About Question enthuware.ocajp.i.v7.2.1303 :

[lovjit]

Thank you.Got it.

[KalpanaAr]

Since we already have a class named 'Other', importing another class named 'Other' should cause compilation problem right?

[admin]

Since we already have a class named 'Other', importing another class named 'Other' should cause compilation problem right?

No, why do you think so?

[chipchipchonkeykonky]

I still can't figure out why the line System.out.print((testPackage.Other.hello == hello) + " "); is valid?
Other class lies in package named "other" not in "testPackage".Please clarify my doubt.What am i missing here?

There is a class Other in testPackage as well. It is there at the bottom of the code listing.

good one

[elit3x]

Im stuck on;

System.out.print((other.Other.hello == hello) + " "); //line 2


the first hello variable is explicitly calling the hello variable in package other.
The second hello is retrieving the hello var local to the main method.

These are two different objects, i would expect == to return false.

What am I missing?

[admin]

That is just how the language is designed. Java keeps only one copy of all strings in a pool, no matter where created unless created using the new keyword.
You should check out section "3.10.5. String Literals" of Java Language Specification for complete details.

HTH,
Paul.

[uqzma1xg]

I'm a little confused with the answer explanation:

5. Strings computed at run time are newly created and therefore are distinct. (So line 4
prints false.)

at oracle, only the concatenation is at runtime, §3.10.5. String Literals

Strings computed by concatenation at run time are newly created and therefore distinct.

[admin]

It doesn't say "only".
The following code print false:

Code: Select all

	String str = "123";
        String str2 = "12345";
        String str3 = str2.substring(0, 3);
        System.out.println(str+" "+str3+" "+(str == str3)); //prints 123 123 false

This proves that even a string created using substring is distinct from the one created using a constant expression.

[uqzma1xg]

It doesn't say "only".
The following code print false:

Code: Select all

	String str = "123";
        String str2 = "12345";
        String str3 = str2.substring(0, 3);
        System.out.println(str+" "+str3+" "+(str == str3)); //prints 123 123 false

This proves that even a string created using substring is distinct from the one created using a constant expression.

So how can i know for sure what will create new Object?
also substring just like concat and replace, by definition should return new String..
so maby they are pointing to different objects?

[admin]

That is why the explanation is given that whenever you compute a string at run time, it will create a new string object.
Computation implies that you don't know the string value at compile time. The value has to be computed (using concatenation) at run time and that will create a new string.
Also, if you ignore method calls and the new operator, the only way to create a new string is using the + operator. It will create a new string object if any of its operand is not a compile time constant.

[uqzma1xg]

That is why the explanation is given that whenever you compute a string at run time, it will create a new string object.
Computation implies that you don't know the string value at compile time. The value has to be computed (using concatenation) at run time and that will create a new string.
Also, if you ignore method calls and the new operator, the only way to create a new string is using the + operator. It will create a new string object if any of its operand is not a compile time constant.

Great, thanks

[Radioactive]

Hello! I think in file Test.java missing public modifier for class Test. In this way program will compile but throw exception when trying to run.

[admin]

Did you try compiling and running it?

[Radioactive]

Did you try compiling and running it?

Yes, but i made a mistake at first attempt compile and run. Now i'm repeated this process, and it work fine.

[dxie1154]

I'm a little confused on:
((hello == ("Hel"+"lo")
and:
(hello == ("Hel"+lo)
If newly created Strings that are created at runtime (like the "Hel") are separate objects, then shouldn't the third line also print false?
I think this will happen because you are creating two new Strings at runtime at line 3. That is why at line 4 it prints false, because of the "Hel", which is a newly created String object that has been created at runtime and is a new String object.
I just don't get why the third line is true, because you're making two completely new String objects, yet it passes the == test.
Why does it print true, instead of false?

Thanks in advance.

[admin]

When you see a string in double quotes, that means it is a compile time constant. That means "hel" and "lo" are compile time constant strings. Since the compiler knows their values, it creates "hel"+"lo" i.e. "hello" at compile time as well, which is same as the one that it created at the line just before line 1.

This is not the case with lo because lo is a variable. Compiler cannot assume the value of this variable. Therefore, it cannot compute "hel"+lo at compile time. The JVM at runtime computes its value and that is why the resulting "hello" is not the same string as the one created by the compiler earlier.

HTH,
Paul.

[dxie1154]

So basically the things that are causing this are the compiler and the JVM?
The compiler adds the two compile time constant strings together, but the variable is added to "Hel" by the JVM, and because the way the two Strings were added together is different, it causes true or false to be returned.

Right?

[admin]

That is correct.

[jkronegg]

Because lo is non-final variable. So the compiler cannot use it as a constant for optimizations. Had it been final, your argument would have been correct.

Actually, the compiler does "constant compile-time Strings" optimization only when it recognizes such "constant compile-time Strings". Having a "final String" may not be sufficient to make the compiler recognize them. For example:

Code: Select all

		boolean c = "ab"==("a"+"b"); // true ("b" is recognized as compile-time constant)
		
		final String b1 = "b";
		boolean c1 = "ab"==("a"+b1); // true (final String b1 is recognized a compile-time constant)
		
		final String b2 = "b"+1;
		boolean c2 = "ab"==("a"+b2); // false (although b2 is final, it is not recognized as a compile-time constant string expression)

		String b3 = "b";
		boolean c3 = "ab"==("a"+b3); // false (although b3 is never modified is not recognized as a compile-time constant string expression)

What counts is what the compiler recognizes, not what we know to be a constant String value.

[admin]

why would you expect "ab"==("a"+b2); to return true? b2 is "b1" as per the code that you've posted. "ab" cannot be equal to "ab1".

[Belcheee]

While trying to understand this, I added the following code in a method:

Code: Select all

String s1 = "Hello";
String s2 = "Hello";
System.out.println("result is " + s1==s2)

Expecting this to output true as per the explanations on this post, but instead it outputs false.

If I remove the "result is" string from the println argument, it returns true (as expected).

I understand that a literal String either side of the + operand converts the other variable to a String, but it's already a String, so what has it changed?

[admin]

+ has higher precedence than ==. "result is hello" is not equal to "hello".

[Belcheee]

+ has higher precedence than ==. "result is hello" is not equal to "hello".

Of course! Thank you.

[Rinkesh]

6. The result of explicitly interning a computed string is the same string as any pre-existing literal string with the same contents. (So line 5 prints true.)
How can 5th one be a true if this statement says that explicitly interning a COMPUTED String?lo is a variable so it will be computed at run-time,right?

[admin]

6. The result of explicitly interning a computed string is the same string as any pre-existing literal string with the same contents. (So line 5 prints true.)
How can 5th one be a true if this statement says that explicitly interning a COMPUTED String?lo is a variable so it will be computed at run-time,right?

Yes, and the intern method will be executed at run time as well. The string "Hel"+lo will be computed at run time and the call to intern will return a reference to the interned string containing "hello", which is same as the one pointed to by the variable hello.