Java Strings whirlpool... oh my god!

[MysticVegeta]

Hello,
I decided to learn Java over the summer, and so i did, although the strings are really giving me some problems. I got the IO Part, I got the Ints, but the strings... damn
ok, now,

code:

int foo; foo = System.in.read(); System.out.println(foo);

Can someone tell me why is it that this returns 49 when i input in 20?

Also

code:

String fi; fi = System.in.read();

Why does this give me an error?

Apparently i thought this was going to be easier than for loops and arrays. but i was wrong, i learned them before knowing how to get a string variable. lol. help?

[wtd]

I think maybe the "read" method isn't doing what you think it's doing?

What are you trying to accomplish? If you just want to read in a String, you should wrap System.in in a BufferedReader instance.

[Aziz]

If he's the experience level I think he is, I think he doesn't know what that means Try the Java Tutorial, here: http://java.sun.com/docs/books/tutorial

[MysticVegeta]

ok what i am trying to do is "get" a string

Turing:

var str : string get str put str

Also

Turing:

var integer : int get integer put integer

Why, if i input 20 as an integer, it outputs 49?

[Aziz]

The System.in.read() method is not for that, I believe.

What you want to do is:

[syntax"java"]BufferedReader in = new BufferedReader(System.in);

String myString = in.readLine();[/syntax]

Let me explain. As much as I know. rizzix or wtd would be better at this, though, but I'll just let you know a least a little bit of whats going on.

BufferedReader is a stream reader. It reads data. We're "creating" (instantiate, i think is the proper word) a BufferedReader object here. When you create a BufferedReader here, we're giving it System.in. System.in is a stream coming from the keyboard (I believe, maybe from somewhere else?) Just like System.out is a stream. When you call System.out.println("hello"), you're telling the program to send the text "hello" through the System.out stream, which goes to the console output. Now readLine(); is a method (function in Turing terms) that returns a String. So you tell your reader (we've named it 'in') to read a line ('readLine()') from it's stream ('System.in') get it? If not, I'm sure the 1337 rizzix or wtd will help you (but hopefully not before the help me )

[1of42]

Yeah, that's basically correct. vegeta, given that I've never used the read() function, I can't help you with that, but the following lines will get an integer from the console:

Java:

int number; BufferedReader reader = new BufferedReader (new InputStreamReader (System.in)); try {    number = Integer.parseInt (reader.readLine ().trim ()); } catch (NumberFormatException ex) {    System.out.println ("Sorry, that was not a valid number"); }

[[Gandalf]]

Java:

import java.io; public class Prog {         public static void GetName ()         {                 String name;                                 BufferedReader reader = new BufferedReader (new InputStreamReader (System.in));                 System.out.println("Enter your name: ");                 name = reader.readLine();         }                 public static void main (String [] args)         {                 System.out.println("Hello person!");                 GetName();         } }

Quote:

---------- Capture Output ----------
> "C:\j2sdk1.4.2_06\bin\javac.exe" Prog.java
Prog.java:1: cannot resolve symbol
symbol : class io
location: package java
import java.io;
^
Prog.java:9: cannot resolve symbol
symbol : class BufferedReader
location: class Prog
BufferedReader reader = new BufferedReader (new InputStreamReader (System.in));
^
Prog.java:9: cannot resolve symbol
symbol : class BufferedReader
location: class Prog
BufferedReader reader = new BufferedReader (new InputStreamReader (System.in));
^
Prog.java:9: cannot resolve symbol
symbol : class InputStreamReader
location: class Prog
BufferedReader reader = new BufferedReader (new InputStreamReader (System.in));
^
4 errors
> Terminated with exit code 1.

[rizzix]

it should be

Java:

import java.io.*;

... you import the classes.. not the package..

[wtd]

[Gandalf] wrote:

Java:

import java.io; public class Prog {         public static void GetName ()         {                 String name;                                 BufferedReader reader = new BufferedReader (new InputStreamReader (System.in));                 System.out.println("Enter your name: ");                 name = reader.readLine();         }                 public static void main (String [] args)         {                 System.out.println("Hello person!");                 GetName();         } }

As rizzix pointed out, your import is a bit goofy, and not at all valid.

Also, method names in Java should always begin with a lower-case letter.

[[Gandalf]]

wtd wrote:

As rizzix pointed out, your import is a bit goofy, and not at all valid.

Yes... I was trying to do it from memory, I was bound to make mistakes.

wtd wrote:

Also, method names in Java should always begin with a lower-case letter.

Yes, it's a habit, I usually have variables like that. You say that variables and methods should be this way? I'm used to thinking of the two quite differently.

Quote:

---------- Capture Output ----------
> "C:\j2sdk1.4.2_06\bin\javac.exe" Prog.java
Prog.java:11: unreported exception java.io.IOException; must be caught or declared to be thrown
name = reader.readLine();

Must I learn up on throwing now, or am I doing something wrong?

[wtd]

[Gandalf] wrote:

wtd wrote:

Also, method names in Java should always begin with a lower-case letter.

Yes, it's a habit, I usually have variables like that. You say that variables and methods should be this way? I'm used to thinking of the two quite differently.

[/quote]

Yes, both should use that naming convention.

Quote:

---------- Capture Output ----------
> "C:\j2sdk1.4.2_06\bin\javac.exe" Prog.java
Prog.java:11: unreported exception java.io.IOException; must be caught or declared to be thrown
name = reader.readLine();

Must I learn up on throwing now, or am I doing something wrong?[/quote]

Exceptions in Java are checked. You either have to explicitly catch the exception in your method, or report that the method is capable of throwing that exception.

Java:

public static void main(String[] args) throws IOException

[MysticVegeta]

thanks a lot for the help guys I have used the bufferedreader before when doing the IO. Thanks a lot

[Guest]

Concerning your first implementation. When you input 20, are you sure it shows 49, not 50? Because System.in.read() reads one character. If you input 20, it will read the first character '2', and store it as an integer. Casting occures automaticaly in this case. So when you print the integer, it is equivalent to printing (int)'2', which is 50.