On Tue, Mar 19, 2013 at 1:02 PM, Andrew Chi <achi@xxxxxxx> wrote: > I'm confused about the parsing of the "L" which should denote a long integer > literal. I'm running GCC 4.5.3 on a 32-bit machine, i.e. long long = 64 > bits and long = 32 bits. Am I misunderstanding the precedence of "L" or how > it's parsed/converted? You didn't really say what your confusion was. I guess that you are wondering why these two lines give you different values. > long long y = (0xffffffffL); > long long z = (long)(0xffffffffL); It's because when the C compiler sees an integer constant, the type depends on the value. For a hex constant with an 'L' suffix, the type given is the first of the following types in which the value can be represented: long, unsigned long, long long, unsigned long long. In the case of 0xffffffffL on a 32-bit machine, the first of those types in which the value can be represented is unsigned long. When an unsigned long value is assigned to a long long variable, it is zero-extended. When an unsigned long value is cast to a long value, the result is, of course, a long, which is signed. When a long is assigned to a long long variable, it is sign-extended. Ian
