# New Ticket Created by Bob Wilkinson
# Please include the string: [perl #43657]
# in the subject line of all future correspondence about this issue.
# <URL: http://rt.perl.org/rt3/Ticket/Display.html?id=43657 >
Hello
I do not think that "type-inference" is a verb. "Infer" is,
however. I have re-worded a sentence.
[EMAIL PROTECTED]:~/src/parrot/docs$ svn diff vtables.pod
Index: vtables.pod
===================================================================
--- vtables.pod (revision 19698)
+++ vtables.pod (working copy)
@@ -29,7 +29,7 @@
To be perfectly honest, this is a slightly flawed example, since it's unlikely
that there will be a distinct "Python scalar" PMC class. The Python compiler
-could well type-inference variables such that C<a> would be a C<PythonString>
+could well infer variables by their type such that C<a> would be a
C<PythonString>
and C<b> would be a C<PythonNumber>. But the point remains - incrementing a
C<PythonString> is very different from incrementing a C<PerlScalar>.
Bob
P.S. Patch enclosed, too
Index: vtables.pod
===================================================================
--- vtables.pod (revision 19698)
+++ vtables.pod (working copy)
@@ -29,7 +29,7 @@
To be perfectly honest, this is a slightly flawed example, since it's unlikely
that there will be a distinct "Python scalar" PMC class. The Python compiler
-could well type-inference variables such that C<a> would be a C<PythonString>
+could well infer variables by their type such that C<a> would be a
C<PythonString>
and C<b> would be a C<PythonNumber>. But the point remains - incrementing a
C<PythonString> is very different from incrementing a C<PerlScalar>.
