On 21 ao�t 2013, at 15:43, Gisela Decuzzi wrote:

Hello, I'm trying to understand how to define a correct pattern and I'm lost with the cases where I use a method pattern and the desired code is inside a block.
In the case where we search (methods) with metavariables and the code is inside a block.

An Example:

This pattern
`a: `aa `b: `bb
| `@temps |
``@.Stats.
``@.some size ``@.more.
``@.Stats2.

Putting a dot after the @ means that you want to match statements.

So '``@.some size' cannot match: 'arg1 size'.

Does not match with:
insideABlock: arg1 part2: arg2
|temp1 temp3|
[arg1 size].

temp1 := arg2 + 3.
^arg2

Then even if you remove the extra dots, the rule won't match the method you give.

This is because this method has no statements matching '``@some size' (even if it has a block that has a statement that can).

So here is a rule that can match your method:

`a: `aa `b: `bb
| `@temps |
`@.stms1.

[ `@rcv size ].

`@.stms2.

After looking while I realize that is not matching with

part1: arg1 part2: arg2
|temp1 temp2 temp3|
arg1 size.
temp1 := arg2 + 3.

^arg2

This one however can be matched by your rule (without the extra dots).

I think that you should try to parse expressions instead of full methods (RBParseTreeSearcher>>#matches:do: instead of RBParseTreeSearcher>>#matchesMethod:do:).

You could just try to match '`@rcv size': it would succeed in both methods.

Until I understand:
``@.some size ``@.more.
This should mean: anything that sends the message size, doesn't matter if we have something before or after, I think that this expression should match both cases.

But maybe I'm missunderstanding the patterns (again).

It should work like that (modulo bugs)

` denotes a metavariable
adding @ means you're looking for a list of elements

ex: `receiver `@msg: `@arg matches every message

adding a . means you're looking for a statement (or a list of statement if you combine it with @)
adding # means you're looking for a literal
adding another ` means that the pattern will be searched recursively in each subexpressions even if the top one already matched

it sometimes leads to infinite loops, but I don't remember exactly when
ex: in 'foo bar bar' :

if you search for '`rcv bar' you'll find one match ('foo bar bar')
if you search for '``rcv bar' you'll find two matches ('foo bar' and 'foo bar bar').

and I think it's all

Any help with this will be really welcome!