> That's a separate issue. I believe (but haven't checked) that the << operator > has lower precedence than +, so the compiler interprets the expression as: > > unsigned int k = 0x80 << (24 + 0x81); > > and it's pretty obvious why this causes an error. Instead, try > compiling: > > unsigned int k = (0x80 << 24) + 0x81; > > You may get an error message about signed-integer overflow, but not about > shift-count overflow. > Hi Alan, Your analysis is correct, I did not check the warning message correctly. Peter
