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So how to create a function to return nearest up so nearest to 9 9+ 1 to given int leaving no remainder when divided by 2 int?
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To round a number down to the nearest even number:floor() method passing it the number divided by 2 . Multiply the result by 2 . The result of the calculation is the number rounded down to the nearest even integer.
If you assume FE_TONEAREST, the result is simply std:nearbyint(x). FE_TONEAREAT is round to even.
If a number is evenly divisible by 2 with no remainder, then it is even. You can calculate the remainder with the modulo operator % like this num % 2 == 0 .
A number is even if, when divided by two, the remainder is 0. A number is odd if, when divided by 2, the remainder is 1. Methods are great!
To round to the nearest int:
number+=(number & 1)
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