Re: Stranger compiler error?

From:
Lew <lew@nowhere.com>
Newsgroups:
comp.lang.java.programmer
Date:
Sun, 12 Nov 2006 10:07:53 -0500
Message-ID:
<U_qdnUcn5P5UpcrYnZ2dnUVZ_uSdnZ2d@comcast.com>
Red Orchid wrote:

Patricia Shanahan <pats@acm.org> wrote or quoted in
Message-ID: <t755h.4869$l25.4327@newsread4.news.pas.earthlink.net>:

Knute Johnson wrote:
...

        for (Enumeration<Integer> e=hash.keys(); e.hasMoreElements();)
            String[] array = hash.get(e.nextElement()); // <---- error

...

You are confusing two very different entities, a statement and a
variable declaration.

The for MUST be followed by a statement.

You have a local variable declaration, which is not itself a statement
but can appear in a compound statement, such as a brace-enclosed block.

Incidentally, it seems a bit pointless because you assign to array and
then immediately leave its scope of declaration.


#0: "for ( ... ) a declaration statement;"

For a moment, let's consider the above '#0' not to be an error.
Now,
Let's take all possible cases into account.

Case 1)
#1_e:
    for (...) int i = 0;

#1_c:
    for (...) { int i = 0; }

Both '#1_e' and '#1_c' are useless because 'i' is never
read and '0' is constant.

Case 2)
#2_e:
    for (...) int i = parse(...);

#2_c:
    for (...) { int i = parse(...); }

Both '#2_e' and '#2_c' are not useless because 'parse(...)'
can change the state of some instances (ex: fields).

Case 3)
#3_e:
    for (...) parse(...);

#3_c:
    for (...) { parse(...); }

Both '#3_e' and '#3_c' are not useless.

From this point of view, '#*_e' are equivalent to '#*_c' .
But,
'#*_e' are errors and '#*_c' do not.

Is it valid that '#*_c' do not be errors ?
If so, what is the advantage of it ?

"Type name = ... ;" is composed of more than one statements
like a brace-enclosed block.

I think that it is unartificial that a compiler give warnings
about '#*_e', not errors.


The Java(tm) language is what it is and it ain't what it ain't.

- Lew

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