I'm a bit confused because I wanted to initialize a variable of type unsigned long whose size is 8 bytes on my system (on every modern system I suppose). When I want to assign 1 << 63 to the variable, I get a compiler warning however and the number is in fact 0. When I do 1 << 30 I get the expected result of 2 ^ 30 = 1073741824. Yet when I do 1 << 31, I get the result of 2 ^ 64 (I think; actually this shouldn't be possible) which prints 18446744071562067968.
Can anyone explain this behaviour to me?
1 << 63 will be computed in int arithmetic, and your int is probably 32 bit.
Remedy this by promoting one of the arguments: 1ULL << 63 will do it.
ULL means the expression will be at least 64 bits.
The expression 1 << 63 has type int. The range of an int is -231 … 231 - 1 on most systems, 263 is too large for that. Try either (unsigned long)1 << 63 or 1UL << 63 to shift a value of type unsigned long left by 63 places.
The 1 here, is called an integer constant. By the norms specified in the standard, C11, chapter §6.4.4.1 the syntax for the same is
integer-constant:
decimal-constant integer-suffixopt
octal-constant integer-suffixopt
hexadecimal-constant integer-suffixopt
and regarding the Semantics,
The type of an integer constant is the first of the corresponding list in which its value can be represented.
and the table says, if there is no suffix, and the value is representable in the int range it should be considered as int. So, 1 here, is considered an int, which is of generally 4 bytes, or 32 bits, also same in your case.
To explicitly specify the 1 as unsigned long (64) bit, we can use the suffix, like
1UL << 63
should solve your issue.
Please note: unsigned long is not guaranteed to be of 64 bits. unsigned long long is guaranteed to have at least 64 bits. However, as long as you're using the platform where unsigned long is of 64 bits , you should be fine