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Move implementation to source for friend operator

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T O P I C R E V I E W
__Daniel__	Posted - Jan 17 2012 : 08:00:41 AM Hello. Latest version of VA. class CTest { public: friend CTest operator( const CTest& crLeft, const CTest& crRight ) { return crLeft crRight; } }; Use "move implementation..." cpp: friend CTest std::operator( const CTest& crLeft, const CTest& crRight ) { return crLeft crRight; } "friend" and "std"? EDIT: When CTest is in namespace in cpp it appears outside namespace
3 L A T E S T R E P L I E S (Newest First)
feline	Posted - Jan 18 2012 : 1:34:05 PM Thank you for the clear description, I am seeing the same problem here, and it does seem to be a friend function specific bug: case=64095 Are you aware that you can use VA Outline to help you move the friend function implementation into the namespace, after it has been created? http://www.wholetomato.com/products/features/vaoutline.asp Hopefully this will help a bit.
__Daniel__	Posted - Jan 18 2012 : 01:50:11 AM quote: For the namespace problem, how is the cpp file set? Are you using a "using namespace" line? Do you have the namespace inside the cpp file? cpp with namespace: namespace Test1 { } friend CTest std::operator( const CTest& crLeft, const CTest& crRight ) { return crLeft crRight; }
feline	Posted - Jan 17 2012 : 3:08:23 PM I am seeing the same effect here. Thank you for the clear description. The problem with "friend" and "std::" is: case=2962 For the namespace problem, how is the cpp file set? Are you using a "using namespace" line? Do you have the namespace inside the cpp file? Using a very simple test, and avoiding friend functions to try and pin down just the namespace effect, I have added the following code to a C++ header file: namespace testMoveImpSpace { class CTestMove { int moveMemberFunc() { return 1; } }; } and when I trigger Move Implementation to Source File on this member function the namespace is correctly added before the class name on the implementation in the cpp file.

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