[pypy-svn] r12549 - pypy/dist/pypy/documentation

[pedronis at codespeak.net]

Author: pedronis
Date: Thu May 19 17:42:44 2005
New Revision: 12549

Modified:
   pypy/dist/pypy/documentation/coding-guide.txt
Log:
include is_w, eq_w in the discussion



Modified: pypy/dist/pypy/documentation/coding-guide.txt
==============================================================================
--- pypy/dist/pypy/documentation/coding-guide.txt	(original)
+++ pypy/dist/pypy/documentation/coding-guide.txt	Thu May 19 17:42:44 2005
@@ -367,7 +367,7 @@
 
 * ``space.is_true(w_x)``: checks if the given wrapped object is considered to be ``True`` or ``False``.  You must never use the truth-value of ``w_x`` directly; doing so (e.g. writing ``if w_x:``) will give you an error reminding you of the problem.
 
-* ``w_x == w_y`` or ``w_x is w_y``: DON'T DO THAT.  The only half-official exception is to check if ``w_x`` contains a wrapped ``None``: you can write ``w_x == space.w_None``.  Follow this rule; the case of ``None`` is easy to fix globally later if we find out that we need to.  The rationale for this rule is that there is no reason that two wrappers are related in any way even if they contain what looks like the same object at application-level.  To check for equality, use ``space.is_true(space.eq(w_x, w_y))``.  To check for identity, use ``space.is_true(space.is_(w_x, w_y))``.
+* ``w_x == w_y`` or ``w_x is w_y``: DON'T DO THAT.  The only half-official exception is to check if ``w_x`` contains a wrapped ``None``: you can write ``w_x == space.w_None``.  Follow this rule; the case of ``None`` is easy to fix globally later if we find out that we need to.  The rationale for this rule is that there is no reason that two wrappers are related in any way even if they contain what looks like the same object at application-level.  To check for equality, use ``space.is_true(space.eq(w_x, w_y))`` or even better the short-cut ``space.eq_w(w_x, w_y)`` returning directly a interpreter-level bool.  To check for identity, use ``space.is_true(space.is_(w_x, w_y))`` or better ``space.is_w(w_x, w_y)``.
 
 * ``space.unpackiterable(w_x)``: this helper iterates ``w_x`` (using ``space.iter()`` and ``space.next()``) and collects the resulting wrapped objects in a list.  Of course, in cases where iterating directly is better than collecting the elements in a list first, you should use ``space.iter()`` and ``space.next()`` directly.