如何在 Java 中处理异常而不必回到 try 块的开头?

Question

我已经在我的程序中实现了异常处理，但现在我遇到的问题是，当异常发生并在 catch 块中处理时，它不是从发生的地方继续，而是回到程序的开头，所以在 catch 块中所做的任何更改都是无用的。

简单示例

public class Example {
public static void main(String[] args) {
int x;
boolean repeat = true;
Scanner input = new Scanner();
do {
try {
x = input.nextInt();
System.out.println("Success!");
repeat = false;
}
catch(InputMismatchException e) {
System.out.println(e.getMesasge());
system.out.println("nYou must enter an integer");
//BTW am I correct in thinking the following clears the buffer?
input.nextLine(); 
x = input.nextInt();
}
} while (repeat);

但是如果我这样做，程序会返回到 do 块的开头，从而重置 X 的值，而不是从成功消息所在的行继续。

我知道这是因为重复布尔值在那一刻为真，因此它会触发 while 条件重新开始，但是如果我在 catch 块内将其设置为 false，我可能会让自己面临未经处理的异常，因为有人仍然可以尝试输入无效的东西。

有没有办法在 catch 块处理异常后引发异常的行之后返回控制权？

Answer 1

如果您的应用程序将有多个用户输入，那么您应该有多个循环来处理这些输入中的每一个，因为每个输入都有可能出错并生成异常。

在大多数情况下，如果由于用户输入而发生异常，则变量中的值无论如何都不需要该输入，因为它显然是错误的，或者至少已经下降到其初始化的默认值。在这种情况下，通常您希望为用户提供提供正确输入的机会。如果没有，则不要一开始就将提示放入循环中。

Scanner input = new Scanner(System.in);
String ls = System.lineSeparator();
// We want the User to enter an inclusive number from 1 to 5.
int x = 0;
while (x < 1 || x > 5) {
System.out.print("Enter an Integer Number from 1 to 5: --> ");
try {
x = input.nextInt();
input.nextLine();   // Consume ENTER.
System.out.println("Success! The nextInt() method accepted your input!");
if (x < 1 || x > 5) {
System.err.println("But...this application does not accept it!");
System.err.println("We asked for a number from 1 to 5! Try again..." + ls);
}
}
catch (InputMismatchException  ex) {
System.out.println(ex.getMessage());
System.err.println("Invalid Input! An Integer number from 1 to 5 only!" + ls);
//BTW am I correct in thinking the following clears the buffer?
/* YES you are since the nextInt() method does not consume the
the newline character provided by the ENTER key like the nextLine()
method does and therefore provides it on the next input request
which in this case ultimately generates an endless loop of exceptions.
Even if this prompt was successful and control is passed on to the 
next prompt and if that prompt was a nextLine() method then it would 
be basically skipped over because it would then consume that ENTER 
key newline character provided in the last nextInt() method.  So
with that in mind, it doesn't hurt to place input.nextline(); 
directly after the x = input.nextInt(); ether.               */
input.nextLine();   // Consume ENTER.
}
}
System.out.println(x + " Was Supplied! - DONE!");

虽然有时它有一个目的，但我个人尽量避免反对例外。我认为如果可以的话，最好避免使用它们，这就是为什么对于控制台应用程序，我更喜欢只使用 Scanner#nextLine(( 方法，该方法接受所有键盘输入，例如：

Scanner input = new Scanner(System.in);
String ls = System.lineSeparator();
// Prompt 1:
// We want the User to enter an inclusive number from 1 to 5.
int x = 0;
String userIN = "";
while (x < 1 || x > 5) {
System.out.print("Enter an Integer Number from 1 to 5 (q to quit): --> ");
userIN = input.nextLine();
if (userIN.toLowerCase().charAt(0) == 'q') {
System.out.println("Quitting!");
System.exit(0);
}
// Is it a signed or unsigned integer number with 1 or more digits?
if (userIN.matches("-?\d+")) {
System.out.println("Success! The nextLine() method accepted your input" + ls
+ "to be a string representation of an Integer value!");
x = Integer.parseInt(userIN);
}
else {
System.err.println("Invalid Input! An Integer number from 1 to 5 only!" + ls);
continue;
}
if (x < 1 || x > 5) {
System.err.println("But...this application does not accept it!");
System.err.println("We asked for a number from 1 to 5! Try again..." + ls);
}
}
System.out.println(x + " Was Supplied! - DONE!" + ls);
// Prompt 2:
// We want the User to enter any float or double type numerical value.
double d = Double.MIN_VALUE;
while (d == Double.MIN_VALUE) {
System.out.print("Enter a float or double type number (q to quit): --> ");
userIN = input.nextLine().toLowerCase().trim();
if (userIN.charAt(0) == 'q') {
System.out.println("Quitting!");
System.exit(0);
} 
// Get rid of the type designator from value if it exists.
if (userIN.endsWith("f") || userIN.endsWith("d")) {
userIN = userIN.substring(0, userIN.length() - 1);
}
// Is it a signed or unsigned integer, float, or double type number?
if (userIN.matches("-?\d+(\.\d+)?")) {
System.out.println("Success! The nextLine() method accepted your input" + ls
+ "to be a string representation of an Integer, float," + ls
+ "or double type value!");
d = Double.parseDouble(userIN);
}
else {
System.err.println("Invalid Input! A Float or Double type numerical value is required!" + ls);
d = Double.MIN_VALUE;
}
}
System.out.println(d + " Was Supplied! - DONE!");

Answer 2

如果在未将repeat设置为 true 的情况下从catch块中删除input.nextLine(); x = input.nextInt();，则执行将继续到do开始，并将要求用户输入值。如果输入了正确的值，则可以将repeat设置为 false 以突破do-while。