For the actual values, i.e. 8.0/(-7.0), the result is roughly -1.143. Your result using integer division is being rounded down toward the more negative value of -2. (This is also known as “Floor division”) This is why you will get the somewhat perplexing answers of: >>> 8/(-7) -2 >>> 8/7 1 Note: This is “fixed” … Read more
You’re dividing integers, which means that you’re using integer division. In integer division the fractional part of the result is thrown away. Try the following: float res = (float) quantity / standard; ^^^^^^^ The above forces the numerator to be treated as a float which in turn promotes the denominator to float as well, and … Read more
Editor’s note: this is not actually what compilers do, and gives the wrong answer for large positive integers ending with 9, starting with div10(1073741829) = 107374183 not 107374182. It is exact for smaller inputs, though, which may be sufficient for some uses. Compilers (including MSVC) do use fixed-point multiplicative inverses for constant divisors, but they … Read more
It’s doing integer division. You can use to_f to force things into floating-point mode: 9.to_f / 5 #=> 1.8 9 / 5.to_f #=> 1.8 This also works if your values are variables instead of literals. Converting one value to a float is sufficient to coerce the whole expression to floating point arithmetic.
Because the conversion to float happens after the division has been done. You need: float percentage = ((float) totalOptCount) / totalRespCount; You should be able to format using something like: String str = String.format(“%2.02f”, percentage);
because in this expression t = (1/100) * d; 1 and 100 are integer values, integer division truncates, so this It’s the same as this t = (0) * d; you need make that a float constant like this t = (1.0/100.0) * d; you may also want to do the same with this k … Read more
This is because you are using the integer division version of operator/, which takes 2 ints and returns an int. In order to use the double version, which returns a double, at least one of the ints must be explicitly casted to a double. c = a/(double)b;
double num = 5; That avoids a cast. But you’ll find that the cast conversions are well-defined. You don’t have to guess, just check the JLS. int to double is a widening conversion. From §5.1.2: Widening primitive conversions do not lose information about the overall magnitude of a numeric value. […] Conversion of an int … Read more