How can "1" != "1"

That's insane! But, very vital to know. It's interesting that

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1. String s = "1";
2. if (s == "1")
3.     MessageBox.Show("C# is not weird!");
4. else
5.     MessageBox.Show("C# is as weird as Java!");
6.  

string comparisons work in C# but not in Java. Thanks for your help, now I'll have to rewrite my homework like I wanted to in the first place!

... it's not 'insane'; it just doesn't work your prejudiced mind thought it should work.

not prejudiced :D I just expect the compiler to say :You don't want to do this, Sam!" At a minimum, it could have said "Illegal Indirection, like C++. ROFL --Sam

I'm trying to keep up with you youngsters, so I'm taking a beginning Java course. As part of a homework assignment, we were to prompt the user for a number between 1 & 3. We also used while loops in the chapter, so I decided to put the prompt in a while loop to force the user to enter 1, 2 or 3. I was very surprised when it did not work, so I have written a 5-liner to show you the problem:

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1. import javax.swing.JOptionPane;
2. public class Weird {
3.  public static void main(String[] args) 
4.  {
5.   String s = "";
6.   while (s != "1" && s!= "2" && s!= "3")
7.   {
8.    s = JOptionPane.showInputDialog(null, "Enter 1, 2 or 3");
9.    if (s == null) break;
10.   }
11.   if (s!= null) JOptionPane.showMessageDialog(null, s);
12.  }
13. }
14.  

I fixed the problem by parsing the string and testing for integers 1, 2, or 3, but the above code should work! The same code in VB works great:

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1. Sub NotWeird()
2.     Dim s As String
3.     Do While (s <> "1" And s <> "2" And s <> "3")
4.         s = InputBox("Enter 1, 2 or 3")
5.     Loop
6.     MsgBox s
7. End Sub
8.  

Looking at the Java code in the debugger, after the user enters 1, the debugger says s = "1" So far, great, but if I inspect s != "1", it says "s != "1"" = true

What is this crazyness?!?

Oh! It's crazy alright! Its' quite the headache...

I also find Java sometimes gives you an error on a line, highlights the line, as result of modifications to completely different lines in the code. And of course the moment I put things back where they belong, Java's happy again...

The compiler can't say that because it's perfectly legal to compare two pointers,
or 'references' as Java calls them; it two pointers compare equal both pointers
point to the same object.

For strings you don't want to compare pointers, you want to compare the *value*
of two (or one?) String object and the String class implements an 'equals' method
for that:

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1. String s1= new String("foo");
2. String s2= new String("foo");
3. if (s1 == s2)
4.    System.out.println("s1 and s2  point to same object");
5. else if (s1.equals(s2))
6.    System.out.println("s1 and s2 have equal value");
7. else
8.    System.out.println("s1 and s2 are different");
9.  

In C++ the overloaded '==' operator in the string class takes care of the 'equals'
functionality. Java doesn't have user defined overloaded operators.

kind regards,

Jos

Good heavens JosAH!

Java IS like JavaScript a lot of repect then. I cannot recall how I wrote this but in a previous reply specific to Sammy's scenario, I wondered whether:

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1.  
2. Var s1 ="Something"
3. Var s2 ="Something"
4.  
5. If (s1 == s2) Then
6. This is good stuff
7. Else
8. MsgBox ("Blaze blaze")
9. End If
10. ...
11.  
12.  

Mind you, I am missing brackets there, but does above signify the similarities in both languages, and if this is a fact, can I look to JavaScrip to get a better understanding of Java, of course to some degree?

But, as Jos pointed out above, this will print good stuff in Java also because s1 and s2 are both pointing to the same string.

Run the following and see if it helps!

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1. import javax.swing.JOptionPane;
2. public class Weird 
3. {
4.  public static void main(String[] args) 
5.  {
6.   String s1 ="Something";
7.   String s2 ="Something";
8.   if (s1 == s2)
9.    JOptionPane.showMessageDialog(null, "You just think this is good stuff");
10.   else
11.    JOptionPane.showMessageDialog(null, "Blaze blaze");
12.   s2 ="Something else";
13.   s2 = s2.substring(0, 9);
14.   if (s1 == s2)
15.    JOptionPane.showMessageDialog(null, "This is not expected");
16.   else
17.    JOptionPane.showMessageDialog(null, "Blaze blaze");
18.   if (s1.equals(s2))
19.    JOptionPane.showMessageDialog(null, "Jos is always right!");
20.   else
21.    JOptionPane.showMessageDialog(null, "Why bother with this message!");
22.  
23.  }
24. }
25.  

I must have missed a step or two but I was stuck in what seemed to be an endless loop, should I get pop ups to enter stuff, should I have pressed ok for in pop up with "{ }" in the box?