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How do I divide two integers to get a double?
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If you want to keep a and b as int s, yet divide them fully, you must cast at least one of them to double: (double)a/b or a/(double)b or (double)a/(double)b . b) c is a double , so it can accept an int value on assignement: the int is automatically converted to double and assigned to c .
Recall that the fraction bar can be read “divided by.” Therefore, 8/2 can be read as “8 divided by 2.” Rule for Dividing Two Integers: TO DIVIDE TWO NUMBERS WITH THE SAME SIGN, divide the absolute values of the numbers. The quotient is positive.
Division of integers involves the grouping of items. It includes both positive numbers and negative numbers. Just like multiplication, the division of integers also involves the same cases. When you divide integers with two positive signs, Positive ÷ Positive = Positive → 16 ÷ 8 = 2.
You want to cast the numbers:
double num3 = (double)num1/(double)num2; Note: If any of the arguments in C# is a double, a double divide is used which results in a double. So, the following would work too:
double num3 = (double)num1/num2; For more information see:
Dot Net Perls
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Complementing the @NoahD's answer
To have a greater precision you can cast to decimal:
(decimal)100/863 //0.1158748551564310544611819235 Or:
Decimal.Divide(100, 863) //0.1158748551564310544611819235 Double are represented allocating 64 bits while decimal uses 128
(double)100/863 //0.11587485515643106 For more details about the floating point representation in binary and its precision take a look at this article from Jon Skeet where he talks about floats and doubles and this one where he talks about decimals.
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