From ISO14882:2011(e) 5.6-4: The binary / operator yields the quotient, and the binary % operator yields the remainder from the division of the first expression by the second. If the second operand of / or % is zero the behavior is undefined. For integral operands the / operator yields the algebraic quotient with any fractional … Read more
If your sh really is sh and not just bash being run as sh then this will work just fine if [ `expr $n % 5` -eq 0 ] then # do something fi If your sh is really bash then put your test in (( )) like so if (( $n % 5 == … Read more
Your expression should be ((x-1) + k) % k. This will properly wrap x=0 around to 11. In general, if you want to step back more than 1, you need to make sure that you add enough so that the first operand of the modulo operation is >= 0. Here is an implementation in C++: … Read more
Try this: if i % 4 == 0 This is called the “modulo operator”.
Try (a % b) * Math.Sign(a) Try this; it works correctly. static int MathMod(int a, int b) { return (Math.Abs(a * b) + a) % b; }
For a datetime.datetime rounding, see this function: https://stackoverflow.com/a/10854034/1431079 Sample of use: print roundTime(datetime.datetime(2012,12,31,23,44,59,1234),roundTo=60*60) 2013-01-01 00:00:00
Modulo function updated to handle boundary cases as noted by aka.nice and arr_sea: static const double _PI= 3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348; static const double _TWO_PI= 6.2831853071795864769252867665590057683943387987502116419498891846156328125724179972560696; // Floating-point modulo // The result (the remainder) has same sign as the divisor. // Similar to matlab’s mod(); Not similar to fmod() – Mod(-3,4)= 1 fmod(-3,4)= -3 template<typename T> T Mod(T … Read more
Ah, the joys of bitwise arithmetic. A side effect of many division routines is the modulus – so in few cases should division actually be faster than modulus. I’m interested to see the source you got this information from. Processors with multipliers have interesting division routines using the multiplier, but you can get from division … Read more
Yep. The timeit module in the standard library is how you check on those things. E.g: $ python -m timeit -s ‘def isodd(x): x & 1’ ‘isodd(9)’ 1000000 loops, best of 3: 0.446 usec per loop $ python -m timeit -s ‘def isodd(x): x & 1’ ‘isodd(10)’ 1000000 loops, best of 3: 0.443 usec per … Read more
To measure is to know (all timings on a Macbook Pro 2.8Ghz i7): >>> import sys, timeit >>> sys.version_info sys.version_info(major=2, minor=7, micro=12, releaselevel=”final”, serial=0) >>> timeit.timeit(‘divmod(n, d)’, ‘n, d = 42, 7’) 0.1473848819732666 >>> timeit.timeit(‘n // d, n % d’, ‘n, d = 42, 7’) 0.10324406623840332 The divmod() function is at a disadvantage here because … Read more