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Old 10-12-2003, 10:42 PM   #1
Laptop2250
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Registered: Oct 2003
Posts: 131

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Question Java does "age" or "Age" matter when declaring an Int??


Hey this is my code, stuff:

public class GMFnTFF
{public static void main (String []args)
{EasyReader r= new EasyReader ();
int age;
System.out.print ("Enter your name: ");
String name= r.readLine ();
System.out.print ("Enter your age: ");
String LineAge= r.readLine ();
int Age= Integer.parseInt (LineAge); ////here i tried doing age why
////doesnt that work? but Age does
System.out.print ("Enter your weight: ");
String weight1= r.readLine ();
double weight= Double.parseDouble (weight1);
System.out.print ("Greetings, "+name+". You are "+Age+" years old and you weight "+weight+" pounds.");
}
}



int Age works
int age doesnt

why? what does age do in java? I am using j2sdk1.4.2 using the simple compiler bluejay and i am new if you cant tell so try to explain it easily. thx

thanks in advance (ey, me crew do yourselves a fava and dont steal my program or else, ali g talk )

also, i was updating this code for the last 1/2hr or so, and i found out that when i would type in

String name= r.readLine ();

it would automatically print the line to the output, does this always happen auto or is there a way to save a string and not print it immediatly? thanks again in advance (also the ali g talk meant dont steal my code, lol)

Last edited by Laptop2250; 10-12-2003 at 11:07 PM.
 
Old 10-13-2003, 12:01 AM   #2
ksgill
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Registered: Apr 2003
Location: Toronto, Canada
Distribution: Ubuntu Jaunty (9.04)
Posts: 1,044

Rep: Reputation: 45
Age and age are 2 different things because variables in java are case sensitive. age is just a variable and doesnt do anything in java.
String name= r.readLine ();
it doesnt print to output, it reads a line and saves it under a string name. ..looks like u just got started with programming.. I think you should read a good book if u want to learn java.
cool
 
Old 10-13-2003, 09:54 AM   #3
Laptop2250
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Registered: Oct 2003
Posts: 131

Original Poster
Rep: Reputation: 15
i try to make age a int, so your telling me all ints have to be lower case? i tried bge (lower case int) and that works, i guess i didn't understand, is age a reserved method(or something) like double?

also, String name= r.readLine (); saves a variable, but why does it print it to the output if i didn't ask it to. the output of this code is

Enter your name: Laptop2250
Enter your age: 2250
Enter you weight: 225
Greetings, Laptop 2250. You are 2250 years old and you weight 225 pounds.

the only thing i understand is that those strings/int/doubles should only appear in the last line (greetings) because i told the program to. why do the ints/strings/doubles appear other places?

thanks again
 
Old 10-13-2003, 12:34 PM   #4
orgcandman
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Registered: May 2002
Location: dracut MA
Distribution: Ubuntu; PNE-LE; LFS (no book)
Posts: 594

Rep: Reputation: 102Reputation: 102
Here's my code that works (I changed the EasyReader class because I have an older JVM)

I think the problem is areound the line that I've flagged with /* >>> */
I think, judging from your code, you had declared age as an int at line /* LINE REF A */ and then
re-declared it as an int on line /* >>> */ which is a nono...
PS: I also used System.out.println because it will cause the output stream buffer to flush() automatically. plus it adds a newline character.


Code:
import java.io.*;
import java.util.*;
import java.lang.String;

public class GMFnTFF{
    public static void main (String []args) throws IOException{
	BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
	/* LINE REF A */ int age;
	System.out.println("Enter your name: ");
	String name= r.readLine ();
	System.out.println("Enter your age: ");
	String LineAge= r.readLine ();
	/* >>> */ age = Integer.parseInt (LineAge);     // I used age and it worked :D
	System.out.println("Enter your weight: ");
	String weight1= r.readLine ();
	double weight= Double.parseDouble (weight1);
	System.out.println("Greetings, "+name+". You are "+age+" years old and you weight "+weight+" pounds.");
    }
}
Aaron
 
  


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