Text in Correct FontSize C++ Api hDC - c++

Question

I'm working with an older C++ application and I need to draw text centered and rotated on an hDC. I've got the centered & rotated part working but the font that gets written to the Image is smaller than I'm expecting. If you take an Image and edit it or a word document and add a line of of Text "COPY" in font size 128 it fills up most of the width of the page. With my code it only covers about 1/3 of the page. What am I missing or doing wrong?
UINT nOptions = 0;//DT_CENTER;
RECT rect = {0,0, BITMAPWIDTH(&WatermarkHandle), BITMAPHEIGHT(&WatermarkHandle)}; //{0,0,FileInfo.Width, FileInfo.Height};
SetMapMode(hdcWatermark, MM_TEXT);
LOGFONT lf;
memset(&lf, 0, sizeof(lf));
lstrcpy(lf.lfFaceName, &sWatermarkFontName[0]);
lf.lfHeight = -MulDiv(lWatermarkFontSize, GetDeviceCaps(hdcWatermark, LOGPIXELSY), 72);
lf.lfEscapement = lWatermarkAngle * 10;
lf.lfOrientation = lf.lfEscapement;
lf.lfClipPrecision = CLIP_LH_ANGLES;// || CLIP_TT_ALWAYS;
lf.lfWeight = FW_THIN;
SetTextColor(hdcWatermark, black);
DrawRotatedText(hdcWatermark, &sWatermarkText[0], &rect, lf, lWatermarkAngle, nOptions);
void DrawRotatedText(HDC hdc, char *str, LPRECT rect, LOGFONT lf, double angle, UINT nOptions)
{
// convert angle to radian
double pi = 3.141592654;
double radian = (angle / 180 * pi);
HFONT hFontText = CreateFontIndirect(&lf);
HFONT hOldFontText = (HFONT)SelectObject(hdc, hFontText);
SIZE TextSize;
GetTextExtentPoint32(hdc, str, strlen(str), &TextSize);
// get the center of a not-rotated text
POINT center;
center.x = TextSize.cx / 2;
center.y = TextSize.cy / 2;
POINT rcenter;
rcenter.x = long(cos(radian) * center.x - sin(radian) * center.y);
rcenter.y = long(sin(radian) * center.x + cos(radian) * center.y);
// finally draw the text and move it to the center of the rectangle
SetTextAlign(hdc, TA_BASELINE);
SetBkMode(hdc, TRANSPARENT);
long lx = rect->left + ((rect->right - rect->left) / 2) - rcenter.x;
long ly = rect->top + ((rect->bottom - rect->top) / 2) + rcenter.y;
SetGraphicsMode(hdc,GM_ADVANCED);
ExtTextOut(hdc, lx, ly, nOptions, rect, str, strlen(str), NULL);
DeleteObject(hFontText);
SelectObject(hdc, hOldFontText);
return;
}

Answer 1

It all depends on what you mean by a "page".
A point is a physical unit of measure, just like an inch or a millimeter. In most modern contexts it's equal to 1/72 of an inch.
A page in Word corresponds to an actual piece of paper with physical dimensions. After subtracting out the margins it might be 6 inches wide, or 432 points. This will be zoomed by an amount to fill the screen.
Sizing a screen is a little more nebulous, since monitor sizes and resolutions may differ. Long ago Windows standardized on 96 pixels per inch for a default, as this was close enough to the typical monitor connected to a Windows system. This means a 1920x1080 monitor is considered by Windows to be 20 inches wide, or 1440 points, no matter how physically large it is.
In Word's case the screen is 432 points wide, while your program considers it to be 1440 points wide. This is very close to the 1/3 ratio you report. If you set Word to an exact 100% zoom the rendered text should be identical because the page will shrink.
The answer is to do as Word does, and zoom in by making everything 3x larger.
Edit: In your case it seems you want to emulate what Word is doing and show a facsimile of a page. That makes it even easier.
The formula given by Microsoft was from a time when floating point arithmetic was slow and avoided whenever possible. That's not the case today. The equivalent is:
lfHeight = -(int)(points * dpi / 72);
The question is what to use for dpi? You can calculate it directly! Since you know the size of the page you're displaying and the width of the window in pixels, it's a simple division:
double dpi = pixelWidth / (double)inches;
If you need to know what the effective zoom is, that's also a simple division: zoom = dpi / GetDeviceCaps(hdc, LOGPIXELSY).