On Sun, 12 Oct 2003 22:25:59 -0700 Jim Bublitz <[EMAIL PROTECTED]> wrote:
> Here is the method referenced from the PyQt docs:
>
> """
> QImage(uchar *data, int w, int h, int depth, QRgb *colorTable,
> int numColors, Endian bitOrder);
>
> The colorTable parameter is a list of QRgb instances or None. (Qt
> v2.1+)
> """
>
>
> This probably won't clarify the uchar* parameter for you, but it
> does indicate that the QRgb* parameter is a Python list of
> QRgb instances or None.
Thanks for your help.
I did notice that documentation. However I though QColor(value, value,
value).rgb() would return a QRgb instance. I notice this also in the
documentation.
"""
void rgb(int *r, int *g, int *b);
This takes no parameters and returns the r, g and b values as a tuple.
"""
So does this mean the
QRgb QColor::rgb () const
function is not implemented because void rgb(int *r, int *g, int *b) which
takes no parameters takes over instead?
I try the following:
grayColorMap = []
for i in range(256):
grayColorMap.append(qRgb(i, i, i))
grayColorMap = []
for i in range(256):
grayColorMap.append(QColor(i, i, i).rgb())
It likes the first one, but not the second one. It gives.
Traceback (most recent call last):
File "./testQColor.py", line 21, in ?
image = QImage(buffer, 2, 2, 8, grayColorMap, 256, QImage.IgnoreEndian)
File "/usr/lib/python2.2/site-packages/qt.py", line 439, in __init__
libqtc.sipCallCtor(123,self,args)
TypeError: an integer is required
I must admit that qRgb(r,g,b) is better than making a QColor(r,g,b) and
calling rgb() on it, so it isn't that big of a deal for me at the moment.
Russell Valentine
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