C and C++ can run on many different architectures, and machine types. Consequently, they can have different representations of numbers: Two’s complement, and Ones’ complement being the most common. In general you should not rely on a particular representation in your program.
For unsigned integer types (
size_t being one of those), the C standard (and the C++ standard too, I think) specifies precise overflow rules. In short, if
SIZE_MAX is the maximum value of the type
size_t, then the expression
(size_t) (SIZE_MAX + 1)
is guaranteed to be
0, and therefore, you can be sure that
(size_t) -1 is equal to
SIZE_MAX. The same holds true for other unsigned types.
Note that the above holds true:
- for all unsigned types,
- even if the underlying machine doesn’t represent numbers in Two’s complement. In this case, the compiler has to make sure the identity holds true.
Also, the above means that you can’t rely on specific representations for signed types.
Edit: In order to answer some of the comments:
Let’s say we have a code snippet like:
int i = -1; long j = i;
There is a type conversion in the assignment to
j. Assuming that
long have different sizes (most [all?] 64-bit systems), the bit-patterns at memory locations for
j are going to be different, because they have different sizes. The compiler makes sure that the values of
Similarly, when we do:
size_t s = (size_t) -1
There is a type conversion going on. The
-1 is of type
int. It has a bit-pattern, but that is irrelevant for this example because when the conversion to
size_t takes place due to the cast, the compiler will translate the value according to the rules for the type (
size_t in this case). Thus, even if
size_t have different sizes, the standard guarantees that the value stored in
s above will be the maximum value that
size_t can take.
If we do:
long j = LONG_MAX; int i = j;
LONG_MAX is greater than
INT_MAX, then the value in
i is implementation-defined (C89, section 22.214.171.124).