integer read using BufferedReader is Lost

[John Eipe]

Hello,

This is a simple code. How come the value 'i' be lost?

public class test {
    static int i;
    public static void main(String[] args) throws IOException{

    BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        
        i = reader.read();
        System.out.print(i);
    }
}

I get stuck at all unexpected places while coding

[Campbell Ritchie]

You need to look through the information about Java available. I would suggest you start with this part of the Java™ Tutorials. You might need a look at the API for BufferedReader and its related classes.

You will find out the read() method does not do what you think it does. It is not at all suitable for reading text, but takes one character at a time. You may need to push enter before you see any results.

[Till Stoschus]

Hi Joe,

reader.read() reads a single character. So, if you enter 1234 +enter, the first number (1) is assigned to i.
I will be 49 because 49 is the ascii-value of 1.

Till

[John Eipe]

Thanks all,

My silly mistake. And you are right Till, it spits out the ASCII values.

I solved it with

 
i = Integer.parseInt(reader.readLine());

Is there any other way to do this. I'm writing a program that takes a lot of cmd line inputs. Maybe I could try with Scanner class too...

Thanks.

[Campbell Ritchie]

Scanner is a lot better and easier to use.