#I have an assignment :'0

1 messages · Page 1 of 1 (latest)

cyan ruinBOT
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<@&987246399047479336> please have a look, thanks.

old hamlet
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What's your specific question? That's your assignment.

wary mist
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Without knowing your specific skillset the answer will vary from 'Done under 5 minutes' to ' Don't panic its to late to do it'

What u can do is break the assigment into parts that u can handle and do them step by step.

Also steer away from GPT. I repeat do not use GPT. It will cause u more trouble to make your code work AND u will NOT understand anything in it, nor will u LEARN anything, which is the key in any course.

shadow juniper
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yeah I don't really use gpt :0 I remember doing this kind of assignment with c++ but idk I just didn't follow java but watching tutorials ain't helping. I need someone to explain this like literally talking to me. I'm attending online class btw and my brain can't stand online classes ToT

old hamlet
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I recommend checking out this course over youtube tutorials:

cyan ruinBOT
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MOOC is a completely free introductory Java course created by the University of Helsinki, it is a great way to learn Java from the ground up.

It consists of two parts, one at beginner, and another at intermediate level. The end of the course is marked by creating your own Asteroids game clone!

Even though the instructions show how to configure and use NetBeans for the course, you can use IntelliJ. To use IntelliJ, simply install the TMC plugin by opening IntelliJ -> File -> Settings -> Plugins and searching for TMC. You will then be able to use IntelliJ to complete MOOC.

Visit MOOC here: https://java-programming.mooc.fi/
(the course is available in both English and Finnish)

simon_verhoeven • used /tag id: mooc ·
About the course - Java Programming

About the course - Java Programming

old hamlet
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And start with first outputting a menu, then capturing input.

#

I'll also already give a nudge:

cyan ruinBOT
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Mixing any nextXXX method with nextLine from the Scanner class for user input, will not ask you for input again but instead result in an empty line read by nextLine.

To prevent this, when reading user input, always only use nextLine. If you need an int, do

int value = Integer.parseInt(scanner.nextLine());

instead of using nextInt.

Assume the following:

Scanner scanner = new Scanner(System.in);

System.out.println("Enter your age:");
int age = scanner.nextInt();
System.out.println("Enter your name:");
String name = scanner.nextLine();

System.out.println("Hello " + name + ", you are " + age + " years old");

When executing this code, you will be asked to enter an age, suppose you enter 20.
However, the code will not ask you to actually input a name and the output will be:

Hello , you are 20 years old.

The reason why is that when you hit the enter button, your actual input is

20\n

and not just 20. A call to nextInt will now consume the 20 and leave the newline symbol \n in the internal input buffer of System.in. The call to nextLine will now not lead to a new input, since there is still unread input left in System.in. So it will read the \n, leading to an empty input.

So every user input is not only a number, but a full line. As such, it makes much more sense to also use nextLine(), even if reading just an age. The corrected code which works as intended is:

Scanner scanner = new Scanner(System.in);

System.out.println("Enter your age:");
// Now nextLine, not nextInt anymore
int age = Integer.parseInt(scanner.nextLine());
System.out.println("Enter your name:");
String name = scanner.nextLine();

System.out.println("Hello " + name + ", you are " + age + " years old");

The nextXXX methods, such as nextInt can be useful when reading multi-input from a single line. For example when you enter 20 John in a single line.

simon_verhoeven • used /tag id: scanner ·
shadow juniper
cyan ruinBOT
#

@shadow juniper

Your question has been closed due to inactivity.

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With enough info, someone knows the answer for sure 👍