Question

Here is a response关于计算Java年龄的question。

/**
 * This Method is unit tested properly for very different cases , 
 * taking care of Leap Year days difference in a year, 
 * and date cases month and Year boundary cases (12/31/1980, 01/01/1980 etc)
**/

public static int getAge(Date dateOfBirth) {

    Calendar today = Calendar.getInstance();
    Calendar birthDate = Calendar.getInstance();

    int age = 0;

    birthDate.setTime(dateOfBirth);
    if (birthDate.after(today)) {
        throw new IllegalArgumentException("Can't be born in the future");
    }

    age = today.get(Calendar.YEAR) - birthDate.get(Calendar.YEAR);

    // If birth date is greater than todays date (after 2 days adjustment of leap year) then decrement age one year   
    if ( (birthDate.get(Calendar.DAY_OF_YEAR) - today.get(Calendar.DAY_OF_YEAR) > 3) ||
            (birthDate.get(Calendar.MONTH) > today.get(Calendar.MONTH ))){
        age--;

     // If birth date and todays date are of same month and birth day of month is greater than todays day of month then decrement age
    }else if ((birthDate.get(Calendar.MONTH) == today.get(Calendar.MONTH )) &&
              (birthDate.get(Calendar.DAY_OF_MONTH) > today.get(Calendar.DAY_OF_MONTH ))){
        age--;
    }

    return age;
}

这段代码工作正常，但为什么会有这样的比较： (birthDate.get(Calendar.DAY_OF_YEAR) - today.get(Calendar.DAY_OF_YEAR) > 3)

到目前为止，我已经创建了一个巨大的电子表格，记录了一年中的所有差异，试图查看它可能涵盖的案例，但我没有看到其他比较没有涵盖的任何内容。任何人都能解释包含这种比较背后的目的吗？它在某种程度上更有效吗？

Answer 1

以下来自ThreetenBP（Java-8的后端）的代码示例支持不需要每年检查日期的声明：

@Override 
public long until(Temporal endExclusive, TemporalUnit unit) { 
LocalDate end = LocalDate.from(endExclusive); 
    if (unit instanceof ChronoUnit) { 
         switch ((ChronoUnit) unit) { 
             case DAYS: return daysUntil(end); 
             case WEEKS: return daysUntil(end) / 7; 
             case MONTHS: return monthsUntil(end); 
             case YEARS: return monthsUntil(end) / 12; 
             case DECADES: return monthsUntil(end) / 120; 
             case CENTURIES: return monthsUntil(end) / 1200; 
             case MILLENNIA: return monthsUntil(end) / 12000; 
             case ERAS: return end.getLong(ERA) - getLong(ERA); 
         } 
         throw new UnsupportedTemporalTypeException("Unsupported unit: " + unit); 
     } 
     return unit.between(this, end); 
} 

[...]     

private long monthsUntil(LocalDate end) { 
   long packed1 = getProlepticMonth() * 32L + getDayOfMonth();  // no overflow 
   long packed2 = end.getProlepticMonth() * 32L + end.getDayOfMonth();  // no overflow 
   return (packed2 - packed1) / 32; 
}

行case YEARS: return monthsUntil(end) / 12;（表达式birthday.until(today, YEARS)和YEARS.between(birthday, today)是等效的 - 一个委托给其他人）利用与OP引用的简化代码相同的算法，并且不引用任何一年中某一天的检查：

age = today.get(Calendar.YEAR) - birthDate.get(Calendar.YEAR);

if (birthDate.get(Calendar.MONTH) > today.get(Calendar.MONTH)) {
    age--;
}else if ((birthDate.get(Calendar.MONTH) == today.get(Calendar.MONTH )) &&
          (birthDate.get(Calendar.DAY_OF_MONTH) > today.get(Calendar.DAY_OF_MONTH ))){
    age--;
}

问题出现了：为什么要进行年度检查？

a）海报最初认真对待了一年一度的想法，然后忘了在以后的版本中清理

b）海报希望“改善”表现

以下Java-8代码演示了基于日期算法的问题，如果认真考虑并且作为完整版本（库的选择与此无关，只有算法很重要）：

LocalDate birthday = LocalDate.of(2001, 3, 6);
LocalDate today = LocalDate.of(2016, 3, 5); // leap year

int age = today.getYear() - birthday.getYear();
if (birthday.getDayOfYear() > today.getDayOfYear()) {
    age--;
}
System.out.println("age based on day-of-year: " + age); // 15 (wrong)
System.out.println("age based on month and day-of-month: " 
  + ChronoUnit.YEARS.between(birthday, today)); // 14 (correct)

<强>结论：

您引用的提议的年度子句只是噪音，因为算法的其余部分与Java-8的作用相对应。也许年度检查来自一些早期的基于日期的建议代码版本，并且还没有被清理过。

为了回答你的上一个问题：这样的不必要的检查不是很好。在性能方面有效（尽管我们在这里谈论微观优化）。

为什么这个Java代码有这个年龄验证日期比较？

1 个答案: