Final variable assignment with try/catch

问题

Because I believe it is a good programming practice, I make all my (local or instance) variables final if they are intended to be written only once.

However, I notice that when a variable assignment can throw an exception you cannot make said variable final:

final int x;
try {
    x = Integer.parseInt(\"someinput\");
}
catch(NumberFormatException e) {
    x = 42;  // Compiler error: The final local variable x may already have been assigned
}

Is there a way to do this without resorting to a temporary variable? (or is this not the right place for a final modifier?)

回答1:

One way to do this is by introducing a (non-final) temporary variable, but you said you didn't want to do that.

Another way is to move both branches of the code into a function:

final int x = getValue();

private int getValue() {
  try {
    return Integer.parseInt("someinput");
  }
  catch(NumberFormatException e) {
    return 42;
  }
}

Whether or not this is practical depends on the exact use case.

All in all, as long as x is a an appropriately-scoped local variable, the most practical general approach might be to leave it non-final.

If, on the other hand, x is a member variable, my advice would be to use a non-final temporary during initialization:

public class C {
  private final int x;
  public C() {
    int x_val;
    try {
      x_val = Integer.parseInt("someinput");
    }
    catch(NumberFormatException e) {
      x_val = 42;
    }
    this.x = x_val;
  }
}

回答2:

No it is not the right place, imagine you got more then 1 Statement in your try and catch block, the first one says : x = 42. After some others Statements the try block fails, and it goes to the catch block, where your Saying x = 30. Now you defined x twice.

来源：https://stackoverflow.com/questions/13604111/final-variable-assignment-with-try-catch