Quote:I really wonder how Word works in this regard. Regular fonts, bold, italicized, underlined. Well, you don't actually need an underlined font set, just position a LINE statement.

Screen _NewImage(900, 600, 32)

nof = 4

ReDim fh(nof), mystring$(nof)



fh(1) = _LoadFont("lmsans10-regular.otf", 21)

fh(2) = _LoadFont("lmsans10-bold.otf", 21)

fh(3) = _LoadFont("lmsans10-oblique.otf", 21)

fh(4) = _LoadFont("lmsans10-boldoblique.otf", 21)



For i = 1 To nof

    If fh(i) <= 0 Then

        Print "Failed to load font."

        End

    End If

Next

_Font fh(1)

For i = 1 To nof

    Restore text_data

    mystring$(i) = LoadUData$

Next

x = 8: j = 1

f = 1: _UPrintString (8, _Height \ 2 - _UFontHeight \ 2 + y - 48), mystring$(f), _Width, 8



For i = 1 To nof * 26 ' Step 4

    f = (i - 1) Mod 4 + 1

    'f = 1: q = q + 1

    If f = 2 Or f = 4 Then adj = 1 Else adj = 0

    _Font fh(f)

    '_UPrintString (x, _Height \ 2 - _UFontHeight \ 2 + y + adj), Mid$(mystring$(f), q, 1), _Width, 8

    _UPrintString (x, _Height \ 2 - _UFontHeight \ 2 + y + adj), Mid$(mystring$(f), j, 1), _Width, 8

    If i = nof / 2 * 26 Then

        x = 8

        y = y + 48

    Else

        If mono Then

            x = x + 11 ' For mono-spaced fonts.

        Else

            x = x + _PrintWidth(Mid$(mystring$(f), j, 1))

        End If

    End If

    If i Mod 4 = 0 Then j = j + 1

Next

_Font fh(1)

End



text_data:

Data 61,62,63,64,65,66,67,68,69,6A,6B,6C,6D,6E,6F,70,71,72,73,74,75,76,77,78,79,7A,7B

Data E2,96,8F

Data E2,89,A1



Function LoadUData$

    Dim As _Unsigned Long i, s

    Dim d As String

    Dim buffer As String



    s = 26

    buffer = Space$(s)



    For i = 1 To s

        Read d

        Asc(buffer, i) = Val("&h" + d)

    Next



    LoadUData = buffer

End Function

(03-10-2024, 09:18 PM)Kernelpanic Wrote: I would have to load this font first, maybe another one would work. But not today, I'm currently im Magic Forest