FontStyle code c# to c++

Question

I wanna to make my bold,italic,underline buttons work with eachother in richtextbox. I found an example, but its in c#. Help pls, how its works in c++:

if(e.Button==toolBarButton1)
{
    int fStyle = (int)FontStyle.Bold + (int)FontStyle.Italic;
    FontStyle f = (FontStyle)fStyle;
    richTextBox1.SelectionFont = 
        new Font(richTextBox1.Font.Name, richTextBox1.Font.Size, f);
}

my version, I want to remake it

     if ( RichTextBox1->SelectionFont != nullptr )
   {
      System::Drawing::Font^ currentFont = RichTextBox1->SelectionFont;
      System::Drawing::FontStyle newFontStyle;
      if ( RichTextBox1->SelectionFont->Bold == true )
      {
         newFontStyle = FontStyle::Regular;
      }
      else
      {
         newFontStyle = FontStyle::Bold;
      }
      RichTextBox1->SelectionFont = gcnew System::Drawing::Font( currentFont->FontFamily,currentFont->Size,newFontStyle );

Are you sure you wanted the sum operation instead of the bitwise or operator? — Machinarius, Mar 31 '13 at 14:07
Be more precise. Win32? MFC? WinForms C++/CLI? What do you have? Do you have any code at all? — David Heffernan, Mar 31 '13 at 14:34
WinForms C++. I added it higher. So I want to check selected text style, that is, if selected text is bold and italic button was pressed its makes text bold and italic at same time. Sorry for my english. — user2227430, Mar 31 '13 at 14:44
The Font class doesn't have a bool property named Bold. Use the Style property instead. — Hans Passant, Mar 31 '13 at 14:49

score 0 · Answer 1 · answered Mar 31 '13 at 18:23

If you want a direct conversion of the C# to C++/CLI, just use:

if (e->Button == toolBarButton1)
{
    int fStyle = safe_cast<int>(FontStyle::Bold) + safe_cast<int>(FontStyle::Italic);
    FontStyle ^f = safe_cast<FontStyle^>(fStyle);
    richTextBox1->SelectionFont = gcnew Font(richTextBox1::Font->Name, richTextBox1::Font::Size, f);
}

FontStyle code c# to c++

1 Answers1