为什么java.util.Date将Year表示为“ year-1900”？

Z时代
2024-01-10
分类：问答

在java.util.Date：

 * In all methods of class <code>Date</code> that accept or return
 * year, month, date, hours, minutes, and seconds values, the
 * following representations are used:
 * <ul>
 * <li>A year <i>y</i> is represented by the integer
 *     <i>y</i><code>-1900</code>.

当然，在Java

1.1中，getYear()不赞成使用方法和类似方法，而推荐使用java.util.Calendar，但仍然具有以下怪异的弃用说明：

 int    getYear() 
    Deprecated. As of JDK version 1.1, replaced by Calendar.get(Calendar.YEAR) - 1900.
 setYear(int year) 
      Deprecated. As of JDK version 1.1, replaced by Calendar.set(Calendar.YEAR, year + 1900).

当然，Month是0基于-的，但我们都知道（尽管您认为他们已经从中消除了这个问题Calendar-他们没有）：

 * <li>A month is represented by an integer from 0 to 11; 0 is January,
 *     1 is February, and so forth; thus 11 is December.

最初的创建者java.util.Date希望通过从中减去1900来存储“年”的数据而获得什么呢？特别是如果它基本上存储了很长时间。

因此：

private transient long fastTime;
@Deprecated
public int getYear() {
    return normalize().getYear() - 1900;
}
@Deprecated
public void setYear(int year) {
    getCalendarDate().setNormalizedYear(year + 1900);
}
private final BaseCalendar.Date getCalendarDate() {
    if (cdate == null) {
        BaseCalendar cal = getCalendarSystem(fastTime);
    ....