CODEWITHSUNDEEP
javajava.util.scanner
Little_idiot
Little_idiot

Reputation: 135

scanner.nextInt() is moving to the next line, how is this possible?

Input

So, I know that nextLine() reads the whole line and moves the cursor to the next line and nextFoo() reads only up till it encounters a space and keeps the cursor there (doesn't move it to a new line).

So initially I tried this:- 

int N = Integer.parseInt(in.nextLine()); //reading the whole line and parsing it to Integer. The cursor pointer is moved to the next line
while(--N>=0){
    p = in.nextInt(); q = in.nextInt(); //reading the two space seperated integers
    in.nextLine();//to move the cursor to the next line
}

But this wasn't working properly, couldn't read the last p and q I was inputting.

Changed it to:

int N = in.nextInt(); 
while(--N>=0){
    p = in.nextInt(); q = in.nextInt();
}

And it is working fine. How is nextInt() going to the next line? All .nextFoo() and .next() except .nextLine() read only up till there is a space and keep the cursor pointer there, right? How is this going to the newline then?

Upvotes: 0

Views: 2221

Answers (1)

slim
slim

Reputation: 41223

From the Scanner Javadoc:

The next() and hasNext() methods and their primitive-type companion methods (such as nextInt() and hasNextInt()) first skip any input that matches the delimiter pattern, and then attempt to return the next token.

So the cursor is before the newline, then when you call nextInt(), skips through the newline and any other whitespace, before consuming the number.

Your code containing nextLine() will work, as long as your input file ends in a newline. It's conventional to end text files with a newline character -- many editors will either force you to have one, or display a warning if it's missing.

Upvotes: 1

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