can get different results depending on the starting order!

a = Date.parse("2008-01-01")
b = Date.parse("2009-10-22")
c = Date.parse("2005-01-04")
d = 0

[a,b,c,d].sort #=> [0, Tue, 04 Jan 2005, Tue, 01 Jan 2008, Thu, 22 Oct 2009]

[b,c,d,a].sort #=> ArgumentError: comparison of Fixnum with Date failed

<=>

Related methods

Comparing Date with Numeric in mixed sort