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.)
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Moritz Schauer
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1Why are you trying to do this? – Ed Heal Mar 21 '15 at 11:17
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Is your x, by chance, a power of 2 ? – manasij7479 Mar 21 '15 at 11:19
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1This is the maximum multiple of x within the range of a `unsigned int` decremented by one. – Moritz Schauer Mar 21 '15 at 11:45
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Your expression simplifies to UINT_MAX, unless you forgot to add some parenthesis. – reader Mar 21 '15 at 13:07
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2In C: "When integers are divided, the result of the / operator is the algebraic quotient with any fractional part discarded." – Moritz Schauer Mar 21 '15 at 13:22
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related: [How to compute 2⁶⁴/n in C?](https://stackoverflow.com/q/55565537/995714) – phuclv Apr 09 '19 at 14:08
3 Answers
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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)
vharavy
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@chmike Thanks for pointing that out. I did not read the question carefully and thought that x is of type unsigned long rather than unsigned int. – vharavy Mar 21 '15 at 16:21
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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.
chmike
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sizeof(unsigned long) == sizeof(unsigned int) == 4 on most modern compilers. You might want to use use unsigned long long.
wik146
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