Why doesn't GCC produce a warning when assigning a signed literal to an unsigned type?

Question

Several questions on this website reveal pitfalls when mixing signed and unsigned types and most compilers seem to do a good job about generating warnings of this type. How

Answer 1

Using (unsigned)-1 is an often used way to set all bits and sometimes even quoted as reason for this (mis)feature of C, even by people who should know better . It's neither obvious nor portable -- the expression you want to use to set all bits is ~0 .