On 02/08/2016 09:16, Jonathan Wakely wrote:
On 2 August 2016 at 07:03, John Emmas wrote:
but when I mentioned it on a popular programming forum, someone
pointed out that if the above was working, that was purely a case
of luck.
Why is it purely a case of luck? Because get_application_name()
returns a std::string by value, which tries to copy the VS2005 type
using the code from VS2015?
On 02/08/2016 09:12, Kjell Ahlstedt wrote:
Glib::ustring::raw() and Glib::ustring::operator std::string() return
string_.
Glib::ustring::c_str() returns string_.c_str().
Glib::ustring::data() returns string_.data().
What more do you want? Or is it the name "raw" that you find misleading?
Oops, you're both right. I was getting mixed up and thinking that
Glib::get_application name() returns std::string, whereas it doesn't -
it returns Glib::ustring.
That explains why calling its 'c_str()' function works, while a simple
assignment doesn't. Glib::ustring's std::string operator returns it's
'string_' member. In my case, this IS a std:;string - but it's the old
type of std::string that was known to VS2005 (which presumably can't be
simply copied to the new type that's known to VS2015). Thanks for
clearing this up guys,
John
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