整数算术：将1加到UINT_MAX并除以n而不溢出

Question

Is there a way to compute the result of ((UINT_MAX+1)/x)*x-1 in C without resorting to unsigned long (where x is unsigned int )? (respective "without resorting to unsigned long long " depending on architecture.)

Answer 1

It is rather simple arithmetic:

((UINT_MAX + 1) / x) * x - 1 =
((UINT_MAX - x + x + 1) / x) * x - 1 = 
((UINT_MAX - x + 1) / x + 1) * x - 1 =
(UINT_MAX - x + 1) / x) * x + (x - 1)

Answer 2

With integer divisions we have the following equivalence

(y/x)*x == y - y%x

So we have

((UINT_MAX+1)/x)*x-1 == UINT_MAX - (UINT_MAX+1)%x

combining this result with the following equivalence

(UINT_MAX+1)%x == ((UINT_MAX % x) +1)%x

we get

((UINT_MAX+1)/x)*x-1 == UINT_MAX - ((UINT_MAX % x) +1)%x

which is computable with an unsigned int .

Answer 3

sizeof(unsigned long) == sizeof(unsigned int) == 4 on most modern compilers. You might want to use use unsigned long long.