Re: initialize reference
* jean:
I have the following code. Basically, I want to insert to different map
depends on m_bDefaultTag . I know I need to initialize updateTags. But
how? Is there a way to achieve my goal? Thanks.
class tag_entry
{
string m_name;
int value;
}
All members private, missing semicolon, inconsistent naming convention
for members, missing indentation.
That means it's probably not your actual code.
It's always a good idea to copy and paste /actual code/. Otherwise
you're bound to misrepresent or completely conceal your actual problem,
and introduce problems that aren't there in your actual code, and thus,
waste everybody's time. Remember to replace tabs with spaces for code
in newsgroup postings.
void MyFunc()
{
std::map<std::string, tag_entry >& customTags =
GetCustomTags();
std::map<std::string, tag_entry >& defineTags =
GetDefineTags();
Here you have used that type twice. At that point it's usually a good
idea to name it, by using a typedef.
std::map<std::string, tag_entry >& updateTags;
std::map<std::string, tag_entry >& updateTags =
(condition? map1 : map2);
Or, you can declare updateTags as a pointer instead of a reference, and
use essentially the logic below.
But a reference is more clear, and more concise.
if( m_bDefaultTag ) {
The m_ naming convention is usually used to indicate a non-static class
data member (other such conventions are "my" prefix, which I prefer, and
"_" suffix, which IIRC is used in Boost). In a free-standing function
like this, you don't have access to non-static class data members. It's
not a good idea to use that naming convention for globals.
updateTags = defineTags;
}
else {
updateTags = customTags;
}
customTags.insert( mapElement );
}
--
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