Make doExport() helper private

One declaration of doExport() was not used, and the other one was only used by
Buffer, so make it private since returning the exported file name is a bit
strange if it can also appear as input argument.
This commit is contained in:
Georg Baum 2014-11-16 20:51:42 +01:00
parent 9d8839f688
commit 4d3f9f3ede

View File

@ -643,20 +643,17 @@ public:
///
ExportStatus doExport(std::string const & target, bool put_in_tempdir) const;
///
ExportStatus doExport(std::string const & target, bool put_in_tempdir,
std::string & result_file) const;
///
ExportStatus preview(std::string const & format) const;
private:
///
ExportStatus doExport(std::string const & target, bool put_in_tempdir,
std::string & result_file) const;
/// target is a format name optionally followed by a space
/// and a destination file-name
ExportStatus doExport(std::string const & target, bool put_in_tempdir,
bool includeall, std::string & result_file) const;
///
ExportStatus doExport(std::string const & target, bool put_in_tempdir,
bool includeall) const;
///
ExportStatus preview(std::string const & format, bool includeall = false) const;
///
void setMathFlavor(OutputParams & op) const;