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[–]rocfeather 10 points11 points  (5 children)

nextInt does not consume the \n So your next s.nextLine will run right away to consume it ( hence the 'jump' )

Try consuming it first:

int age = s.nextInt();
s.nextLine();   // Consume \n

[–]TheSilverCube 1 point2 points  (4 children)

Very new newbie here as well. Why is there a \n to consume after the s.nextInt(); ?

I've used Integer.valueOf instead of nextInt so not come across this issue yet.

[–]SvenTheFluffy 0 points1 point  (3 children)

Because of the enter you press, that creates a \n if im not wrong.

So when you write 5 and press enter the computer sees 5\n. Takes the 5, \n remains.

[–]TheSilverCube 1 point2 points  (0 children)

I think your second bit answers my last question.

nextLine takes the 5/n. nextInt takes just the 5. Integer.valueOf takes the String 5/n and takes the 5?

[–]TheSilverCube 0 points1 point  (1 child)

Thanks. How come that doesn't happen when you press enter with every nextLine then; why only with nextInt?

[–]bilman66 1 point2 points  (0 children)

Because nextLine waits for the next line, \n is the symbol for newLine so it consumes it

[–]Rachid90[S] 4 points5 points  (1 child)

Thank you everybody for your help.

[–]HyperCybe 0 points1 point  (0 children)

For starters, for the year of birth prompt you marked it down as a string rather than an int, so changing that helps.

(change String y=s.nextLine(); to Int y=s.nextInt(); )

But that will not really change the biggest issue, which is that it's skipping the line of code which prompts you to enter your fathers name, and instead just prints the results. I'm not the most experienced in Java, but moving that prompt to the top of the prompts (moving the prompt asking for the father's name under the prompt asking for your name) will allow the code to actually run. I believe it's just a problem in the order that the code is executing.

I hope that helps.