我必須捕獲3個異常并循環，直到輸入有效的輸入?，F在，我的程序在輸入有效輸入時結束，但只是連續循環無效輸入。它永遠不會捕獲異常。如果我取出循環并輸入一個“m”，它會捕獲算術異常，但會打印輸入不匹配“僅數字”的錯誤消息// Get input from user and perform integer divsion.do{ try{ System.out.print("Please enter a positive value for the numerator: "); numerator = keyboard.nextInt(); System.out.print("Please enter a positive value for the denominator: "); denominator = keyboard.nextInt(); } catch(InputMismatchException e){ System.out.println("Enter only numbers."); } catch(NegativeValueException e){ System.out.println("no negative values."); } catch(ArithmeticException e){ System.out.println("Division by zero."); } if (numerator >=0 && denominator >= 0){ quotient = numerator / denominator; }else if(numerator<0 && denominator <0){ throw new NegativeValueException(); }else invalidInput = true; System.out.println(); System.out.print("The result of integer division is: "); System.out.println(quotient); System.out.println();}while (invalidInput); 我只需要捕獲異常并讓它打印出異常和循環的相應錯誤消息，直到輸入有效輸入。如果沒有循環，它將捕獲負值，但不會打印錯誤并以構建失敗結束。任何幫助將不勝感激！
3 回答
料青山看我應如是
TA貢獻1772條經驗 獲得超8個贊
既然你已經表現出了努力，但你似乎也是Java的新手，我正在嘗試糾正你的代碼。通常，不建議用于這種低水平的校正。
我們需要在捕獲它之前拋出一個異常 - 這就是你周圍的情況 - 。NegativeValueException
我假設變量 &類型 - ， &是聲明和定義的。NegativeValueExceptioninvalidInputkeyboard
我已經提供了示例代碼，對您提出的代碼進行了最小的更改，盡管您提出的問題可以通過更好的代碼來解決。
do{
invalidInput=false;
try{
System.out.print("Please enter a positive value for the numerator: ");
int numerator = keyboard.nextInt();
System.out.print("Please enter a positive value for the denominator: ");
int denominator = keyboard.nextInt();
if(numerator<0 && denominator <0){
throw new NegativeValueException();
}else if (numerator >=0 && denominator >= 0){
int quotient = numerator / denominator;
System.out.println();
System.out.print("The result of integer division is: ");
System.out.println(quotient);
System.out.println();
}
} catch(InputMismatchException e){
System.out.println("Enter only numbers.");
invalidInput = true;
}catch(NegativeValueException e){
System.out.println("no negative values.");
invalidInput = true;
}catch(ArithmeticException e){
System.out.println("Division by zero.");
invalidInput = true;
}
}while (invalidInput);
慕尼黑8549860
TA貢獻1818條經驗 獲得超11個贊
Scanner s = new Scanner(System.in);
do {
try {
System.out.print("Please enter a positive value for the numerator: ");
int numerator = s.nextInt();
System.out.print("Please enter a positive value for the denominator: ");
int denominator = s.nextInt();
int quotient = numerator / denominator;
} catch (InputMismatchException e) {
System.out.println("InputMismatchException " + e);
System.exit(0);
} catch (ArithmeticException e) {
System.out.println("ArithmeticException " + e);
System.exit(0);
} catch (Exception e) {
System.out.println("NegativeValueException " + e);
System.exit(0);
}
} while (true);
HUX布斯
TA貢獻1876條經驗 獲得超6個贊
實際上，問題是你在外面做數學運算，比如說如果它拋出一個程序通過這些檢查：try-blockexception
if (numerator >=0 && denominator >= 0) // false as there was exception
else if(numerator<0 && denominator <0) // false as there was exception
else
invalidInput = true; //gets executed
執行每個異常并因此中斷 .將數學移入內部invalidInput = truedo-while looptry-block
試試這個代碼：
do
{
try
{
System.out.print("Please enter a positive value for the numerator: ");
numerator = keyboard.nextInt();
System.out.print("Please enter a positive value for the denominator: ");
denominator = keyboard.nextInt();
if (numerator >=0 && denominator >= 0)
{
quotient = numerator / denominator;
}
else if(numerator<0 && denominator <0)
{
throw new NegativeValueException();
}
else
invalidInput = true;
System.out.println();
System.out.print("The result of integer division is: ");
System.out.println(quotient);
System.out.println();
}
catch(InputMismatchException e)
{
System.out.println("Enter only numbers.");
}
catch(NegativeValueException e)
{
System.out.println("no negative values.");
}
catch(ArithmeticException e)
{
System.out.println("Division by zero.");
}
}while (invalidInput);
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