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Home/ Questions/Q 378571
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Editorial Team
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Editorial Team
Asked: May 12, 20262026-05-12T14:48:27+00:00 2026-05-12T14:48:27+00:00

I got the following code: int nnames; String names[]; System.out.print(How many names are you

  • 0

I got the following code:

int nnames;
String names[];

System.out.print("How many names are you going to save: ");
Scanner in = new Scanner(System.in);
nnames = in.nextInt();
names = new String[nnames];

for (int i = 0; i < names.length; i++){
  System.out.print("Type a name: ");
  names[i] = in.nextLine();
}

And the output for that code is the following:

How many names are you going to save:3 
Type a name: Type a name: John Doe
Type a name: John Lennon

Notice how it skipped the first name entry?? It skipped it and went straight for the second name entry. I have tried looking what causes this but I don’t seem to be able to nail it. I hope someone can help me. Thanks

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-12T14:48:27+00:00Added an answer on May 12, 2026 at 2:48 pm

    The reason for the error is that the nextInt only pulls the integer, not the newline. If you add a in.nextLine() before your for loop, it will eat the empty new line and allow you to enter 3 names.

    int nnames;
    String names[];
    
    System.out.print("How many names are you going to save: ");
    Scanner in = new Scanner(System.in);
    nnames = in.nextInt();
    
    names = new String[nnames];
    in.nextLine();
    for (int i = 0; i < names.length; i++){
            System.out.print("Type a name: ");
            names[i] = in.nextLine();
    }
    

    or just read the line and parse the value as an Integer.

    int nnames;
    String names[];
    
    System.out.print("How many names are you going to save: ");
    Scanner in = new Scanner(System.in);
    nnames = Integer.parseInt(in.nextLine().trim());
    
    names = new String[nnames];
    for (int i = 0; i < names.length; i++){
            System.out.print("Type a name: ");
            names[i] = in.nextLine();
    }
    
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