Why does QColor use 32-bit signed int to represent e.g. rgba values?

Question

QColor can return rgba values of type int (32-bit signed integer). Why is that? The color values range from 0-255, don't they? Is there any situation where this might not be the case?

I'm considering to implicitly cast each of the rgba values returned by QColor.red()/green()/blue()/alpha() to quint8. It seems to work but I don't know if this will lead to problems in some cases. Any ideas?

Answer 1

I assume you are talking about QColor::rgba() which returns a QRgb.

QRgb is an alias to unsigned int. In these 32 bits all fours channels are encoded as #AARRGGBB, 8 bits each one (0-255, as you mentioned). So, a color like alpha=32, red=255, blue=127, green=0 would be 0x20FF7F00 (553615104 in decimal).

Now, regarding your question about casting to quint8, there should be no problem since each channel is guaranteed to be in the range 0..255 (reference). In general, Qt usually uses int as a general integer and do not pay too much attention to the width of the data type, unless in some specific situations (like when it is necessary for a given memory access, for example). So, do not worry about that.

Now, if these operations are done frequently in a high performance context, think about retrieving the 32 bits once using QColor::rgba and then extract the components from it. You can access the individual channels using bitwise operations, or through the convenience functions qAlpha, qRed, qBlue and qGreen.

For completeness, just to mention that the sibbling QColor::rgb method returns the same structure but the alpha channel is opaque (0xFF). You also have QColor::rgba64, which returns a QRgba64. It uses 16 bits per channel, for higher precision. You have the 64 bits equivalents to qAlpha, etc, as qAlpha64 and so on.