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Old 04-16-2004, 03:39 PM   #1 (permalink)
Psycho
 
[java] "if" problem

Please ignore my poor formatting and my bad way of doing things. At the time of writing I have about an hours experience with java, and I am a tad stuck with this code I have written.


class maths
{
public static void main (String args[])
{
int x;
x = Integer.parseInt(args[0]) ;
int y;
y = Integer.parseInt(args[2]) ;
System.out.println (x*y) ;
String f;
f = (args[1]);
if (f == "x")
System.out.println (x*y);
else
System.out.println ("Sorry, I only do multiplication");
}
}


OK, the idea is I enter in the command line
java maths 3 x 4
and it returns
12 (as a test, to make sure that my casting and maths were right)
12 (because I have an "x" as args[1])

Or I enter
java maths 3 / 4
and it returns
12 (as the test)
Sorry, I only do multiplication

What happens is that it compiles perfectly, then:

% java maths 3 x 3
9
Sorry, I only do multiplication

% java maths 3 / 3
9
Sorry, I only do multiplication

I am at a loss.

Also, is there a better way of taking a string from args and turning it into an int?

Thanks all.

Last edited by TheBrit; 04-17-2004 at 02:20 AM..
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Old 04-16-2004, 07:47 PM   #2 (permalink)
Crazy
 
The big error is that in your if statement, you are using == instead of .equals to do the comparison. == only works for primitive data types. Since a String is a referential data type, you have to use .equals.

So,

if( args[ 1 ].equals( "x" ) )
...


As far as taking a String argument and casting it to an int, Integer.parseInt works pretty well.
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Old 04-16-2004, 10:43 PM   #3 (permalink)
Free Mars!
 
feelgood's Avatar
 
Location: I dunno, there's white people around me saying "eh" all the time
Just to add to firebirdta's solution. He is right about the primitive data types and the referential data types.

What you did was trying to do it in the middle way.

When I first saw your if statement, I thought you were evaluating chars but were actually doing strings.

Another possible solution is
if(args[1].charAt(0) == 'x')

Either way works, it just a matter of what kind of data you're dealing with.

The equal method actually deals with object which is another type of referential data types.

Other possible solutions are

if(args[1].equalsIgnoreCase("x"))

In this case, 2 strings are compared regardless of case size.

But one problem with firebirdta's solution. What if the user decides to use the bigger case X? (eg: "java maths 3 X 4" instead of "java maths 3 x 4") Then your if statement would return false since it is only looking for the lower case x. In this case, I suggest you use the if(args[1].equalsIgnoreCase("x"))

One of the fundamentals of data structuring and alogrithms is the ability to predict human behavoior involved when it comes to running programs. Programmers are actually expected to build foolproof programs, meaning that we try to anticptate all possible situation that occurs when the user use the program.

A elite programmer once said

"Programming is an race between the universe and programmers. Programmers build bigger, better and more idiot-proof programs while the universe produce more idiots. So far, the universe is winning"
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Old 04-17-2004, 02:23 AM   #4 (permalink)
Psycho
 
Ah-hah!
It all works perfectly now, I think I am starting to like this language enough to pick up the O'Reilly book, or whatever book my local library has.
Thanks all.
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Old 04-17-2004, 11:06 AM   #5 (permalink)
Once upon a time...
 
Quote:
Originally posted by TheBrit
Ah-hah!
It all works perfectly now, I think I am starting to like this language enough to pick up the O'Reilly book, or whatever book my local library has.
Thanks all.
try this one:
http://www.mindview.net/Books/TIJ/
It's free
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