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Refactor to make new Type

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T O P I C R E V I E W
Greg	Posted - Jun 19 2008 : 10:17:52 PM Had a quick search of the forum and couldn't seem to locate a current thread on the topic. Sorry if this is a duplicate. Sometimes I'll make a quick inline template while identifying a member variable. std::vector<wxString> m_captions; I'd like to be able to refactor the type into a new named type so that I can work with it more easily. Is there a standard refactoring for this - basically turn it into a typedef typedef std::vector<wxString> Captions; Captions m_captions;
4 L A T E S T R E P L I E S (Newest First)
sean	Posted - Jun 23 2008 : 9:57:57 PM Nice snippet - but you need a prompt for the new name: typedef $selected$ $newTypeName$; $newTypeName$
feline	Posted - Jun 23 2008 : 08:50:27 AM Assuming you have added the VA Snippet select the type in your code: then press the "Insert VA Snippet" button on the VA toolbar and select the snippet "convert to typedef". You cannot simply define a new refactoring command by adding a snippet since VA would not know what the refactoring command was supposed to do.
Greg	Posted - Jun 22 2008 : 11:37:55 PM Thanks Feline, mostly there, I was thinking something more like the following - though I don't know enough to get this to work as a selection, if I put this in the Encapsulate field refactoring I get more like what I'm after. typedef $SymbolType$ $GeneratedPropertyName$; $GeneratedPropertyName$ $SymbolName$; $end$ It appears you can't just add new refactorings?
feline	Posted - Jun 20 2008 : 4:07:37 PM You can do this with VA Snippets. I have just created a new snippet with the code: typedef $selected$; $selected$ $end$ In your original line select the type: std::vector<wxString> m_captions; and then just insert this new "convert to typedef" snippet.

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