error Identifier expected?

[chandresh]

Hi

Some 10 years ago I used to write Java programs effieciently, then stopped due to some circumstances.
Now trying to recap java again.

My PC has Win 7 (32 bit) with JDK 1.7.0_03 installed and have also put the path of "bin" folder in the PATH variable, other wise javac was not working.

Anyways, I wrote a simple code to test and it gave me some error which I could not understand WHY?
Please advice as it gives me "error <identifier> expected at line 9

1. public class tst2 {
2. public static void main(String[] args) {
3. System.out.println("HI");
4. }
5.
6.
7. public class tstinner {
8. int a1, b1;
9. a1 = b1 = 6;
10. }
11.
12.}

I have put the type "int" and then initialise the var a1 & b1 in a single line which is valid I guess, then why is it giving me the error?
thanks

[pbrockway2]

The following compiles OK:

public class Tst2 {
    public static void main(String[] args) {
        System.out.println("HI");
    }

    public class TstInner {
        int a1, b1;
        
        TstInner() {
            a1 = b1 = 6;
        }
    }
}

The difference is that the initialisation of a1 and b1 occurs within a constructor. You cannot try and assign the result of evaluating b1=6 to a1 except in a method, constructor or initialisation block. Of these a constructor is the most obvious place.

I guess the main point is that foo=bar=42 does not mean "assign 42 to foo and bar" as you might have supposed. It means something like "evaluate bar=42 and assign the result to foo".

-----

When you post code don't include the line numbers. And use the "code" tags: put [code] at the start of the code and [/code] at the end. Your code will also be more readable if you follow java coding standards and use capital letters to begin classes - like TstInner.

[awinston]

The following compiles OK:
I guess the main point is that foo=bar=42 does not mean "assign 42 to foo and bar" as you might have supposed. It means something like "evaluate bar=42 and assign the result to foo".

Does bar equal 42 after that statement is executed?

[pbrockway2]

Also at javaprogrammingforums.com

Please, if you are going start a discussion in multiple places, post links so that everyone who might want to contribute to the discussion can see what else is being said. Also be aware that some will not respond for fear of wasting their time on a question that has been adequately dealt with elsewhere.

[pbrockway2]

Does bar equal 42 after that statement is executed?

Yes. Provided it is executed (ie it is part of valid Java code that compiles). bar gets initialised as a side effect of foo getting initialised.

Initialising things outside of constructors and methods is rather restricted. For instance you also cannot say:

foo = 40 + 2;

except in a method or constructor. There are probably exact specifications for such things in the JLS, but as a general rule: put code - outside of simple this_variable=that_value assignments - into methods and constructors.

[chandresh]

The following compiles OK:

public class Tst2 {
    public static void main(String[] args) {
        System.out.println("HI");
    }

    public class TstInner {
        int a1, b1;
        
        TstInner() {
            a1 = b1 = 6;
        }
    }
}

The difference is that the initialisation of a1 and b1 occurs within a constructor. You cannot try and assign the result of evaluating b1=6 to a1 except in a method, constructor or initialisation block. Of these a constructor is the most obvious place.

I guess the main point is that foo=bar=42 does not mean "assign 42 to foo and bar" as you might have supposed. It means something like "evaluate bar=42 and assign the result to foo".

-----

When you post code don't include the line numbers. And use the "code" tags: put [code] at the start of the code and [/code] at the end. Your code will also be more readable if you follow java coding standards and use capital letters to begin classes - like TstInner.

Got it now, thanks for the help, much appreciated..

[pbrockway2]

You're welcome.

[JosAH]

I guess the main point is that foo=bar=42 does not mean "assign 42 to foo and bar" as you might have supposed. It means something like "evaluate bar=42 and assign the result to foo".

Huh? Did they change the language overnight? As far as I know it does mean assign 42 to both variables.

kind regards,

Jos

[chandresh]

Huh? Did they change the language overnight? As far as I know it does mean assign 42 to both variables.

kind regards,

Jos

lol... I am sure the world hasnt moved that fast to change the language that drastically fast.
What was mentioned did surprise me but then I was sticking to my learning that it does mean foo & bar gets the value of 42.
thanks clearing the cofusion...

[Tolls]

Huh? Did they change the language overnight? As far as I know it does mean assign 42 to both variables.

kind regards,

Jos

You could have anything on the right hand side, though.
The expression bar = 42 is no different to any other expression that could legitimately occupy that space. Taking operator precedence foo is assigned the result of bar = 42 in much the way it would be assigned the result of bar * 42 below:
foo = bar * 42;

It's nitpicking, though, as the end result is the same.

[awinston]

You could have anything on the right hand side, though.
The expression bar = 42 is no different to any other expression that could legitimately occupy that space. Taking operator precedence foo is assigned the result of bar = 42 in much the way it would be assigned the result of bar * 42 below:
foo = bar * 42;

It's nitpicking, though, as the end result is the same.

So the compiler has no problem with an uninitialized int variable being used in a multiplication operation?

[Tolls]

Nothing in my example implies bar is not initialised.
After all, 'a' and 'b' are both initialised in the original post at declaration.

[awinston]

Nothing in my example implies bar is not initialised.
After all, 'a' and 'b' are both initialised in the original post at declaration.

My fault, I misinterpreted what you were trying to prove with your example.

[pbrockway2]

Did they change the language overnight?

No, nothing's changed.

Within a constructor or method "foo=bar=42" has the effect of assigning 42 to both variables: the same effect is achieved with "foo=42; bar=42;".

The meaning of "foo=bar=42;" - and the meaning of "foo=<complex expression>;" generally - is that the result of the complex expression is assigned to foo. In the context of the original post there is no restriction on assigning a value to two variables, but there is a restriction on assigning the result of a complex expression to a variable. The compiler will discriminate between "foo=bar=42;" and "foo=42; bar=42;".

I don't much care whether you call the difference one of meaning or not. But, either way, the compiler will discriminate between them: allowing one, and forbidding the other.

[awinston]

No, nothing's changed.

Within a constructor or method "foo=bar=42" has the effect of assigning 42 to both variables: the same effect is achieved with "foo=42; bar=42;".

The meaning of "foo=bar=42;" - and the meaning of "foo=<complex expression>;" generally - is that the result of the complex expression is assigned to foo. In the context of the original post there is no restriction on assigning a value to two variables, but there is a restriction on assigning the result of a complex expression to a variable. The compiler will discriminate between "foo=bar=42;" and "foo=42; bar=42;".

I don't much care whether you call the difference one of meaning or not. But, either way, the compiler will discriminate between them: allowing one, and forbidding the other.

So outside of a method or constructor, [foo = bar = 42;] is not permitted because [bar = 42] constitutes as a complex expression. [foo = bar = 42;] is permitted inside methods and constructors because complex expressions can be assigned to variables within a method or constructor.

[pbrockway2]

So outside of a method or constructor, [foo = bar = 42;] is not permitted because [bar = 42] constitutes as a complex expression.

Yes, you should try it. It's a complex expression whose value is being assigned to the variable. (Reading what I've posted again I think I've erroneously suggested it's the complexity that matters. But it's the assignment. And you can't escape assigning with foo=bar=42.)

Contrast the following where the problem is reduced to its simplest:

class Good {
      // initialization
    int foo = 42;
}

class Bad {
    int foo;
        // assignment
    foo = 42;
}

[awinston]

What still bothers me about this is that an assignment statement can be evaluated. Does the compiler just take the value of the variable on the left of the assignment operator when evaluating an assignment expression? That seems to be what is happening.

[pbrockway2]

Yes, the value of the assignment expression is the value being assigned. It gets "passed" continually to the left, assigning as it goes.

But you can't assign where the OP was trying to - you can only initialise.

[pbrockway2]

Yes, the value of the assignment expression is the value being assigned. It gets "passed" continually to the left, assigning as it goes.

But you can't assign where the OP was trying to - you can only initialise.