On 2014/06/11 14:33:37, Yang wrote:
https://codereview.chromium.org/264333007/diff/150001/src/debug.cc
File src/debug.cc (left):

https://codereview.chromium.org/264333007/diff/150001/src/debug.cc#oldcode2705
src/debug.cc:2705: if (was_in_scope && (after_compile_flags &
SEND_WHEN_DEBUGGING) == 0) return;
Is it guaranteed that when this method is called from CompileToplevel, we are
not in the debug scope? If not, this will change behavior.
There are next three lines of code (2267-2671) in base version:
if (in_debug_scope() || ignore_events()) return;
HandleScope scope(isolate_);
bool was_in_scope = in_debug_scope();

If in_debug_scope() is true, then we will leave the function and do nothing.
Otherwise, was_in_scope is false, and condition (was_in_scope &&
(after_compile_flags & SEND_WHEN_DEBUGGING) == 0) is false, too.
Variable was_in_scope is always false.

Could behavior was changed earlier patches?


https://codereview.chromium.org/264333007/diff/150001/src/parser.cc
File src/parser.cc (right):


https://codereview.chromium.org/264333007/diff/150001/src/parser.cc#newcode3863
src/parser.cc:3863: isolate()->Throw(*result, &location);
Won't Debug::OnException trigger as well? The call to Debug::OnCompileError doesn't actually contain the syntax error location, right? So you would need
the
following Debug::OnException to get the actual syntax error location, right? A
split API like that is kind of weird.
Yes, I need the following Debug::OnException. But it is important to distinguish between an Exception like SyntaxError and compilation errors. Event CompileError dual event AfterCompile and provides information about the script to debugger.




https://codereview.chromium.org/264333007/

--
--
v8-dev mailing list
[email protected]
http://groups.google.com/group/v8-dev
--- You received this message because you are subscribed to the Google Groups "v8-dev" group.
To unsubscribe from this group and stop receiving emails from it, send an email 
to [email protected].
For more options, visit https://groups.google.com/d/optout.

Reply via email to