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Greg
New Member

Australia
6 Posts
Posted - Jun 19 2008 :  10:17:52 PM  Show Profile  Reply with Quote
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;


feline
Whole Tomato Software

United Kingdom
18950 Posts
Posted - Jun 20 2008 :  4:07:37 PM  Show Profile  Reply with Quote
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.
zen is the art of being at one with the two'ness
Edited by - feline on Jun 20 2008 4:08:23 PM
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Greg
New Member

Australia
6 Posts
Posted - Jun 22 2008 :  11:37:55 PM  Show Profile  Reply with Quote
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?
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feline
Whole Tomato Software

United Kingdom
18950 Posts
Posted - Jun 23 2008 :  08:50:27 AM  Show Profile  Reply with Quote
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.
zen is the art of being at one with the two'ness
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sean
Whole Tomato Software

USA
2817 Posts
Posted - Jun 23 2008 :  9:57:57 PM  Show Profile  Reply with Quote
Nice snippet - but you need a prompt for the new name:


typedef $selected$ $newTypeName$;
$newTypeName$

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