使用開關語句的效率，Java

我最近在一個程序中詢問用戶年齡：年，月，日。使用開關語句的效率，Java

接收輸入之後，它必須計算和打印

a）年齡在秒（totalAgeInSecs）和

B）秒的量可活。 b將基於平均壽命秒數(avgLifeSpan = 250000000000l. So secondsLeft = avgLifeSpan - totalAgeInSecs)。

無論如何，爲了簡單起見，我能夠使用（switch）語句來工作，而不必編寫一堆if/else語句，但是我覺得在這樣做的時候，我最終編寫了重複行，我希望能夠不必重複計算或打印報表。

我知道有些類和數組可以與循環結合使用，但爲了簡單和邏輯理解，我沒有使用它們來理解這個項目的無鞍骨骼和邏輯在「英語」中。哈哈。

無論如何，檢查下面的代碼，讓我知道你如何簡化重複線或更好的方法來解決這個問題的想法。謝謝。

import java.util.*; 

public class AgeInSeconds { 

    static Scanner kbd = new Scanner(System.in); 
    public static void main(String[] args) { 

     int totalNumDays, daysInMonth, daysToHours; 
     int yrsToDays,minsInHr, secsInMin; 

     long timeRemaining, avgLifeSecs; 

     System.out.println("Enter your age in years months and days: "); 

     System.out.print("Years: "); 
     int years = kbd.nextInt(); 

     System.out.print("Months: "); 
     int months = kbd.nextInt(); 

     System.out.print("Days: "); 
     int days = kbd.nextInt(); 

     yrsToDays = years * 365; 
     avgLifeSecs = 2500000000l; 

     switch (months){ 
     case 1: 
      daysInMonth = 31; 
      totalNumDays = yrsToDays + daysInMonth + days; 
      daysToHours = totalNumDays * 24; 
      minsInHr = daysToHours * 60; 
      secsInMin = minsInHr * 60; 

      timeRemaining = avgLifeSecs - secsInMin; 

      System.out.printf("You have been alive for %,d seconds.\n", secsInMin); 
      System.out.printf("The average human life is %,d seconds.\n", avgLifeSecs); 
      System.out.printf("You have %,d seconds.\n", timeRemaining);   
      break; 
     case 2: 
      daysInMonth = 59; 
      totalNumDays = yrsToDays + daysInMonth + days; 
      daysToHours = totalNumDays * 24; 
      minsInHr = daysToHours * 60; 
      secsInMin = minsInHr * 60; 

      timeRemaining = avgLifeSecs - secsInMin;  

      System.out.printf("You have been alive for %,d seconds.\n", secsInMin); 
      System.out.printf("The average human life is %,d seconds.\n", avgLifeSecs); 
      System.out.printf("You have %,d seconds.\n", timeRemaining); 
      break;  
     case 3: 
      daysInMonth = 90; 
      totalNumDays = yrsToDays + daysInMonth + days; 
      daysToHours = totalNumDays * 24; 
      minsInHr = daysToHours * 60; 
      secsInMin = minsInHr * 60; 

      timeRemaining = avgLifeSecs - secsInMin; 

      System.out.printf("You have been alive for %,d seconds.\n", secsInMin); 
      System.out.printf("The average human life is %,d seconds.\n", avgLifeSecs); 
      System.out.printf("You have %,d seconds.\n", timeRemaining); 
      break;  
     case 4: 
      daysInMonth = 120; 
      totalNumDays = yrsToDays + daysInMonth + days; 
      daysToHours = totalNumDays * 24; 
      minsInHr = daysToHours * 60; 
      secsInMin = minsInHr * 60; 

      timeRemaining = avgLifeSecs - secsInMin; 

      System.out.printf("You have been alive for %,d seconds.\n", secsInMin); 
      System.out.printf("The average human life is %,d seconds.\n", avgLifeSecs); 
      System.out.printf("You have %,d seconds.\n", timeRemaining); 
      break;  
     case 5: 
      daysInMonth = 151; 
      totalNumDays = yrsToDays + daysInMonth + days; 
      daysToHours = totalNumDays * 24; 
      minsInHr = daysToHours * 60; 
      secsInMin = minsInHr * 60; 

      timeRemaining = avgLifeSecs - secsInMin; 

      System.out.printf("You have been alive for %,d seconds.\n", secsInMin); 
      System.out.printf("The average human life is %,d seconds.\n", avgLifeSecs); 
      System.out.printf("You have %,d seconds.\n", timeRemaining); 
      break;  
     case 6: 
      daysInMonth = 181; 
      totalNumDays = yrsToDays + daysInMonth + days; 
      daysToHours = totalNumDays * 24; 
      minsInHr = daysToHours * 60; 
      secsInMin = minsInHr * 60; 

      timeRemaining = avgLifeSecs - secsInMin; 
      System.out.printf("You have been alive for %,d seconds.\n", secsInMin); 
      System.out.printf("The average human life is %,d seconds.\n", avgLifeSecs); 
      System.out.printf("You have %,d seconds.\n", timeRemaining);   
      break;  
     case 7: 
      daysInMonth = 212; 
      totalNumDays = yrsToDays + daysInMonth + days; 
      daysToHours = totalNumDays * 24; 
      minsInHr = daysToHours * 60; 
      secsInMin = minsInHr * 60; 

      timeRemaining = avgLifeSecs - secsInMin; 
      System.out.printf("You have been alive for %,d seconds.\n", secsInMin); 
      System.out.printf("The average human life is %,d seconds.\n", avgLifeSecs); 
      System.out.printf("You have %,d seconds.\n", timeRemaining);  
      break;  
     case 8: 
      daysInMonth = 243; 
      totalNumDays = yrsToDays + daysInMonth + days; 
      daysToHours = totalNumDays * 24; 
      minsInHr = daysToHours * 60; 
      secsInMin = minsInHr * 60; 

      timeRemaining = avgLifeSecs - secsInMin; 
      System.out.printf("You have been alive for %,d seconds.\n", secsInMin); 
      System.out.printf("The average human life is %,d seconds.\n", avgLifeSecs); 
      System.out.printf("You have %,d seconds.\n", timeRemaining); 

      break; 

     case 9: 
      daysInMonth = 273; 
      totalNumDays = yrsToDays + daysInMonth + days; 
      daysToHours = totalNumDays * 24; 
      minsInHr = daysToHours * 60; 
      secsInMin = minsInHr * 60; 

      timeRemaining = avgLifeSecs - secsInMin; 
      System.out.printf("You have been alive for %,d seconds.\n", secsInMin); 
      System.out.printf("The average human life is %,d seconds.\n", avgLifeSecs); 
      System.out.printf("You have %,d seconds.\n", timeRemaining);   
      break;  
     case 10: 
      daysInMonth = 304; 
      totalNumDays = yrsToDays + daysInMonth + days; 
      daysToHours = totalNumDays * 24; 
      minsInHr = daysToHours * 60; 
      secsInMin = minsInHr * 60; 

      timeRemaining = avgLifeSecs - secsInMin; 
      System.out.printf("You have been alive for %,d seconds.\n", secsInMin); 
      System.out.printf("The average human life is %,d seconds.\n", avgLifeSecs); 
      System.out.printf("You have %,d seconds.\n", timeRemaining);   
      break;   
     case 11: 
      daysInMonth = 334; 
      totalNumDays = yrsToDays + daysInMonth + days; 
      daysToHours = totalNumDays * 24; 
      minsInHr = daysToHours * 60; 
      secsInMin = minsInHr * 60; 
      timeRemaining = avgLifeSecs - secsInMin; 

      System.out.printf("You have been alive for %,d seconds.\n", secsInMin); 
      System.out.printf("The average human life is %,d seconds.\n", avgLifeSecs); 
      System.out.printf("You have %,d seconds.\n", timeRemaining); 
      break;  
     case 12: 
      daysInMonth = 365; 
      totalNumDays = yrsToDays + daysInMonth + days; 
      daysToHours = totalNumDays * 24; 
      minsInHr = daysToHours * 60; 
      secsInMin = minsInHr * 60; 

      timeRemaining = avgLifeSecs - secsInMin; 

      System.out.printf("You have been alive for %,d seconds.\n", secsInMin); 
      System.out.printf("The average human life is %,d seconds.\n", avgLifeSecs); 
      System.out.printf("You have %,d seconds.\n", timeRemaining); 

     default: 

     } 

     kbd.close();  

    } 


}

這是當輸出：年= 24個月= 5天= 8

Enter your age in years months and days: 
Years: 24 
Months: 5 
Days: 8 
You have been alive for 770,601,600 seconds. 
The average human life is 2,500,000,000 seconds. 
You have 1,729,398,400 seconds.

來源

2015-09-28 mlopman

這是一個應該調用方法並將整個代碼放入其中的情況。 – SomeJavaGuy

閏年呢？順便說一下，從統計的角度來看，程序的結果是不正確的。如果我100歲，該怎麼辦？我有負秒數？那沒有意義。 –

這確實重複了很多代碼。另外還有一個叫閏年的東西。 – laune

你的整數daysInMonth將轉移過去switch語句。因此，您可以在 switch語句後利用重複代碼。

作爲一個經驗法則：當你有重複的代碼把它放在它自己的方法中，或者合併代碼，所以你只需要在一個地方調用它。

import java.util.*; public class AgeInSeconds { static Scanner kbd = new Scanner(System.in); public static void main(String[] args) { int totalNumDays, daysInMonth, daysToHours; int yrsToDays,minsInHr, secsInMin; long timeRemaining, avgLifeSecs; System.out.println("Enter your age in years months and days: "); System.out.print("Years: "); int years = kbd.nextInt(); System.out.print("Months: "); int months = kbd.nextInt(); System.out.print("Days: "); int days = kbd.nextInt(); yrsToDays = years * 365; avgLifeSecs = 2500000000l; switch (months){ case 1: daysInMonth = 31; break; case 2: daysInMonth = 59; break; case 3: daysInMonth = 90; break; case 4: daysInMonth = 120; break; case 5: daysInMonth = 151; break; case 6: daysInMonth = 181; break; case 7: daysInMonth = 212; break; case 8: daysInMonth = 243; break; case 9: daysInMonth = 273; break; case 10: daysInMonth = 304; break; case 11: daysInMonth = 334; break; case 12: daysInMonth = 365; break; default: daysInMonth = 0; } totalNumDays = yrsToDays + daysInMonth + days; daysToHours = totalNumDays * 24; minsInHr = daysToHours * 60; secsInMin = minsInHr * 60; timeRemaining = avgLifeSecs - secsInMin; System.out.printf("You have been alive for %,d seconds.\n", secsInMin); System.out.printf("The average human life is %,d seconds.\n", avgLifeSecs); System.out.printf("You have %,d seconds.\n", timeRemaining); kbd.close(); } }

上面是「快速和骯髒」的解決方案，以你的代碼，所以你可以看到你是如何傳輸變量過去的switch語句。

更好的做法是使用模運算符來檢查它是奇數還是偶數，如果它不是第二個月，則給出正確的值。

import java.util.*; public class AgeInSeconds { static Scanner kbd = new Scanner(System.in); public static void main(String[] args) { int totalNumDays, daysInMonth, daysToHours; int yrsToDays,minsInHr, secsInMin; long timeRemaining, avgLifeSecs; System.out.println("Enter your age in years months and days: "); System.out.print("Years: "); int years = kbd.nextInt(); System.out.print("Months: "); int months = kbd.nextInt(); System.out.print("Days: "); int days = kbd.nextInt(); yrsToDays = years * 365; avgLifeSecs = 2500000000l; /** predifine with 0 so we always have a value **/ daysInMonth = 0; /** Our months here. Please consider using calendar **/ int[] legaldays = {31,28,31,30,31,30,31,31,30,31,30,31}; /** Looping through all months **/ for(i=0;i<legaldays.length;i++) { /** check if we didn't pass our max limit **/ if(i+1 > daysInMonth) { break; } /** add the days to our tally **/ daysInMonth += legaldays[i]; } totalNumDays = yrsToDays + daysInMonth + days; daysToHours = totalNumDays * 24; minsInHr = daysToHours * 60; secsInMin = minsInHr * 60; timeRemaining = avgLifeSecs - secsInMin; System.out.printf("You have been alive for %,d seconds.\n", secsInMin); System.out.printf("The average human life is %,d seconds.\n", avgLifeSecs); System.out.printf("You have %,d seconds.\n", timeRemaining); kbd.close(); } }

查看評論，瞭解我是如何改進它的。通過循環，您可以擺脫頭痛的難題，即必須對月份的值進行硬編碼，併爲您提供了一定的靈活性。爲了更可靠的一天數而不是硬編碼，我建議你看看Number of days in a month of a particular year，這樣你就可以靈活地用閏年等等。

儘可能嘗試不對無形或動態數據值進行硬編碼，但要儘可能準確地扣除它們。日期很難保持秩序。

來源

2015-09-28 08:01:16 Tschallacka

曾經使用過一個數組來存儲基於1..12（或0..11）的值嗎？ – laune

我認爲，但他無效輸入的情況。再加上我仍在編輯一些最佳做法。 – Tschallacka

還要考慮OP提出的方法是無稽之談。如果我今天是x年和5個月，那麼第二個月肯定沒有28天。 – laune

要正確計算用戶存活的天數，您應該首先根據提供的數據和今天的日期計算他的出生日期。例如：

用戶1個月大，當前日期爲2015年9月28日，因此用戶出生於2015年8月28日，他已經31天。
用戶爲1個月大，當前日期爲2015年3月2日，因此用戶出生於2015年2月2日，他的年齡爲28歲。

之後，您可以計算以秒爲單位的差異。 Java API中有準備好的類和方法來完成這些步驟。最簡單的是使用Java 8時間API：

import java.time.LocalDateTime; 
import java.time.Period; 
import java.time.temporal.ChronoUnit; 
import java.util.Scanner; 

public class AgeInSeconds { 

    public static void main(String[] args) { 
     try (Scanner kbd = new Scanner(System.in)) { 

      System.out.println("Enter your age in years months and days: "); 

      System.out.print("Years: "); 
      int years = kbd.nextInt(); 

      System.out.print("Months: "); 
      int months = kbd.nextInt(); 

      System.out.print("Days: "); 
      int days = kbd.nextInt(); 

      Period period = Period.of(years, months, days); 
      LocalDateTime now = LocalDateTime.now(); 
      LocalDateTime birthDate = now.minus(period); 
      long seconds = birthDate.until(now, ChronoUnit.SECONDS); 
      long avgLifeSecs = 2500000000l; 
      long timeRemaining = avgLifeSecs - seconds; 

      System.out.printf("You have been alive for %,d seconds.\n", seconds); 
      System.out.printf("The average human life is %,d seconds.\n", avgLifeSecs); 
      System.out.printf("You have %,d seconds.\n", timeRemaining); 
     } 
    } 
}

我沒有解決這裏的統計問題。要計算估計剩餘壽命（假設我是一般人），您應該平均比我年長的人壽命。

來源

2015-09-28 08:14:33

我想你可能會超過OP的頭。我認爲他在第一章或第二章中學習如何編碼;-)但答案很好，+1 – Tschallacka

使用開關語句的效率，Java

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