Other than what people already said about the try…catch…finally block, I believe what you’re looking for is try { file = new FileStream(“example.txt”, FileMode.Open); file.open(); } catch (Exception ex) { MessageBox.Show(ex.Message); } But you need to add reference to System.Windows.Forms in your project and add the using statement to your class using System.Windows.Forms; Or, if … Read more
Simply throw-catch an Exception such as proposed ArithmeticException or create your own: try { if(num2 == 0){ throw new ArithmeticException(); } } catch (ArithmeticException e) { System.out.println(“Please input another integer than zero”); } solved 0 cant be divided using try and catch java [closed]
The first line that doesn’t make sense to me and shouldn’t complie is this. Names = Names [Y] = myScanner.next(); You are setting the String[] Names to a single String value. 7 solved War – Card Game – With Betting – Java
Most probably your code runs into an exception. Add a e.printStackTrace() to your catch clause and execute your application again to see what is printed on the console. try { // your code } catch (Exception e) { e.printStackTrace(); } That should help. 1 solved Try/catch doesn’t run full code
It is simple: Address = Address.concat(String.valueOf(i)); Address initailly had a whatever value: 5.20.86.0. In your for loop you are appending the i current value, but you don’t delete the previous value! So it is going to 5.20.86.012345678910 as 5.20.86. and 012345678910 than 012345678910 11 12 13 14 and so on. I hope it is a … Read more
If just want to handle “quit” or a float, that’s easy: total = 0.0 while True: value = raw_input(“enter a number or quit to finish”)) if value == “quit”: break total += float(value) print total But if the user types, say, qiut or 2..137, the program will bail out with a ValueError. How do you … Read more
That is reinventing the wheel, use str.isalpha You could use assert and AssertionError from string import ascii_letters value = None while True: try: value = input(“Give a value: “) assert all(c in ascii_letters for c in value) break except AssertionError: print(“Invalid input, try again”) print(“Valid input:”, value) Give a value: aa! Invalid input, try again … Read more
You can’t put reader.close() between the try and the catch. Either put it in a finally block, or use a try-with-resources. Like, try (BufferedReader reader = new BufferedReader(new FileReader(filenameIn))) { reader.readLine(); for (int i = 0; i < personArray.length; i++) { String[] data = reader.readLine().split(“/t”); // <– should be \\t for tab. personArray[i] = new … Read more
You are catching InputMismatchException but typing a letter instead of a number raise a NumberFormatException. You can use a generic catchexception: try { } catch(Exception e){ //<– Generic exception } Or use multiple catch block: try { } catch(InputMismatchException | NumberFormatException e ){ //<– Multiple exceptions } Updated to explain why InputMismatchException doesn’t work: That’s … Read more
The problem in your current code is that you’re always trying to read an int so when receiving a non-integer input you can’t handle the error in the right way. Modify this to always read a String and convert it into an int: int numberOfPeople = 0; while (numberOfPeople <= 0) { try { System.out.print(“Enter … Read more
Though question seems to be very broad, but you need understanding on Java Exception framework + bit of design patterns 1.) If you are exposing some public methods in a service, it is always better to catch exception Log Exception -> Return proper error code/message 2.) throwing exception means everyone who is calling that method … Read more
To wrap up this discussion. From @Erresen, “you’re currently catching ALL exceptions with your current catch.” This is fine if you don’t want / need to do something specific based on the specific exception received. However, as @Johny Mopp mentioned, ObjectDisposedException and IOException are potential exceptions. Perhaps, if you have an IOException, you’ll want to … Read more
The functions are not being executed “automatically,” they’re being executed because you’re explicitly calling them: const arr = [ { ‘Some text’: this.f(‘a’) }, // ^^^^^^^^^^^−−−−−−−−−−−−−− here { ‘Some other text’: this.f(‘b’) } // ^^^^^^^^^^^−−−−−−−− and here ] The result of the above is an array with two objects in it, where the first object … Read more