Possible Duplicate:
Why is floating point arithmetic in C# imprecise?
I have been dealing with some numbers and C#, and the following line of code results in a different number than one would expect:
double num = (3600.2 - 3600.0);
I expected num to be 0.2, however, it turned out to be 0.1999999999998181. Is there any reason why it is producing a close, but still different decimal?
This is because double
is a floating point datatype.
If you want greater accuracy you could switch to using decimal
instead.
The literal suffix for decimal
is m, so to use decimal
arithmetic (and produce a decimal
result) you could write your code as
var num = (3600.2m - 3600.0m);
Note that there are disadvantages to using a decimal
. It is a 128 bit datatype as opposed to 64 bit which is the size of a double
. This makes it more expensive both in terms of memory and processing. It also has a much smaller range than double
.
There is a reason.
The reason is, that the way the number is stored in memory, in case of the double data type, doesn't allow for an exact representation of the number 3600.2. It also doesn't allow for an exact representation of the number 0.2.
0.2 has an infinite representation in binary. If You want to store it in memory or processor registers, to perform some calculations, some number close to 0.2 with finite representation is stored instead. It may not be apparent if You run code like this.
double num = (0.2 - 0.0);
This is because in this case, all binary digits available for representing numbers in double data type are used to represent the fractional part of the number (there is only the fractional part) and the precision is higher. If You store the number 3600.2 in an object of type double, some digits are used to represent the integer part - 3600 and there is less digits representing fractional part. The precision is lower and fractional part that is in fact stored in memory differs from 0.2 enough, that it becomes apparent after conversion from double to string
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