我正在努力正確循環(huán)我編寫的代碼以將整數(shù)轉(zhuǎn)換為羅馬數(shù)字。我嘗試實(shí)現(xiàn)一個(gè) do while 循環(huán)來運(yùn)行代碼，從“請(qǐng)輸入一個(gè)整數(shù)”開始，在我的 switch 語句之后結(jié)束，while 部分是：while（大小寫 “y” ||“Y” == true ） 任何幫助將不勝感激。我已經(jīng)搜索了幾個(gè)小時(shí)的堆棧溢出的先前帖子，并且無法找到任何有用的東西。公共類項(xiàng)目8 {/** * Constructor for objects of class Project4 */public static void main(String[] args) { System.out.println("Welcome to my integer Roman numeral conversion program"); System.out.println("------------------------------------------------------"); System.out.println(" "); Scanner in = new Scanner (System.in); System.out.print("Enter an integer in the range 1-3999 (both inclusive): "); int input = in.nextInt(); if (input < 0 || input > 3999){ System.out.println("Sorry, this number is outside the range."); System.out.println("Do you want to try again? Press Y for yes and N for no: "); String userInput = in.next(); switch (userInput) { case "N": case "n": System.exit(0); break; case "Y": case "y": break; } } else if (input > 0 && input < 3999); { System.out.println(Conversion.Convert(input)); } }}
1 回答
蝴蝶不菲
TA貢獻(xiàn)1810條經(jīng)驗(yàn) 獲得超4個(gè)贊
1）您的條件是多余的。您可以使用簡單，因?yàn)檩斎胫荒茉谠摲秶鷥?nèi)進(jìn)行。 僅當(dāng)您要檢查兩個(gè)或多個(gè)范圍時(shí)才有意義，例如if - else ifif - elseelse if
if(input > 0 && input < 3999){
...
}
else if (input > 4000 && input < 8000){
...
}
else {
...
}
2） 您不需要開關(guān)塊，而是在 while 條件下使用用戶輸入，因?yàn)槟Ｍ谟脩糨斎霝?Y/y 時(shí)繼續(xù)循環(huán)，即while(userChoice.equals("Y"))
3） 使用循環(huán)，因?yàn)槟Ｍ麘?yīng)用程序至少按時(shí)運(yùn)行do - while
public static void main(String[] args) {
System.out.println("Welcome to my integer Roman numeral conversion program");
System.out.println("------------------------------------------------------");
System.out.println(" ");
Scanner in = new Scanner (System.in);
String choice;
do{
System.out.print("Enter an integer in the range 1-3999 (both inclusive): ");
int input = in.nextInt();
if(input > 0 && input < 3999){
System.out.println(Conversion.Convert(input));
}
else{
System.out.println("Sorry, this number is outside the range.");
}
System.out.println("Do you want to try again? Press Y for yes and N for no: ");
choice = in.next();
}while(choice.equals("Y") || choice.equals("y"));
}
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