Iterable arrays

[Roedy Green]

You can use arrays in for:each. I presume that means somehow arrays
must have an iterator method that produces an Iterator.

What are the methods of arrays?

Or is this just fudged as a special case by the compiler?

 

[Chris ( Val )]

You can use arrays in for:each. I presume that means somehow arrays
must have an iterator method that produces an Iterator.

What are the methods of arrays?

Or is this just fudged as a special case by the compiler?

I'm not sure of the internal workings of it, but I read
recently that it was specifically designed to work with
arrays and collections.

 

[Eric Sosman]

Roedy Green wrote On 10/25/07 09:19,:

You can use arrays in for:each. I presume that means somehow arrays
must have an iterator method that produces an Iterator.

What are the methods of arrays?

Or is this just fudged as a special case by the compiler?

Special-cased by the compiler. (How do I know?
I wrote foreach loops for a List<String> and for
a String[], and looked at the generated bytecode.)

For the List<String>, the generated code calls
the .iterator() method and then uses .hasNext() and
..next() on the Iterator<String>.

For the String[], the generated code is roughly
equivalent to

for (int n = array.length, i = 0; i < n; ++i)

.... which may be a hair more efficient than

for (int i = 0; i < array.length; ++i)

 

[Roedy Green]

For the String[], the generated code is roughly
equivalent to

for (int n = array.length, i = 0; i < n; ++i)

that's odd. I've never noticed for:each iterating in reverse order.

 

[Eric Sosman]

Roedy Green wrote On 10/25/07 13:48,:

For the String[], the generated code is roughly
equivalent to

for (int n = array.length, i = 0; i < n; ++i)

that's odd. I've never noticed for:each iterating in reverse order.

"Reverse order?"

 

[Daniel Pitts]

For the String[], the generated code is roughly
equivalent to

for (int n = array.length, i = 0; i < n; ++i)

that's odd. I've never noticed for:each iterating in reverse order.

I think you misread that.
it iterates i from 0 to n. I recall hearing that array.length is a
no-no on JavaME, so perhaps they just use the same output for both cases?

I wonder what happens when you do something like this:
Object[] arr = new Object[2];
for (Object o: arr) {
arr = new Object[0];
}

(I know I could test that, but I will in a bit.)
Anyone want to speculate before I test?

 

[Roedy Green]

"Reverse order?"

sorry. I read the first few chars and presumed you wrote:

for (int n = array.length-1; n>=0; n--)

 

[Mike Schilling]

For the String[], the generated code is roughly
equivalent to

for (int n = array.length, i = 0; i < n; ++i)

that's odd. I've never noticed for:each iterating in reverse order.

I think you misread that.
it iterates i from 0 to n. I recall hearing that array.length is a no-no
on JavaME, so perhaps they just use the same output for both cases?

I wonder what happens when you do something like this:
Object[] arr = new Object[2];
for (Object o: arr) {
arr = new Object[0];
}

(I know I could test that, but I will in a bit.)
Anyone want to speculate before I test?

I'll speculate that the generated bytecode makes a copy of "arr" into a
temporay, so that nothing untoward happens.

 

[Mike Schilling]

For the String[], the generated code is roughly
equivalent to

for (int n = array.length, i = 0; i < n; ++i)

that's odd. I've never noticed for:each iterating in reverse order.

I think you misread that.
it iterates i from 0 to n. I recall hearing that array.length is a no-no
on JavaME, so perhaps they just use the same output for both cases?

I wonder what happens when you do something like this:
Object[] arr = new Object[2];
for (Object o: arr) {
arr = new Object[0];
}

(I know I could test that, but I will in a bit.)
Anyone want to speculate before I test?

I'll speculate that the generated bytecode makes a copy of "arr" into a
temporary, so that nothing untoward happens.

Copy of the reference "arr", of course, not of the array itself.

 

[Patricia Shanahan]

For the String[], the generated code is roughly
equivalent to

for (int n = array.length, i = 0; i < n; ++i)
that's odd. I've never noticed for:each iterating in reverse order.

I think you misread that.
it iterates i from 0 to n. I recall hearing that array.length is a no-no
on JavaME, so perhaps they just use the same output for both cases?

I wonder what happens when you do something like this:
Object[] arr = new Object[2];
for (Object o: arr) {
arr = new Object[0];
}

(I know I could test that, but I will in a bit.)
Anyone want to speculate before I test?

I'll speculate that the generated bytecode makes a copy of "arr" into a
temporay, so that nothing untoward happens.

The JLS gives the equivalent basic for code. See
http://java.sun.com/docs/books/jls/third_edition/html/statements.html#14.14.2

And, yes, it does call for the array reference expression to be copied
to a temporary.

Patricia

 

[Wayne]

You can use arrays in for:each. I presume that means somehow arrays
must have an iterator method that produces an Iterator.

What are the methods of arrays?

Or is this just fudged as a special case by the compiler?

It has nothing to do with fudge. It's the caffeine in Java
that makes the arrays iterable.

(Okay, no more puns, I promise.)

-Wayne

 

[Eric Sosman]

The JLS gives the equivalent basic for code. See
http://java.sun.com/docs/books/jls/third_edition/html/statements.html#14.14.2


And, yes, it does call for the array reference expression to be copied
to a temporary.

Makes sense when you think about it, in light of

for (String s : expensiveMethodReturningArray())

or even

for (String s : methodWithSideEffectsReturningArray())