Read a String after an Integer

Question

Hello, my program throws an error... i think it is because I put first an Integer and then a String... but im not sure. Can anyone help?

https://code.sololearn.com/ca8A15A24a13/?ref=app

Answer 1

Sandeep Kushwaha
Diego Becerril
Martin Taylor

Better change Scanner delimiters regex

Scanner s = new Scanner(System.in)
// ADDED
.useDelimiter("\n")
;

and to recover original, call reset()

https://docs.oracle.com/en/java/javase/14/docs/api/java.base/java/util/Scanner.html#useDelimiter(java.lang.String)

Added to my code.

https://code.sololearn.com/c1UK7uiKcrxp/?ref=app

Answer 2

That's fine as long as you input a number on the first line hit enter and input a string on the second line then run, it should be ok.

Answer 3

Diego Becerril you pass your funtion a string and the strings length

Your first if statment sees this and if the string length does not equal 0 then it falls to the else statement and its repeating its self there.

Answer 4

Martin Taylor, yeah i was thinking that... how can i solve it?

Answer 5

Martin

I agree with you, the program also must treat errors produced by a bad input to retrofit user about. Eg. Showing error and request same input again using a do-while and catching exceptions, or infer defaults...

In Scanner: NoSuchElementException when EOF and InputMismatchException on bad type.

In Integer:
NullPointerException y
NumberFormatException when no match

Due to this it's wellknown you should have in one hand your program and the other the api docs 😂😂

Resuming... Input all in line e.g.

9 Hola amigos del Java

It works with my patch

Answer 6

Yeah, the problem is when I enter the number 9 the program stops giving me a StackOvdrFlowError

Answer 7

Yes, but what can i do? I mean... all seems to be good, why the error?

Answer 8

Easy... swap condition... and code seems cleaner

Pseudocode:
==========
If i > 0 {
despR(s, i-1);
}
Print (c at i)

Answer 9

Yeah, i think all its good, but the error is that when I enter the 9 the program doesnt even allow me to enter the String

Answer 10

Martin Taylor I think that it only happens here in Playground because Scanner read by default any of Character.isWhitespace() including CRLF

Also there are a particularity about nextInt respect parseInt. First only consume digits while parseInt accepts formatted in e-notation. Even, if first part are valid num and found other non valid... the rest it's ignored.

a bit of light

- https://docs.oracle.com/en/java/javase/14/docs/api/java.base/java/util/Scanner.html
- https://docs.oracle.com/en/java/javase/14/docs/api/java.base/java/lang/Integer.html#parseInt(java.lang.String)

**** input popup

Diego Becerril here some optimizations in the line I explained you

https://code.sololearn.com/c1UK7uiKcrxp/?ref=app