Got prob with if else

Question

Hi, i'm a newbie to java and while i'm studying it, i got one problem with the code. Can you help me please because i don't know how to interpret correctly. I just know that when i excute and run it, then enter any number larger than 200, it still gimme the same result which is "you are ok". Here is the code:
public class IQ
public static void main(String[] args) throws Exception
int iq;
System.out.println("Enter your IQ number: ");
iq = (int)System.in.read();
if (iq > 200)
System.out.println("Abnormal");
else
System.out.println("you are ok");
}

Answer 1

Looks like a troll (there is no problem with if/else).
But if it's not a troll (and for the benefit of other n00bs).
The OP had int iq = (int)System.in.read();We look up the System class and find it containspublic static final java.io.Inputstream in;so we look up the InputStream class and see that int read() reads a single byte and returns it (or -1 for EOF). [Also int read(byte[]) returns bytes in the array passed.]
Maybe it's worth noting that the ascii values for '0' thru '9' is 48 thru 57.
If the user attempts to enter <1><0><7><Enter> then read() returns the ascii value for '1' which is 49 and the rest of the user input is discarded.
Now to the post made by PhHein.
BufferedReader keyboard = new BufferedReader(new InputStreamReader(System.in));
int iq = Integer.parseInt(keyboard.readLine());We look up the InputStreamReader class and find that it converts byte[] to char[] (not strictly true, but that explanation will have to suffice in this context).
We look up the BufferedReader class and find that readLine() very handily reads until it encounters a line terminator (user pressed enter), and returns the input as a String.
We look up the Integer class and find that parseInt() gets the numeric value of a number stored in a String.
If the user enters <1><0><7><Enter> then readLine() returns a String containing "107" and parseInt() returns 107. Neat, eh.