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Java Stdin and Stdout II

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  • KumKum9
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    Unlike the scanner.nextLine() method, the scanner.nextInt() method only consumes the integer part and leaves the enter or newline character in the input buffer.

    When the third scanner.nextLine() is called, it finds the enter or newline character still existing in the input buffer, mistakes it as the input from the user, and returns immediately.

    There are two ways to solve this problem. You can either consume the newline character after the scanner.nextInt() call takes place, or you can take all the inputs as strings and parse them to the correct data type later on.

  • dorisgreenwell
     + 0 comments

    Java is such a versatile and powerful language—it's amazing how it's still going strong after all these years. betguru247.net

  • Damian_GDO_92
     + 0 comments
    import java.util.Scanner;

public class Solution {

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        
        int i = scan.nextInt();
        double d = scan.nextDouble();
        
        scan.nextLine();
        String s = scan.nextLine();

        System.out.println("String: " + s);
        System.out.println("Double: " + d);
        System.out.println("Int: " + i);
        
        scan.close();
    }
}

  • charlespatch06
     + 0 comments

    Java is a powerful, object-oriented programming language known for its portability, security, and scalability. betguru247net

  • kc30krishna
     + 0 comments

    This problem solution